HCF–LCM/e-l-i- -y-l-i-

(CLASSROOM SHEET)
1. Find the LCM of 15, 24, 35 & 54. 8. The least number which is exactly divisible
15] 24] 35 vkSj54 dk y?kqÙke lekioR;Z
(LCM) by 5, 6, 8, 10 and 12 is:
Kkr dhft,A lcls de la[;k tks 5] 6] 8] 10 vkSj 12 ls iw.kZ
SSC CPO 03/10/2023 (Shift-01) foHkkT; gS%
(a) 7650 (b) 7560 SSC CPO 23/11/2020 (Shift-03)
(c) 6570 (d) 5670 (a) 240 (b) 180
2. Find the HCF of 110, 180 and 540. (c) 150 (d) 120
100] 180 vkSj 540 dk e-l-i- Kkr dhft,A 1 3 5 7 9

r
DP Head Constable 13/10/2022 (Shift- 02) 9. Find the HCF of , , , ,
2 4 6 8 10

si
(a) 10 (b) 1
1 3 5 7 9
(c) 5 (d) 3 , , , , dk e-l-i Kkr djksa
2 4 6 8 10
3.
an by
Find the HCF of 240, 280 and 560.
240] 280 vkSj 560 dk egÙke lekiorZdHCF 1 1

n
(a) (b)
Kkr dhft,A 120 7

ja
SSC CPO 04/10/2023 (Shift-02) 1 1
R s
(a) 40 (b) 30 (c) (d)
32 40
a th

(c) 20 (d) 10
10. What is the Highest Common Factor (HCF)
4. Find LCM of 48, 50, 98, 54 and 72.
7 21 49
48, 50, 98, 54 vkSj72 dk y?kqÙke lekioR;Z Kkr of ,
16 32
and
8
?
dhft,A
ty a

7 21 49
(a) 24 × 33 × 52 × 72 (b) 23 × 33 × 52 × 72 , rFkk dk egÙke lekiorZd (HCF)
16 32 8
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4 2 2 2 3 3
(c) 2 × 3 × 5 × 7 (d) 2 × 3 × 5 × 7
fdruk gS\
5. Find the LCM of 15, 24, 32 & 45
SSC MTS 06/08/2019 (Shift-01)
15, 24, 32 vkSj45 dk y?kqÙke lekioR;Z Kkr dhft,A
(a) 1448 (b) 1436 7 147
(a) (b)
64 32
(c) 1435 (d) 1440
6. What is the LCM of 3.6, 1.8 and 0.144? 147 7
(c) (d)
3-6] 1-8 vkSj 0-144 dk y?kqÙke lekioR;Z D;k gS\ 8 32

SSC CGL 19/04/2022 (Shift-02) 1 7 5 4 8

11. The LCM of , , , and is:
(a) 36 (b) 360 6 27 9 15 3
A

(c) 3.6 (d) 3600 1 7 5 4 8

7. Find the LCM of 3 × 5 × 7² × 11, 33 × 5 ×
4 , , , vkSj dk y?kqÙke lekioR;Z
(LCM)
6 27 9 15 3
11² and 3² × 5³ × 114.
Kkr djsaA
34 × 5 × 7² × 11, 33 × 5 × 11² vkSj 3² × 5³ DP Head Constable 14/10/2022 (Shift - 03)
× 114 dk y-l-i- Kkr dhft,
280 3
DP Head Constable 20/10/2022 (Shift - 03) (a) (b)
3 280
(a) 3² × 5 × 11 (b) 34 × 5³ × 7² × 114 280 15
(c) 34 × 5 × 7 ×114 (d) 32 × 5 × 7 ×11 (c) (d)
15 280
12. Find the difference between the LCM and 18. What is the HCF of 329 – 9 and 338 – 9?
4 5 9 329 – 9 vkSj338 – 9 dk HCF D;k gS\
HCF of , and .
7 14 35 UPSC CDS 2023 (1)
(a) 39 – 1 (b) 311 – 1
4 5 9
, vkSj ds LCM vkSj HCF ds chp dk (c) 311 – 3 (d) 311 – 9
7 14 35
varj Kkr dhft,A 19. Find HCF of 33333 + 1 and 33334 + 1?
CRPF HCM 26/02/2023 (Shift - 02) 33
333
+ 1 and 33334 + 1 dk HCF Kkr djsaA
(a) 19.5 (b) 20.7
(a) 33333  1 (b) 33334  1
(c) 22.3 (d) 25.7
13. If x is the HCF and y is the LCM of (c) 3333  1 (d) 3334  1

3 6 9 27 20. What is the HCF of the polynomials (x 4

, , , then which one of the – 1), (x³ – 3x² + 3x – 1) and (x³ – 2x² + x)?
5 25 20 50
following is correct? cgqinksa
(x4 – 1), (x³ – 3x² + 3x – 1) vkSj (x³ –
2x² + x) dk egÙke lekiorZd HCF D;k gksxk\

r
3 6 9 27
;fn , , , dk HCF, x gS vkSj
LCM, DP Head Constable 20/10/2022 (Shift - 02)

si
5 25 20 50
(a) (x – 4) (b) (x – 2)
y gS] rks fuEufyf[kr esa ls dkSu&lk lgh gS\
(c) (x – 8) (d) (x – 1)

(a) y = 90x an by UPSC CDS 2022 (2)

(b) y = 180x
21. What is the HCF of

n
(x8 – y8) and (x7 – y7 + x5y2 – x2y5)?
(c) y = 270x (d) y = 360x
(x8 – y8) vkSj (x7 – y7 + x5y2 – x2y5) dk HCF

ja 3 81 9 D;k gS\
R s
14. Find the LCM of , and
2 16 8 UPSC CDS 2022 (2)
a th

3 81 9 2
(a) (x + y ) 2
(b) (x2 – y2)
, vkSj dk y?kqÙke lekioR;Z
(LCM) Kkr
2 16 8 (c) (x3 – y3 – x2y +xy2)
djsaA
(d) (x3 – y3 + x2y – xy2)
SSC CPO 03/10/2023 (Shift-02)
ty a

22. Let p(x) = x4 + x2 + 1,

111 91 q(x) = x4 – 2x3 + 3x2 – 2x + 1. If HCF of
(a) (b)
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2 2 p(x) and q(x) is x2 – x + 1, then what is

81 101 their LCM?
(c) (d)
2 2 eku yhft, p(x) = x4 + x2 + 1,
15. Find the HCF of (3 –1) and (335–1)?
45
q(x) = x4 – 2x3 + 3x2 – 2x + 1 gSA ;fnp(x)
45
(3 –1) vkSj(3 –1) dk HCF Kkr djsa\
35
vkSjq(x) dk HCF, x – x + 1 gS] rks budk
2
LCM
(a) 243 (b) 242
D;k gS\
(c) 245 (d) 244
UPSC CDS 2022 (2)
16. What is the largest number which divides
(a) (x + x + 1) (x – x + 1)2
2 2
both 235 – 1 and 291 – 1?
(b) (x4 + x2 + 1)2 (x2 – x + 1)
235 – 1vkSj291 – 1 nksuksa dk foHkkftr djus okyh
A

(c) (x4 + x2 + 1) (x2 + x + 1)2

lcls cM+h la[;k D;k gS\ 4 2 2 2
(d) (x + x + 1) (x – x + 1)
UPSC CDS 2023 (1)
23. The HCF and LCM of two numbers is 6 and
(a) 34 (b) 90 5040 respectively. If one of the numbers
(c) 127 (d) 129 is 210, then the other number is:
17. Find HCF of 4 63
+ 1 and 445 + 1 ? nks la[;kvksa dk egÙke lekiorZd vkSj y?kqÙke le
4 63
+ 1 and 4 45
+1 dk HCF Kkr djsaA Øe'k% 6 vksj 5040 gSA ;fn buesa ls ,d la[;k 21
(a) 4 + 1 5
(b) 49 + 1 gS] rks nwljh la[;k Kkr djsaA
(c) 47 + 1 (d) 49 – 1 SSC CPO 14/03/2019 (Shift-01)
 (a) 256 (b) 144 29. The sum of two numbers is 528 and their
(c) 30 (d) 630 H.C.F is 33. The number of pairs of numbers
are possible :
24. The product of HCF and LCM of two
numbers is 3321. If one of the numbers is nks la[;kvksa dk ;ksx 528 gS vkSj mudk e-l-i- 3
369. The HCF of the number is la[;kvksa ds ;qXeksa dh la[;k laHko gS%
(a) 4 (b) 6
;fn nks la[;kvksa HCF
ds vkSjLCM dk xq.kuiQy
(c) 8 (d) 12
3321 gSA ;fn buesa ls ,d la[;k 369 gSAHCF
rks 30. The sum of two numbers is 1224 and their
Kkr dhft,\ HCF is 68. The number of pairs of numbers
SSC CPO 16/03/2019 (Shift-01) satisfying the above condition is:
(a) 21 (b) 9 nks la[;kvksa dk ;ksxiQy 1224 gS vkSj mudk
(c) 3 (d) 27 (HCF) 68 gSA la[;kvksa ds dqy fdrus ;qXe mi
25. The LCM of two numbers is 2079. And
fLFkfr dks larq"V djrs gSa\
their HCF is 27. If one of the numbers is SSC CPO 09/11/2022 (Shift-02)
189, then what is the second number? (a) 3 (b) 4
nks la[
;kvksa dk y?kqÙke lekioR;Z 2079 gSA vkSj mudk (c) 6 (d) 2

r
egÙke lekiorZd 27 gSA rnuqlkj] ;fn muesa ,d la[;k
31. The HCF of two natural number A & B is 39

si
and their product is equal to 15210. How
189 gks] rks nwljh la[;k fdurh gS\ many sets of values of A and B are possible?
(a) 297 (b) 528 nks çkÑr la[;kvksa
A vkSjB dk e-l-i- 39 gS vkSj
(c) 189
an by (d) 216 mudk xq.kuiQy 15210 gSA
fdrus lsV laHko gSa\
A vkSjB ds ekuksa d

n
26. Two numbers are in the ratio of 5 : 7. The
product of their LCM and HCF is 12635. (a) 2 (b) 3
then the sum of the numbers will be.

ja (c) 4 (d) 8
R s
nks la[;k,a 5 % 7 ds vuqikr esa gSaA muds e-l-32. rFkkProduct of two numbers is 2028 and their
y-l- dk xq.kuiQy 12635 gSA la[;kvksa dk ;ksxiQy gksxkAH.C.F. is 13. The number of such pairs is:
a th

(a) 252 (b) 228

nks la[;kvksa dk xq.kuiQy 2028 gS vkSj mudk
H.C.F.
13 gSA ,sls ;qXeksa dh la[;k gS%
(c) 304 (d) 380
(a) 1 (b) 2
27. The sum of and difference between the
ty a

(c) 3 (d) 4
LCM and HCF of two numbers are 512 and 33. HCF and LCM of two 3-digit numbers are
496, respectively. If one number is 72, 16 and 2640 respectively. Find the
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then the other number is: numbers?

nksla[;kvksa ds y?kqÙke lekioR;Z
LCM vkSjegÙke 3 vadksa dh nks la[;kvksa dk e-l- rFkk y-l- Øe
lekiorZd HCF dk ;ksx rFkk muds chp dk varj 16 vkSj 2640 gS] la[;k,a Kkr djsa\
(a) 264, 120 (b) 240, 176
Øe'k% 512 vkSj 496 gSA ;fn muesa ls ,d la[;k 72 (c) 300, 160 (d) 192, 176
gS] rks nwljh la[;k dkSu&lh gS\ 34. HCF and LCM of two 3-digit numbers are
SSC CGL MAINS 29 Jan 2022 23 and 1771 respectively. Find the sum of
their numbers:
(a) 80 (b) 40
3 vadksa dh nks la[;kvksa dk e-l- rFkk y-l- Øe
(c) 64 (d) 56 23 vkSj 1771 gS] mu la[;kvksa dk ;ksx Kkr dhft
28. When product of two numbers, is divided by (a) 414 (b) 391
A

its HCF then we get 5775, but when it is (c) 460 (d) 322
divided by LCM, we get 25. If one number 35. Suppose p and q are the LCM and HCF
is 525, what is the second number? respectively of two positive numbers. If p
nksla[;kvksa ds xq.kuiQy esa tc mlds e-l- ls Hkkx : q = 14 : 1 and pq = 1134, then what is
the difference between the two numbers?
fn;k tkrk gS rc 5775 izkIr gksrk gS ysfdu tc y-l-
ls Hkkx fn;k tkrk gS rc 25 izkIr gksrk gSA ;fn muesaeku yhft,] nks èkukRed la[;kvksaLCM ds v kSj
HCF Øe'k%p v kSjq gSaA ;fn p : q = 14 : 1
ls ,d la[;k 525 gks] rks nwljh la[;k D;k gS\
v kSjpq = 1134, r ks mu nksukas la[;kvksa d
(a) 275 (b) 325
dk varj D;k gS\
(c) 405 (d) 210
UPSC CDS 2022 (1)
 (a) 27 (b) 35 (a) 30 (b) 20
(c) 45 (c) 24 (d) 25
(d) Cannot be determined due insufficient 41. The LCM of the two numbers is 12 times
data/vi;kZIr vk¡dM+ksa ds dkj.k fuèkkZfjr ugha fd;k
their HCF. If the sum of LCM and HCF is
tk ldrk 169 and the sum of the numbers is 91, find
36. The L.C.M of two numbers is 495 and their the difference between the numbers.
H.C.F is 5. If the sum of the numbers is
100, then their difference is
nks la[;kvksa dk y?kqÙke lekioR;Z muds e-l-i-
xquk gSA ;fn
LCM vkSj HCF dk ;ksx 169 gS vkSj
nks la[;kvksa dk y?kqÙke lekioR;Z 495 gS vkSj mudk
e-l-i- 5 gSA ;fn la[;kvksa dk ;ksx 100 gS] rks mudk la[;kvksa dk ;ksx 91 gS] rks la[;kvksa ds chp dk
varj gS Kkr dhft,A
(a) 10 (b) 46 CRPF HCM 11/03/2023 (Shift - 02)
(c) 70 (d) 90 (a) 13 (b) 58
37. LCM and HCF of two numbers are 90 and
(c) 72 (d) 66
15, respectively. If the sum of the two
numbers is 75, then find the greater 42. The HCF and the LCM of two numbers are 5

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number. and 120, respectively. If the sum of the two

si
nks la[;kvksa dk y-l- vkSj e-l- Øe'k% 90 vkSj 15 numbers is 55, then the sum of the
reciprocals of these two numbers is equal to:
gSA ;fn nksuksa la[;kvksa dk ;ksx 75 gS] rks cM+h l[a;k
Kkr dhft,A
an by
SSC CGL 21/04/2022 (Shift-03)
nks la[;kvksa dk egÙke lekiorZd
y?kqÙke lekioR;Z
(LCM) 120 gSA ;fn
(HCF) 5 vkSj
bu nksuksa la[;

n
(a) 60 (b) 45 dk ;ksx 55 gS] rks buds O;qRØeksa dk ;ksx Kkr
(c) 75

ja (d) 90 SSC MTS 05/07/2022 (Shift- 3)

R s
38. The difference of two numbers is 14. Their
LCM and HCF are 441 and 7. Find the two 55 11
(a) (b)
a th

numbers ? 601 120

nks la[;kvksa dk varj 14 gSA mudk y?kqÙke lekioR;Z
120 601
vkSj e-l-i- 441 vkSj 7 gSA nks la[;k,¡ Kkr dhft,\ (c) (d)
11 55
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(a) 63 and 49 (b) 65 and 48

(c) 64 and 50 (d) 64 and 49 43. The sum of two numbers is 1215 and their
di M

39. The LCM of two numbers is five times their HCF is 81. If the numbers lie between 500
HCF. If the product of the two numbers is and 700, then the sum of the reciprocals
20480,then find their HCF and LCM, of the numbers is .....
respectively. nks la[;kvksa dk ;ksxiQy 1215 gS rFkk mudk e-
nks la[;kvksa LCM
dk mudsHCF dk ikap xquk gSA gSA ;fn la[;k,a 500 vkSj 700 ds chp esa gSa] rks l
;fn nksuksa la[;kvksa dk xq.kuiQy 20480 gS] rks Øe'k%
ds O;qRØeksa dk ;ksx-------gksxkA
mudkHCF vkSjLCM Kkr djsaA
5 5
SSC CPO 05/10/2023 (Shift-3) (a) (b)
702 378
(a) 64 & 320 (b) 56 & 280
(c) 48 & 240 (d) 46 & 230 5 5
(c) (d)
40. LCM of two numbers is 22 times their 1512 1188
A

HCF. If one of the numbers is 132 and the 44. Two numbers are in the ratio 7 : 11. If their
sum of LCM and HCF is 276, then what is HCF is 28, then the sum of the two numbers is:
the other number?
nks la[;kvksa dk vuqikr
7 : 11 gSA ;fn mudk egÙk
nks la[;kvksa dk y?kqÙke lekioR;Z
(LCM) muds egÙke
lekiorZd (HCF) 28 gS] rks nksuksa la[;kvksa dk ;ksx
lekiorZd (HCF) dk 22 xquk gSA ;fn ,d la[;k
SSC CPO 25/11/2020 (Shift-1)
132 gS vkSj LCM vkSj HCF dk ;ksx 276 gS] rks
nwljh la[;k D;k gS\ (a) 196 (b) 504
(c) 112 (d) 308
SSC CGL 18/04/2022 (Shift-03)
 1 2 3 rhu la[;k,¡ 3%8%15 ds vuqikr esa gSa vkSj mudk
45. Three numbers are in the ratio : : .
2 3 4 lekioR;Z 8280 gSA mudk egÙke lekiorZd D;k g
If the difference between the greatest SSC CGL 19/04/2022 (Shift-01)
number and the smallest number is 33, (a) 60 (b) 69
then HCF of the three numbers is: (c) 75 (d) 57
1 2 3 51. Three numbers are in the ratio of 3 : 4 :
rhu la[;k,¡ : : ds vuqikr esa gSaA ;fn lcls 5 and their HCF is 18. What is the sum
2 3 4
of these numbers?
cM+h la[;k vkSj lcls NksVh la[;k dk 33
varjgS] rks rhu la[;kvksa dk vuqikr 3 % 4 % 5 gSa vkSj
rhuksa la[;kvksa dk egÙke lekiorZd
(HCF) fdruk egÙke lekiorZd 18 gSA bu la[;kvksa dk ;ksx D;k
gksxk\ MTS 05/09/2023 (Shift- 01)
SSC CGL MAINS 03 Feb 2022 (a) 216 (b) 198
(a) 9 (b) 5 (c) 180 (d) 234
(c) 13 (d) 11 52. Find the greatest number that divides 556,
46. The ratio of two numbers 9 : 14 and their 763 and 349 and leaves 4 as remainders

r
LCM is 1008. The numbers are: respectively.
og vf/dre la[;k Kkr djsa ftlls 556] 763 rFkk
nks la[;kvksa dk vuqikr 9 % 14 vkSj mudk y?kqÙke

si
lekiorZd 1008 gsA la[;k,¡ gSa%& 349 esa Hkkx nsus ij izR;sd fLFkfr esa 4 'ks"k c
(a) 63, 98
(c) 81, 126 an by
(b) 72, 112
(d) 54, 105
(a) 69
(c) 36
(b) 92
(d) 54

n
47. The HCF of two numbers is 29 and the 53. Find the greatest number that divides
other two factors of their LCM are 15 and 797, 1085 and 1232 and leaves 16, 20, 25

ja
13. The larger of the two numbers is: as remainders respectively.
R s
nks la[;kvksa dk e-l-i- 29 gS] vkSj muds y-l-i- ds og vf/dre la[;k Kkr djsa ftlls 797] 1085 rFkk
a th

vU; nks xq.ku[kaM 15 vkSj 13 gSaA mu nks la[;kvksa 1232esaesa Hkkx nsus ij izR;sd fLFkfr esa 16]
cM+h la[;k Kkr djsaA 'ks"k cprk gS\
(a) 435 (b) 377 (a) 69 (b) 71
ty a

(c) 406 (d) 464 (c) 65 (d) 91

48. The sum of LCM and HCF of two numbers
54. Let x be the greatest number which when
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is 4956. Those numbers have a ratio of 11

divides 955, 1027, 1075, the remainder in
: 16. What's the difference between LCM
each case is the same. Which of the
and HCF of two numbers? following is NOT a factor of x?
nks la[;kvksa ds y-l- rFkk e-l- dk ;ksx 4956 gSA mu eku ys fd x lcls cM+h la[;k gS tks 955] 1027] 107
la[;kvksa esa 11 % 16 dk vuqikr gSA mu la[;kvksa ds dks foHkkftr djrh gS] rks izR;sd ekeys esa 'ks"k l
y-l- rFkk e-l- dk varj D;k gS\ fuEufyf[kr essa ls dkSu
x dk xq.ku[kaM ugha gS\
(a) 4480 (b) 4620
SSC CGL Tier-II (16/11/2020)
(c) 4780 (d) 4900
49. The LCM of 165, 176, 385 and 495 is k. (a) 6 (b) 16
When k is divided by the HCF of the (c) 4 (d) 8
numbers, the quotient is p. What is the 55. When 7897, 8110 and 8536 are divided by
A

value of p? the greatest number, then the remainder

165] 176] 385 vkSj 495 dk y-l- k gSA tck dks in each case is the same. The sum of the
la[;kvksa ds e-l- }kjk foHkkftr fd;k tkrk gS] rks digits of x is:
HkkxiQyp izkIr gksrk gSA
p dk eku crkb,A tc 7897] 8110 vksj 8536 dks lcls cM+h la[;kx
(a) 2520 (b) 5040 ls foHkkftr fd;k tkrk gS] rks izR;sd fLFkfr es
(c) 6720 (d) 3360 leku gksrk gSAxrks
ds vadksa dk ;ksx gS%
50. Three numbers are in the proportion of 3 (a) 6 (b) 9
: 8 : 15 and their LCM is 8280. What is (c) 14 (d) 5
their HCF?
56. When 2388, 4309 and 8151 are divided by (a) 2520 (b) 842
a certain 3-digit number, the remainder in
(c) 2522 (d) 840
each case is the same. The remainder is:
;fn 2388] 4309 vkSj 8151 dks rhu vadksa okyh ,d 62. What is the sum of digits of the least
multiple of 13, which when divided by 6,
fuf'pr la[;k ls foHkkftr fd;k tkrk gS rks izR;sd 8 and 12 leaves 5, 7 and 11 respectively
ckj leku 'ks"kiQy cprk gSA 'ks"kiQy Kkr dhft,A as the remainders?
(a) 15 (b) 19
(c) 39 (d) 23 13 ds ml y?kqre xq.kt ds] ftls ;fn 6] 8 vkSj 12
57. When 1062, 1134 and 1182 are divided by ls foHkkftr fd;k tk, rks Øe'k% 5] 7 vkSj 11 'ks"kiQ
the greatest number x, the reminder in cprs gSa] vadksa dk ;ksxiQy D;k gS\
case is y. What is the value of (x-y)?
(a) 5 (b) 6
tc 1062] 1134 vkSj 1182 dks lcls cM+h la[;k x
(c) 7 (d) 8
ls foHkkftr fd;k tkrk gS] rks ckj 'ks"kiQy
y gksrk gSA
63. When the smallest number x is divided by
(x – y) dk eku D;k Kkr dhft,\
5,6,8,9 and 12, it gives remainder 1 in
SSC CGL Tier - II (15/11/2020) each case. But x is divisible by 13. What

r
(a) 19 (b) 17 will be the remainder when x will be
(c) 16 (d) 18 divided by 31 ?

si
58. If r is the remainder when each of 4749, 5601 tc 5] 6] 8] 9 vkSj 12 ls lcls NksVh la[;k x dks

an by
and 7092 is divided by the greatest possible
number d(>1), then the value of (d + r) will be:
foHkkftr fd;k tkrk gS] rks izR;sd ekeys esa 1
izkIr gksrk gS] fdUrq
x, 13 ls foHkkT; gSA xtcdks

n
;fn 4749] 5601 vkSj 7092 esa ls izR;sd dks cM+h ls 31 ls foHkkftr fd;k tk,xk rks izkIr 'ks"k fdruk gks
cM+h laHkkfor la[;k
d ls foHkkftr fd;k tkrk gS rks

ja
izR;sd ckj 'ks"kiQy
r cprk gS] rks(d + r) dk eku
SSC MTS 20/08/2019 (Shift-02)
R s
Kkr djsaA (a) 1 (b) 5
a th

(a) 276 (b) 271 (c) 3 (d) 0

(c) 137 (d) 149 64. Let x be the least number which when
59. When 4546, 5398 and 6889 are divided by divided by 16, 24, 30, 36 and 45, the
ty a

the greatest number m, the remainder in remainder in each case is 4, and x is

each case is n. What is the value of (2m - 3n)? divisible by 28. If the HCF of x and 3193
is y, then what is the sum of the digits of y?
di M

tc 4546] 5398 vkSj 6889 dks lcls cM+h la[;k m

eku yhft, x lcls NksVh la[;k gS ftls 16] 24] 30]
ls foHkkftr fd;k tkrk gS] rks izR;sd fLFkfr esa 'ks"kiQy
n gksrk gSA
(2m – 3n) dk eku D;k gS\ 36 vkSj 45 ls foHkkftr djus ij çR;sd fLFkfr esa '
(a) 189 (b) 140
4 gS] vkSjx] 28 ls foHkkT; gSA ;fn x vkSj 3193
(c) 207 (d) 134
dk HCF y gS] rksy ds vadksa dk ;ksx D;k gS\
60. Let the least number which when divided (a) 4 (b) 9
by 4, 6, 10, 15 leaves in each case same (c) 5 (d) 10
remainder 2 be N. the sum of digits in N is. 65. Let x be the least number which when
eku fy;k tk, fd og U;wure la[;k N gS] ftlesa 4] divided by 12, 18, 20, 27 and 30, the
6] 10 rFkk 15 ls Hkkx nsus ij çR;sd fLFkfr esa 2 'ks"kremainder in each case is 2 and x is
divisible by 47. If the HCF of x and 1932
cprk gS rksN la[;k ds vadksa dk ;ksx Kkr dhft,A
A

is y, then the sum of the digits of y is?

(a) 3 (b) 5
eku yhft, x lcls NksVh la[;k gS ftls 12] 18] 20]
(c) 4 (d) 6
27 vkSj 30 ls foHkkftr djus ij çR;sd fLFkfr esa '
61. The least multiple of 13, when divided by
2 gS vkSj
x] 47 ls foHkkT; gSA ;fnx vkSj 1932 dk
4, 5, 6, 7 and 8 leaves remainder 2 in each
HCF] y gS] rks y dk vadksa dk ;ksx gS\
case?
(a) 7 (b) 10
13 dk og lcls NksVk xq.kt ftls 4] 5] 6] 7 vkSj 8
ls Hkkx nsus ij çR;sd n'kk esa 2 'ks"k cprk gSA
& fuEu gS(c) 5 (d) 11
66. Six bells begin to toll together and toll, 70. There are three traffic signals. Each signal
respectively, at intervals of 3,4,6,7,8 and changes colour from green to red and then
12 seconds. After how many seconds, will from red to green. The first signal takes 25
they toll together again? seconds, the second signal takes 39
6 ?kafV;k¡] tks fd Øe'k% 3] 4] 6] 7] 8 vkSj 12 seconds and the third signal takes 60
lsdaM ds varjky ij ctrh gSa] ,d lkFk ctuk 'kq: seconds to change the colour from green
to red. The durations for green and red
djrh gSaA fdrus lsdaM ds ckn] os fiQj ls ,d lkFk colours are same. At 2:00 p.m, they
ctsaXkh\ together turn green. At what time will they
SSC CGL 12/04/2022 (Shift-02) they change to green next,
(a) 167 (b) 168 simultaneously?
(c) 176 (d) 186 rhu VªSfiQd flXuy gSaA izR;sd flXuy dk jax
67. The traffic lights at 3 different road
yky vkSj fiQj yky ls gjk cnyrk gSA gjs ls yky ja
crossings change after every 48 sec, 72 sec
and 108 sec, respectively. If they all cnyus esa igys flXuy dks 25 lsdaM] nwljs flX
change simultaneously at 8:20 a.m., then dks 39 lsd.M vksj rhljs flXuy dks 60 lsd.M yxrs
at what time will they next change again gSaA gjs o yky jaxksa dh vof/;k¡ leku gSA 2%

r
simultaneously? vijkß dks] os ,d lkFk gjs gks tkrs gSaA vxyh
fdlh lM+d dh 3 vyx&vyx ØkWflax ij VªSfIkQd ykbVsafdl le; ij os ,d lkFk gjs gksaxs\

si
Øe'k% izR;sd 48 lsdaM] 72 lsdaM vkSj 108 lsdaM ds UPSC CSE 28/05/2023 (CSAT)
ckn cny tkrh gSA ;fn os lHkh8 : 20 a.m. ij

an by
,d LkkFk cnyrh gS] rks vxyh ckj os fdlh le;
(a) 4 : 00 p.m.
(c) 4 : 20 p.m.
(b) 4 : 10 p.m.
(d) 4 : 30 p.m.

n
fiQj ls ,d lkFk cnysaxh\ 71. What is the least number of soldiers that
SSC MTS 07/07/2022 (Shift- 1) can be drawn up in troops of 10, 12, 15,
(a) 8:27:12 a.m.
ja (b) 8:33:32 a.m. 18 and 20 soldiers, and also in form of
R s
(c) 9:12:18 am. (d) 8:40:14 a.m. a solid square?
a th

68. Six bells commence tolling together at 7:59

mu lSfudksa dh U;wure la[;k Kkr djas]10,
ftUg
am. They toll at intervals of 3,6,9,12 and 15
seconds respectively. How many time will 12, 15, 18 vkSj20 lSfudksa dh VqdM+h vkS
they toll together till 8:16 am? (excluding the oxZ ds :i esa rS;kj fd;k tk ldrk gSA
ty a

toll at 7:59 am) SSC CPO 24/11/2020 (Shift-2)

Ng ?kafV;ka lqcg 7 % 59 cts ,d lkFk ctus yxrh (a) 400 (b) 625
di M

gSaA os Øe'k% 3] 6] 9] 12 vkSj 15 lsdaM ds varjky (c) 900 (d) 180

ij ctrh gSaA os lqcg 8 % 16 cts rd ,d lkFk 72. A fruit vendor brings 1092 apples and 3432
fdruh ckj ctsaxh\ (lqcg 7 % 59 cts ctus dks oranges to a market. He arranges them in
NksM+dj) heaps of equal number of oranges as well as
SSC MTS 25/07/2022 (Shift- 3) apples such that every heap consists of the
(a) 5 (b) 6 maximum possible number of the fruits. What
(c) 3 (d) 4 is this number?
69. There are four bells which ring at an ,d iQy foØsrk ,d cktkj esa 1092 lsc vkSj 3432
interval of 15 minutes, 25 minutes, 35
larjs ykrk gSA og mUgsa larjs ds <sjksa vkSj
minutes and 45 minutes respectively. If
all of them ring at 9 A.M., how many more esa bl izdkj ls O;ofLFkr djrk gS fd gj <sj esa iQ
dh la[;k vf/dre gksA ;g la[;k Kkr djsa\
A

times will they ring together in the next

72 hours? SSC MTS 05/10/2021(Shift - 01)
pkj ?kafV;k¡
(bells) gSa tks Øe'k% 15 feuV] 25 (a) 78 (b) 312
feuV] 35 feuV vkSj 45 feuV rjky esa ctrh gSaA (c) 39 (d) 156
;fn lHkh ?kafV;k¡
9 A.M. ij ctrh gSa] rks vxys 73. Four ropes of lengths 102m, 119m, 153m
72 ?kaVksa esa os vkSj fdruh ckj ,d lkFk ctsaxh\ and 204 m are to be cut into parts of equal
UPSC CDS 2023 (1) length. Each part must be long as possible.
(a) 0 (b) 1 What is the maximum number of pieces
that can be cut?
(c) 2 (d) 3
 102 ehVj] 119 ehVj] 153 ehVj vkSj 204 ehVj yach
77. Swapnil, Aakash and Vinay begin to jog
pkj jfLl;ksa dks leku yackbZ ds Hkkxksa esa dkVk around
tkuk a circular stadium. They complete
their revolutions in 36 seconds 48 seconds
gSA çR;sd Hkkx ;FkklaHko yack gksuk pkfg,A dkVs tk
and 42 seconds respectively. After how
ldus okys VqdM+ksa dh vf/dre la[;k D;k gS\ many seconds will they be together at the
CRPF HCM 27/02/2023 (Shift - 01) starting point?
(a) 42 (b) 36 LofIuy] vkdk'k vkSj fou; ,d xksykdkj LVsfM
(c) 252 (d) 34 ds pkjksa vksj nkSM+uk 'kq: djrs gSaA os viu
74. A gardener had some plants for planting. Øe'k% 36 lsdaM 48 lsdaM vkSj 42 lsdaM es
He tried planting them in rows of 7, 8, 9
and 12. But he always had one plant left.
djrh gSaA fdrus lsdaM ds ckn os çkjafHkd fc
When he tried to plant 11 in a row, he had ,d lkFk gksaxs\
none left. Find the minimum number of (a) 504 seconds (b) 940 seconds
plants with the gardener. (c) 1008 seconds (d) 470 seconds
,d ekyh ds ikl yxkus ds fy, dqN ikS/s FksA mlus78. A floor of a big hall has dimensions 30 m
mUgsa 7] 8] 9 vkSj 12 dh iafÙkQ;ksa esa yxkus dh dksf'k'k
60 cm and 23m 40 cm. It is to be paved
dhA ysfdu mlds ikl ges'kk ,d ikS/k cpk jgrk FkkA with square tiles of same size. What is the

r
minimum number of tiles required?
tc mlus yxkrkj 11 ikS/s yxkus dh dksf'k'k dh] rks

si
,d fo'kky gkWy ds iQ'kZ ds vk;ke 30 ehVj 6
mlds ikl ,d Hkh ugha cpkA ekyh ds ikl ikS/ksa dh
lseh vkSj 23 ehVj 40 lseh gSaA blesa leku v
U;wure la[;k Kkr dhft,A
an by
CRPF HCM 11/03/2023 (Shift - 01)
dh oxkZdkj Vkbysa yxkbZ tkuh gSaA de ls de
Vkbyksa dh vko';drk gS\

n
(a) 4631 (b) 1804
(c) 506 (d) 3025 UPSC CDS 2023 (I)

ja 1 (a) 30 (b) 36
R s
75. Three pieces of cakes of weight 4 kg, (c) 169 (d) 221
2
a th

3 1 79. The dimensions of the floor of a

6 kg and 7 kg respectively are to be rectangualr room are 3m 60 cm × 5 m 40
4 5
divided into parts of equal weight. Each cm. It has to be covered with square tiles.
part must be as heavy as possible. If one What is the dimension of the largest square
ty a

such part is served to each guest, then what tile that can be fitted? How many such
is the maximum number of guests that can tiles are required to cover the floor?
be entertained?
,d vk;rkdkj dejs ds iQ'kZ dh foek,¡ 3eh60lseh
di M

1 3 1
4 fdxzk
, 6 fdxzk vkSj
7fdxzk otu ds dsd × 5eh40lseh gSaA bls oxkZdkj Vkbyksa ls <
2 4 5
ds rhu VqdM+s Øe'k% leku otu ds Hkkxksa esa foHkkftr fiQV dh tk ldus okyh lcls cM+h oxkZdkj Vk
dh foek D;k gS\ iQ'kZ dks <dus ds fy, ,slh fdru
fd, tkus gSaA çR;sd Hkkx ftruk laHko gks mruk Hkkjh gksuk
pkfg,A ;fn çR;sd vfrfFk dks ,slk ,d Hkkx ijkslk tk,] Vkbyksa dh vko';drk gksxh\
rks esgekuksa dh vf/dre la[;k fdruh gks ldrh gS\ (a) 180×180, 6 (b) 180×180, 9
(a) 20 (b) 54 (c) 120×120, 12 (d) 90×90, 8
80. A servant was hired for fixed days for
(c) 72 (d) 41
which he had to pay Rs 3239. He was
76. Three runners running around a circular absent for some days, and only Rs 2923
track, can complete one revolution in 2, was paid, what was his maximum daily
A

6 and 6.5 hours respectively when they wages?

will meet at the starting point?
,d ukSdj dks fuf'pr fnukssa ds fy, ds fy, dke ij
,d o`Ùkkdkj VªSd ds pkjksa vksj nkSM+rs gq, rhuj[kk / x;k ftlds fy, mls 3239 :i;s fn;k tkuk Fkk
kod Øe'k% 2] 6 vkSj 6-5 ?kaVs esa ,d pDdj iwjk og dqN fnuksa ds fy, vuqifLFkr jgk ,ao mls ds
dj ldrs gSa] izkjafHkd fcanq ij os igyh ckj fdrus 2923 :i;s fn;k x;k] mldh vf/dre nSfud etwnjh
?kaVs cknfeysaxs\ D;k Fkh\
(a) 36 (b) 39 (a) 80 (b) 79
(c) 41 (d) 78 (c) 78 (d) 81
81. If the greatest common factor (HCF) of x (a) 3.5 (b) 2
and y is 15, then the HCF of 36x² – 81y² (c) 5 (d) 4.5
and 81x² – 9y² is divisible by _________? 83. In finding the HCF of two numbers by
;fn x vkSj y dk egÙke lekiorZd (HCF) 15 gS] division method four successive quotient
rks 36x² – 81y² vkSj 81x² – 9y² dk HCF are 4, 3, 6 and 5 respectively and final
divisor is 12. What are two numbers?
_________ ls foHkkT; gksxk\
foHkktu fof/ }kjk nks la[;kvksa ds Kkr djus esa
HCF
SSC CGL TIER- II 06/03/2023
pkj Øfed HkkxiQy Øe'k% 4] 3] 6 vkSj 5 gSaA
(a) 135 (b) 120
vafre Hkktd 12 gS] nks la[;k,¡ D;k gSa\
(c) 90 (d) 180 (a) 1226, 5376 (b) 1116, 4836
82. 13 a,b,c are four distinct numbers and the (c) 1056, 4596 (d) 1176, 5076
HCF of each pair of numbers (13,a) ; (13,b); 84. In finding the HCF of two numbers by the
(13,c) is 13, where a,b,c are each less than division method, the last divisor is 28 and
a+c the quotient are 1, 30, 1 and 3
60 and a<b<c. What is the value of ?
b respectively. What is the sum of the two
13 a,b,c pkj vyXk&vyXk la[;k,a gSa vkSj la[;kvksanumbers?

r
ds izR;sd tksM+s
(13,a) ; (13,b); (13,c) dk e-l- 13 Hkkx fof/ }kjk nks la[;kvksaHCF
dk Kkr djus esa]
gS] tgkaa,b,c izR;sd 60 ls de gS vkSja<b<c gSA vafre Hkktd 28 gS vkSj HkkxiQy Øe'k% 1] 3

si
a+c
vkSj 3 gSA nks la[;kvksa dk ;ksx D;k gS\
dk eku D;k gS\
b
an by
SSC CGL 13/04/2022 (Shift-01)
(a) 7000
(c) 6860
(b) 6944
(d) 7140

n
ja
R s
ANSWER KEY
a th

1.(b) 2.(a) 3.(a) 4.(a) 5.(d) 6.(c) 7.(b) 8.(d) 9.(a) 10.(d)
ty a

11.(a) 12.(d) 13.(d) 14.(c) 15.(b) 16.(c) 17.(b) 18.(d) 19.(a) 20.(d)
di M

21.(c) 22.(a) 23.(b) 24.(b) 25.(a) 26.(b) 27.(d) 28.(a) 29.(a) 30.(a)

31.(c) 32.(b) 33.(b) 34.(a) 35.(c) 36.(a) 37.(b) 38.(a) 39.(a) 40.(c)

41.(a) 42.(b) 43. (c) 44. (c) 45.(d) 46.(b) 47.(a) 48.(d) 49.(b) 50.(b)

51.(a) 52.(a) 53.(b) 54.(b) 55.(a) 56.(a) 57.(d) 58.(a) 59.(c) 60.(b)

61.(d) 62.(d) 63.(b) 64.(a) 65.(b) 66.(b) 67.(a) 68.(a) 69.(c) 70.(b)
A

71.(c) 72.(d) 73.(d) 74.(d) 75.(d) 76.(d) 77.(c) 78.(d) 79.(a) 80.(b)

81.(a) 82.(b) 83.(d) 84.(a)

 LCM & HCF/y-l-i- vkSj e-l-i
1. The product of two co-prime numbers is 201. Find 10. Find the least number that when divided by 6, 15
the value of 45% of their LCM. 36 and 75 leaves the same remainder 3 in each cas
nks lg&vHkkT; vksa
la[;k dk xq.kuiQy 201 gSALCM
muds
dk and is divisible by 11.
45» dk eku Kkr dhft,A og lcls NksVhla[;k Kkr dhft, ftls 6] 15] 36 vkSj 7
(a) 201 (b) 90.45 foHkkftr djus ij çR;sd fLFkfr esa leku 'ks"k 3 cps rF
(c) 162 (d) 128 foHkkT; gksA
2. What is the smallest number which is exactly divisible (a) 6033 (b) 6330
by 9, 12, 15 and 16? (c) 6303 (d) 6003
og lcls NksVh la[;k D;k gS tks 9] 12] 15 vkSj 16 ls iw.kZr%11.
foHkkT;
Let xgks\
be the smallest four digit number, such tha
(a) 720 (b) 700 when it is divided by 5, 6 and 7, it leaves th
(c) 680 (d) 1440 remainders 2, 3 and 4 respectively. Find the valu
3. What is the least number, which when increased by of K, if K = (3x + 4).

r
17, is completely divisible by 7, 12, 13 and 5? eku yhft, x pkj vadksa dh lcls NksVh la[;k gS] ftls 5]
og lcls NksVh la[;k D;k gS] ftls 17 ls c<+kus ij og 7] 12] 13 7 ls foHkkftr djus ij Øe'k% 2] 3 vkSj 4 'ks"k cprs Kd

si
vkSj 5 ls iw.kZr% foHkkT; gks tkrh gS\ eku Kkr dhft,] ;fn K = (3x + 4) gSA
(a) 5460 (b) 2730

y
(a) 1047 (b) 3452
(c) 4200 (d) 5443 (c) 1081 (d) 3145
4. The least common multiple of m and n is 21. The LCM 12. Find the sum of digits of a smallest number whic

n
of 8m and 5n is: when divided by 7, 8 and 9 leaves remainder 2, 3 an
m vkSjn dk y?kqÙke lekioR;Z 21mgSA vkSj8n 5 dk LCM gS 4 respectively but exactly divisible by 5.
,d lcls NksVh [;kla ds vadksa dk ;ksx Kkr dhft, ftls

ja
(a) 40 (b) 840
(c) 1680 (d) 2720 vkSj 9 ls foHkkftr djus ij Øe'k% 2] 3 vkSj 4 'ks
5. ABBAB is a five digit number, which is exactly
ysfdu og 5 ls iw.kZr% foHkkT; gSA
a th
divisible by five consecutive numbers 11, 12, 13, 14
and 15. Find the value of (A+B). (a) 8 (b) 13
(c) 7 (d) 6
ABBAB ,d ik¡p vadksa dh la[;k gS] tks ik¡p Øekxr la[;kvksa
13. When a number is divided by 11, the remainder is 5
11] 12] 13] 14 vkSj 15 ls iw.kZr% foHkkT;
(A+B) dk gSA
ekuKkr dhft,A When the same number is divided by 13, th
ty a

(a) 60 (b) 0 remainder is 2. How many such numbers are ther

(c) 6 (d) 14 between 400 and 700?
6. What is the difference between largest four digit tc fdlh la[;k dk s 11 ls foHkkftr fd;k tkrk gS] rk
di M

number and largest five digit number, which is exactly

divisible by 12, 14, 32 and 40?
cprk gSA tc mlh la[;k dks 13 ls foHkkftr fd;k tkrk g
vadh; la[;k vkSj lcls cM+h ikap vadh; la[;k ds 2 cprk gSA 400 vkSj 700 ds chp ,slh fdruh la[;k,¡ g
lcls cM+h pkj
(a) 3 (b) 5
chp D;k varj gS] tks 12] 14] 32 vkSj 40 ls iw.kZr% foHkkT; gS\
(c) 7 (d) 1
(a) 92559 (b) 90720 14. If the LCM of two numbers 1728 and K is 5184. Fin
(c) 95414 (d) 90,000 the number of possible values of k.
7. What is the sum of the numbers between 400 and
700 such that when they are divided by 5, 15 and
;fn nks la[;kvks K dk LCM 5184 gSA
a 1728 vkSj K ds laHk
25, it leaves remaindar as 3 always? ekuksa la[;k
dh Kkr dhft,
400 vkSj 700ds chp dh la[;kvksa dk ;ksx D;k gS tc mUgsa (a) 5 (b) 6
(c) 7 (d) 8
5] 15 vkSj 25 ls foHkkftr fd;k tkrk gS] rks 'ks"k ges'kk 315.
cprk gS
Let N be the least number, which when divided b
(a) 1622 (b) 2226 15, 25, 35, 40 and 42. the remainder in each cas
(c) 2262 (d) 6222 is 1 and N is divisible by 271. what is the sum o
8. Let K be the least number divisible by 12, 18, 21 and digits of N?
15 and K is also a perfect square. What is the
remainder, when K is divided by 231?
eku yhft, N lcls NksVh[;k
la gS] ftls 15] 25] 35] 40 vk
eku yhft, K] 12] 18] 21 vkSj 15 foHkkT;
ls lcls NksVh la[;k ls foHkkftr djus ij çR;sd fLFkfr esa 'ks"kiQy 1 cprN
K ,d iw.kZ oxZh Hk
gS vkSj gSA tcK dks 231 ls foHkk ftr fd;k ds vadksa dk ;ksx
271 ls foHkkT; NgSA D;k gS
(a) 8401 (b) 12
tkrk gS] rks 'ks"kiQy D;k gksxk\ (c) 1841 (d) 13
(a) 512 (b) 210 16. The sum and LCM of two numbers are 493 and 2088
(c) 221 (d) 141 Find the difference between the numbers.
9. What least number must be subtracted from 5000 so nks la[;kvksa;ksx
dk vkSj y?kqÙke lekioR;Z 493 vk
that the resulting number when divided by 7, 12, 15,
the remainder in each case is 6? la[;kvksa ds chp varj Kkr dhft,A
ls dkSu lh lcls NksVh la[;k ?kVkbZ tk, fd ifj.kkeh la[;k(a)
5000 esa 17
(c) 29
(b) 72
(d) 92
dks 7] 12] 15 ls foHkkftr djus ij çR;sd fLFkfr esa 'ks"k 617.
cps\ Three bells commence tolling together and toll a
(a) 374 (b) 376 intervals of 9 sec, 11 sec and 12 sec respectively. I
(c) 378 (d) 716 30 mins, how many times do they toll together?
 11 lsdaM vkSj 12 lsdaM ds varjky ij ctrh gSaA 30 feuV esa]xbZ
os 'krZ dks larq"V djus okys ;qXeksa dh la[;k Kk
,d lkFk fdruh ckj ctrh gSa\ (a) 8 (b) 10
(a) 8 (b) 4 (c) 11 (d) 12
(c) 7 (d) 6 25. Three sets of Hindi, Science and Social Science boo
18. The HCF of four numbers is 31. Find the difference containing 672, 480 and 192 books respective
between the second and fourth numbers, if they are have to be stacked in such a way that all the book
in ratio 3 : 5 : 4 : 7. are stored subject wise and the height of each stack
pkj la[;kvksa dkHCF 31 gSA nwljh vkSj pkSFkh la[;k ds chp dk
is the same. Total number of stackes will be:
varj Kkr dhft,] ;fn os 3 % 5 % 4 % 7 ds vuqikr esa gSa fganh] foKku vkSj lkekftd foKku dh iqLrdksa ds
(a) 62 (b) 31 Øe'k% 672] 480 vkSj 192 iqLrdsa gSa] mUgsa bl
(c) 93 (d) 115 gS fd lHkh iqLrdsa fo"k;okj j•h tk,a vkSj çR;sd <
19. The sum of two numbers is 1904 and their HCF is leku gksA <sjksa dh dqy la[;k gksxh%
56. Find the number of such pairs. (a) 14 (b) 42
nks la[;kvksa dk ;ksx 1904 gS vkSj mudk
HCF 56 gSA ,sls ;qXeksa (c) 44 (d) 96
dh la[;k Kkr dhft,A 26. The traffic light at three different road crossing
(a) 4 (b) 9 change after 24 seconds,36 seconds and 54 second
(c) 3 (d) 8 respectively. If they all change simultaneously a

r
20. When 589, 823 and 1057 are divided by a certain 11:00 Am, then at what time will they change for th
three digit number aab, then remainder in each case third time simultaneously.

si
is 4. Find the value of (b-a). rhu vyx&vyx lM+d ØkWflax ij VªSfiQd ykbV Øe'k%
tc 589] 823 vkSj 1057 dks ,d fuf'pr rhu vadksa dh la[;k
a lsdaM vkSj 54 lsdaM ds ckn cnyrh gSaA ;fn os 00lHkh
cts ,d

an by
ls foHkkftr fd;k tkrk gS] rks çR;sd fLFkfr esa 'ks"kiQy 4 gksrk gSA
lkFk cnyrh gSa] rks os rhljh ckj ,d lkFk fdl le; cnysaxh
(b-a) dk eku Kkr dhft,A (a) 11:03:36 AM (b) 11:05:12 AM

n
(a) 8 (b) 5 (c) 11:05:36 AM (d) 11:07:12 AM
(c) 6 (d) 7 27. When 4546, 5398 and 6889 are divided by th
21. When we multiply a two digit number by sum of greatest number m, the remainder in each case is n

ja
its digits, number 280 is obtained. If we reverse What is the value of (2m-3n)?
R s
the number and multiply it by sum of its digit, tc 4546]5398vkSj 6889 dks lcls cM+h la[;k m ls foHk
we get 820. What is the sum of squares of digits fd;k tkrk gS]rksçR;sdfLFkfr
esa'ks"kiQy
n gksrk gSA
(2m-3n) d
a th
of original number?
eku D;k gS\
tc ge nks vadksa dh la[;k dks mlds vadksa ds ;ksx ls xq.kk djrs
(a) 189 (b) 140
gSa] rks la[;k 280 çkIr gksrh gSA ;fn ge la[;k dks myV nsa vkSj
(c) bls
207 (d) 134
mlds vadksa ds ;ksx ls xq.kk djsa] rks gesa 820 çkIr 28.
gksrk AgSA ewyof wood 4.25 m
plank long and 3.4 m wide to b
ty a

la[;k ds vadksa ds oxks± dk ;ksx D;k gS cut into square pieces of equal size. How man
(a) 100 (b) 68 squares pieces of largest size can be cut from th
di M

(c) 64 (d) 70 plank, if no wastage is allowed?

22. Find the value of (x + y). ydM+h dk ,d r[rk 4-25 ehVj yack vkSj 3-4 ehVj p
(x + y) dk eku Kkr dhft,A cjkcj vkdkj ds oxkZdkj VqdM+ksa esa dkVk tkuk gS
1 2 3 4 rks r[rs ls lcls cM+s vkdkj ds fdrus oxkZdkj VqdM+s
x = LCM of  , , ,  (a) 45 (b) 10
2 3 4 5
(c) 40 (d) 20
1 2 3 4
y = HCF of  , , ,  29. A shopkeeper has three types of rice weighing 126 k
2 3 4 5 189 kg and 315 kg respectively. What will be th
(a) 72 (b) 60 minimum number of bags of equal size in which a
721 60 the rice can be kept without mixing them?
(c) (d)
60 721 ,d nqdkunkj ds ikl rhu çdkj ds pkoy gSa ftudk otu
23. The HCF of two numbers is 67 and their LCM is 437 126 fdxzk] 189 fdxzk vkSj 315 fdxzk gSA leku vk
times the HCF. If one of the numbers lie between 1250
and 1300, then find the other number- dh U;wure la[;k D;k gksxh ftuesa
Hkh pkoyksa
l dks fcuk f
tk lds\
A

nks la[;kvksaHCF
dk 67 gS rFkk mudk LCM] HCF dk 437
xquk gSA ;fn muesa ls ,d la[;k 1250 rFkk 1300 ds chp gS] (a)
rks 10 (b) 12
(c) 15 (d) 8
nwljh la[;k Kkr dhft, 30. What is the difference between LCM and HCF of 9
(a) 1541 (b) 1273
8.1, 0.27 and 0.09?
(c) 1261 (d) 1316
24. The sum of two positive numbers is 437 and their 9] 8-1] 0-27 vkSj 0-09HCF
ds vkSjLCM esa D;k varj gS
HCF is 19. Find the number of pairs Satisfying the (a) 81 (b) 9
given condition. (c) 82 (d) 80.91

ANSWER KEY
1. (b) 2. (a) 3. (d) 4. (b) 5. (c) 6. (b) 7. (c) 8. (b) 9. (a) 10. (c)
11. (d) 12. (b) 13. (a) 14. (c) 15. (d) 16. (c) 17. (b) 18. (a) 19. (d) 20. (c)

21. (b) 22. (c) 23. (a) 24. (c) 25. (a) 26. (d) 27. (c) 28. (d) 29. (a) 30. (d)
 SOLUTIONS
1. (b) LCM of two co-prime number is equal to their n = 5 will give the value of smallest four digi
multiplication. x = 1050 – 3 = 1047
LCM = 201 K = 3x + 4 = 3 × 1047 + 4 = 3145
45 12. (b) (7 – 2), (8 – 3), (9 – 4) = 5
45% of LCM = × 201 = 90.45 Number = LCM of (7, 8 and 9) × n – 5
100
2. (a) 9 = 32 = 504n – 5
12 = 22 × 3 Number is divisible by 5
15 = 3 × 5 Number = 504 × 5 – 5 = 2515
16 = 24 Sum of digits = 2 + 5 + 1 + 5 = 13
LCM = 24 × 32 × 5 = 16 × 9 × 5 = 720 13. (a) Dividend = Quotient × Divisor + Remainder
3. (d) LCM of 5, 7, 12 and 13 = 5 × 7 × 12 × 13 = 5,460 11a + 5 = 13b + 2
Required number = 5460 – 17 = 5443  11a = 13b – 3

r
4. (b)  11a = 11b + 2b – 3
m × n = 3 × 7 = 21 (2b – 3) should be divisible by 11.

si
LCM of 8m and 5n = 8 × 5 × m × n b=7
= 40 × 21 = 840 Number = 13b + 2 = 91 + 2 = 93

an by
5. (c) LCM of 11, 12, 13, 14 and 15 = 60,060 Number of multiples of 93 between 400 and 700 = 3
On comparing with ABBAB, 14. (c) A = 1728 = 26 × 33

n
A = 6, B = 0 B=K
A+B=6+0=6 LCM = 5184 = 26 × 34

ja
6. (b) LCM of 12, 14, 32 and 40 = 3360 B = n × 34
R s
Largest four digit number divisible by 3360 Possible values of n = 20, 21, 22, 23, 24, 25, 26
Total 7 values.
a th
 9999 
=  3360  = 3279 LCMof 15,25,35,40,42  K  1
R 15. (d) Required no. =
271
Number = 9999 – 3279 = 6720
4200K  1
=
ty a

 99999  271
 
3360  R = 2,559 K = 2 (satisfy)
di M

Largest five digit number divisible by 3360 N = 4200 × 2 + 1 = 8401

= 99999 – 2559 = 97440 Sum of digits = 8 + 4 + 1 = 13
Difference = 97440 – 6720 = 90720 16. (c) Let, 1st no. = x, 2nd no. = y
7. (c) LCM of 5, 15 and 25 = 75 (x + y) = 493
Required numbers = (450 + 3), (525 + 3), (600 + 3), (xy) = 2088
(675 + 3) HCF of 493 and 2088 = 29
= 453, 528, 603, 678 493
Sum = 453 + 528 + 603 + 678 = 2262 x+y= = 17
29
8. (b) LCM of 12, 15, 18 and 21 = 22 × 32× 5 × 7
K, which is a perfect square = 22 × 32 × 52 × 72 2088
xy = = 72
= 44, 100 29
x = 9, y = 8
 44100  Difference between number = HCF(x – y)
 
231  R = 210
A

= 29(9 – 8) = 29
9. (a) LCM of 7, 12 and 15 = 420 17. (b) LCM of 9, 11 and 12 = 396 sec
Greatest number which is divisible by 7, 12 and 15 30  60
leaving remainder 6, and less than 5000 = 4626 No. of times they will toll together = =4
396
Required number = 5000 – 4626 = 374
18. (a) Let number be 3x, 5x, 4x and 7x
10. (c) LCM of 6, 15, 36 and 75 = 900
HCF = x = 31
Number which is when divided by 6, 15, 36 and 75
leaves remainer 3 and divisible by 11, would be Required difference = 7x – 5x = 2x = 2 × 31 = 62
(900k + 3) 19. (d) Sum of no. = (x + y) = 1904
K = 7 (satisfy) 1904
Sum of pair = x + y = = 34
Required number = 6303 56
11. (d) (5 – 2), (6 – 3), (7 – 4) = 3 Possible pairs = (11, 23) (13, 21), (15, 19), (25, 9
Required number (x) = LCM of (5, 6, 7) × n – 3 (27, 7), (29, 5), (31, 3), (33, 1)
= 210n – 3 Total 8 such pairs are possible
 (589 – 4), (823 – 4), (1057 – 4) 672 480 192
= 585, 819, 1053 Total no. of stacks =  
585 = 117 × 5 96 96 96
819 = 117 × 7 = 7 + 5 + 2 = 14
1053 = 117 × 9 26. (d) LCM of 24, 36 and 54 = 216 seconds
HCF = 117 First change = 11:00 AM
aab = 117 Third change = 11:00 AM + 2 × 216 seconds
b–a=7–1=6 = 11:07:12 AM
27. (c)
21. (b) 10x  y x  y   280  28 4546 5398 6889
10y  x x  y  820 82
diff.  852 1491
 10x + y = 28
x = 2, y = 8 HCF of 852 and 1491 = 213
x2 + y2 = 4 + 64 = 68 m = 213

22. (c) x = LCM of 1,2,3,4  12 = 12  4546 

n
HCF of 2,3,4,5 1  213  R = 73

r
2m – 3n = 426 – 219 = 207
HCF of 1,2,3,4 1
y = LCM of 2,3,4,5  60 28. (d) HCF of 425 and 340 = 85
 

si
425  340
1 721 No. of pieces = = 5 × 4 = 20
85  85

an by
x + y = 12 + =
60 60 29. (a) HCF of 126, 189 and 315 = 63
23. (a) HCF = 67
126 189 315

n
LCM = 437 × HCF = HCF × x × y  
Minimum no. of bags =
 x × y = 437 63 63 63
= 2 + 3 + 5 = 10

ja
R s
19 23 30. (d) 9, 8.1, 0.27, 0.09
x = 67 × 19 = 1273 (in range)
9 81 27 9
a th
y = 67 × 23 = 1541 = , , ,
24. (c) Let, 1st no. = x, 2nd no. = y 1 10 100 100
Sum of no. = (x + y) = 437 LCM of 9,81,27,9 81
437 LCM = HCF of 1,10,100,100  1 = 81
Sum of pair = (x + y) = = 23  
ty a

19
x + y = 23 HCF of 9,81,27,9 9
HCF = LCM of 1,10,100,100  100 = 0.09
(x, y) = (1, 22), (2, 21), (3, 20), (4, 19), (5, 18), (6,  
di M

17), (7, 16), (8, 15), (9, 14), (10, 13), (11, 12)
Total 11 pairs are possible. Difference = 81 – 0.09 = 80.91
A
 Algebra Sheet-01/ chtxf.kr
(CLASSROOM SHEET)

1 1 1
5. If x   –6 , what will be the value of x 5  5
1. If x + = 3 Find x x
x
1 1
1 1 ;fn x   –6 , gS rksx 5  dk eku D;k gksxk\
3 x x5
(i) x2 + 2 (ii) x + 3
x x SSC CGL 17/07/2023 (Shift-03)
(a) –7776 (b) –6726
4 1 5 1
(iii) x + (iv) x + (c) –6730 (d) –6732
x4 x5

r
1 1

si
6
6 1 7 1 6. If a + = 3, then a + 6 is equal to :
(v) x + (vi) x + a a
x6 x7

2. If x –
1
= 4 , Find an by (a) 319
SSC CHSL 21/07/2019 (Shift-02)
(b) 322

n
x (c) 780 (d) 730

 1
1

ja 2 1 7. If  x –   10, , what is the value of

R s
(i) x2 + (ii) x – x
x2 x2
a th

1
1 1 x4  ?
3
(iii) x – 3
4
(iv) x + 4 x4
x x
  1 1
;fn  x – x   10 gS] rks
x 4  4 dk eku D;k gS\
ty a

5 1 6 1 x
(v) x – (vi) x +
x5 x6
SSC CGL 26/07/2023 (Shift-03)
di M

1 1 (a) 10404 (b) 10402

3. If a + = 3, then the value of a4 + 4 is: (c) 10406 (d) 10400
a a

1 1 1 3 1
;fn a + =3 gS] rksa4 + dk eku D;k gksxk\ 8. If x – = 10, then x – 3 is equal to :
a a4 x x
SSC CPO 04/10/2023 (Shift-01) SSC CHSL 4/07/2019 (Shift-02)
(a) 970 (b) 1000
(a) 27 (b) 81
(c) 1030 (d) 1100
(c) 48 (d) 47
 3  2 1 
4. If a 
1 1
 7 , then a 5  5 is equal to: 9. If  3y –  = 5, find the value of  y  2  .
a a  y  y 
A

1 1  3  2 1 
;fn a   7 gS] rksa 5  fuEu esa ls fdlds cjkcj ;fn  3y – y  = 5 gS] rks y  y 2  dk eku
a a5
gS\ Kkr djsaA
SSC CGL 17/07/2023 (Shift-02) SSC CGL 19/07/2023 (Shift-02)
(a) 15127 47 49
(b) 13127 (a) (b)
9 9
(c) 14527 41 43
(d) 11512 (c) (d)
9 9
 1 1 1 1
10. If x – = 13 , what will be the value of x 4 + 4 ? 16. If a – = 4, then the value of a + is:
x x a a
1 1
;fn x – = 13 gS] rksx 4 + 4 dk eku D;k gksxk\ 1 1
x x ;fn a – =4 gS] rks
a+ dk eku gS%
a a
SSC CGL 03/12/2022 (Shift-01)
SSC CPO 05.10.2023 (Shift-2)
(a) 28561 (b) 29243
(c) 27887 (d) 29239
(a) 5 5 (b) 4 5

(c) 2 5 (d) 3 5
 1  5 1 
11. If  x –  =5, then the value of  x – 5  is :
x x
 1
1 1  17. If  x –  = 0.4, and x>0, what is the value
  5 x
;fn  x –  =5] rks  x – 5 
x x
dk eku gS%
 1 
SSC CHSL 07/08/2023 Shift-02 of  x 2 – 2  ?
 x 

r
(a) 3775 (b) 3740
 1   1

si
(c) 3715 (d) 3725 ;fn  x – x  = 0.4, vkSjx > 0 rks x 2 – 2  dk
x
1
eku D;k gS\
12. If x –

1
x
an by
= – 6, what will be the value of
SSC CHSL 07/08/2023 Shift-02

n
x5 – ?
x5 4 2
(a) 26 (b) 29
;fn x –
1

ja
= – 6 gS] rks
1
x5 – 5 dk eku D;k gksxk\
25 25
R s
x x 7 3
(c) 26 (d) 29
a th

SSC CGL 18/07/2023 (Shift-02) 25 25

(a) – 8898 (b) – 8896
 2 1 
(c) – 8886 (d) – 8892 18. If  x + 2  = 6 and 0 < x < 1, what is the
x
ty a

 1  1  4 1
13.
6
If  y    4, find the value of  y  6  . value of x – 4 ?
 y  y  x
di M

  1 1
 1  6 1  ;fn  x 2 + 2  = 6 vkSj0 < x < 1 gS] rk 4
sx – 4
;fn  y    4, gS rks y  y 6  dk eku Kkr djsa  x  x
 y
dk eku D;k gksxk\
SSC CGL 25/07/2023 (Shift-01) SSC CGL 26/07/2023 (Shift-01)
(a) 5774 (b) 4096 (a) 24 2 (b) 24 2
(c) 5776 (d) 5778
(c) 12 10 (d) 12 10
1 1
14. If x – = 4 , then x 2 + is equal to : 1
x x2 19. If x 2 – = 4 2 , what is the value of
x2
A

SSC CGL 10/06/2019 (Shift-01)

1
(a) 192 (b) 326 x4 – ?
x4
(c) 322 (d) 256
1 1
1 ;fn x 2 – =4 2 gS] rksx 4 – dk eku D;k gS\
x –
1
= 6 , then x 2 + 2 is equal to : x2 x4
15. If
x x
SSC CGL 27/07/2023 (Shift-02)
SSC CGL 11/06/2019 (Shift-02)
(a) 16 2 (b) 8 2
(a) 62 (b) 40
(c) 54 (d) 66 (c) 24 2 (d) 32 2
  1 (a) 3.5 (b) 4.5
20. If  x +  = 5 2 , and x > 1, what is the value (c) 2.5 (d) 5.5
x
 6 1 
of  x – 6  ? 1
x 25. If x + = 5, x  0 then the value of
x
 1   6 1 
;fn  x + x  = 5 2 , vkSjx > 1, rks x – x 6  1
x4 +
x2
dk eku D;k gS\ 2 is equal to :
x - 3x +1
SSC CHSL 02/08/2023 Shift-02
SSC CHSL 20/10/2020 (Shift- 01)
(a) 22970 23 (b) 23030 23
(a) 55 (b) 60
(c) 23060 23 (d) 22960 23 (c) 65 (d) 50
 1
21. If  x –   2 2 , and x > 1, what is the 1
x 26. If x – = 5, x  0, then what is the value
x

r
 6 1 
value of  x – 6  . x 6 + 3x 3 – 1
x of ?

si
x 6 – 8x 3 – 1
 1  6 1 
;fn  x –   2 2 vkSj x > 1 gS] rks x – 6 
x
dk eku D;k gksxk\ an by x
13
SSC CGL TIER II 16/11/2020

11

n
SSC CGL 20/07/2023 (Shift-03) (a) (b)
12 13
(a) 372 6 (b) 384 6

ja
R s
3 4
(c) 396 6 (d) 420 6 (c) (d)
8 9
a th

 1 27. If x2 + 3x + 1 = 0, then what is the value of

22. If  x   = 5, and x >1, what is the value
x 1
x6 + 6 ?
 8 1  x
ty a

of  x – 8  ? SSC CGL 3/03/2020 (Shift- 02)

x
 1  8 1  (a) 324 (b) 322
di M

;fn  x   = 5 vkSj x >1 gS] rks x – 8  dk (c) 318 (d) 327

x x
eku D;k gksxk\ 28. If x 2 – 2 5x +1 = 0 , then what is the value
SSC CGL 18/07/2023 (Shift-03) 5 1
of x + 5 ?
(a) 60605 21 (b) 60615 21 x
SSC CGL 5/03/2020 (Shift- 01)
(c) 60705 21 (d) 60725 21
(a) 610 5 (b) 408 5
1
23. If x + = 8, then find the value of (c) 612 5 (d) 406 5
x
5x
. 6x
x 2 +1 - 6x 29. If = 1, x > 0, then the value
2x 2 + 5x - 2
A

SSC CHSL 14/10/2020 (Shift- 03)

(a) 2.5 (b) 6 3 1
(c) 5 (d) 6.5 of x + is :
x3
1 SSC CGL 07/06/2019 (Shift- 03)
24. If x   10 , then find the value of
x
3 5
(a) 17 (b) 17
7x 8 8
.
x 2  1 – 8x 5 3
(c) 17 (d) 17
SSC CHSL 03/06/2022 (Shift- 01) 16 4
30. If x² – 5x + 1 = 0, then the value of
174 144
x6 + x4 + x2 +1 (a) (b)
=?
125 125
5x 3
SSC CGL 01/12/2022 (Shift-01)
114 119
(c) (d)
(a) 30 (b) 25 25 25
2
(c) 23 (d) 28 37. If 20x – 30x + 1 = 0, then what is the value
31. If x² – 3x + 1 = 0, then the value of 2 1
of 25x + is :
16x 2
1 SSC CGL 05/03/2020 (Shift-02)
x4 
x 2 is:
2 1 1
x  5x  1 (a) 53 (b) 58
2 2
SSC CGL MAINS 03/02/2022
3 3
(c) 53 (d) 58
9 27 4 4
(a) (b)
4 8 1
38. If 4a + = 4, then the value of
5a

r
5 1
(c) (d) 2 25a2 + is :

si
2 16a2
32. If x2 – 3x – 1 = 0, then the value SSC CHSL 21/10/2020 (Shift-03)

an by
o f ( x 2 + 8 x – 1) (x3 + x –1) –1 is :
SSC CGL 07/06/2019 (Shift- 02) (a)
45
2
(b)
55
2

n
3
(a) (b) 8 43 45
8 (c) (d)

ja 2 4
R s
(c) 1 (d) 3 39. If 2x² + 5x + 1 = 0, then one of the values
a th

1 2 1 1
33. If 5x + = 5, find the value of 9x + of x – ?
3x 25x 2 2x
(a) 0 (b) 5 SSC CGL MAINS 29/01/2022
ty a

(c) 4 (d) 7.8

17 13
1 1 (a) (b)
di M

3
34. If 3x + = 5 then find 8x + 2 2
2x 27x 3
5 13
1 10 (c) (d)
(a) 118 (b) 30 2 2
2 27
(c) 0 (d) 1 1
40. If 5x – = 6, x > 0, then find the value
1 4x
35. If x + = 3, then the value of
16x 1
of 25x ² – .
1 16 x ²
16x 3 + is :
256x 3 SSC CGL 21/04/2022 (Shift 01)
A

SSC CGL TIER II 12/09/2019 (a) 6 41 (b) 36

(a) 423 (b) 441
(c) 246 (d) 6 31
(c) 432 (d) 414
1  2 1 
36. If 5x + = 4, then the value of 41. If  4x + 4 x 2  = 2,than what is the value of
3x
 3 1 
1  8x + ?
9x 2 + is : 8x 3 
25x 2
(a) +2 (b) –2
SSC CGL 04/03/2020 (Shift- 01)
(c) +4 (d) –4
  1 1 3 1
42. If  0.4x   = 5, what is the value of 50. If x4 + = 322, find x – 3
 x x4 x
(a)  76 (b) 76
 3 1  (c) –76 (d) 95
 0.064x  3  ?
 x 
1
51. If x + = 2, find
SSC CGL 20/04/2022 (Shift 01) x
(a) 119 (b) 125 (i) x20 + x19 + x18 + ........+ x2 + x + 1
(c) 110 (d) 105 (ii) x49 – x48 + x47 – x46 + ....... + x3 – x2 + x + 1
1 1 1
43. If x2 + = 27 , then find (iii) x12 + (iv) x17 +
x2 x12 x9
1 1
(i) x + (ii) x – 1
x x (v) x15 –
x17
1
44. If x4 + = 23, find 1 

r
x4
52. If p +  p  = 2, then find the value of p × p × p.
1 1

si
3
(i) x + (ii) x + 3 SSC CHSL 18 /03/2020 (Shift-02)
x x
(a) 4 (b) 8
45.

of x –
1
is : an by
If x8 – 1442x4 + 1 = 0, then a possible value
(c) 1 (d) 2

n
x 1
53. If x + = –2
SSC CGL TIER-II (11/09/2019) x
(a) 5
ja (b) 8 (i) x39 + x38 + ......... + x2 + x + 1
R s
(c) 4 (d) 6 (ii) x100 – x99 + x98 – x97 + ....... + x2 – x + 1
a th

1
46. If x > 0 and x4 + = 142, what is the value 1
x4 (iii) x5 –
x5
1
of x7 + ? 1 1
x7 7
ty a

(iv) x18 + (v) x –

SSC CGL 20/07/2023 (Shift-02) x18 x9
1
di M

(a) 1561 14 (b) 1563 14 = 1 , find

54. If x +
x
(c) 1560 14 (d) 1562 14
(i) x18 + x12 + x6 + 1
4 1 (ii) x99 + x96 + x93 + x90 + ....... + x6 + x3 + 1
47. If x > 0, and x + = 2207, what is the
x4 (iii)x45 – x42 + x39 – x36 + ...... + x9 – x6 + x3 + 1
7 1 (iv) x32 + x46 + x90 + 2
value of x + ?
x7 1
SSC CGL 02/12/2022 (Shift-01)
55. If x + = 3 , find
x
(a) 710649 (b) 710647 1 92 1
(i) x90 + 90 (ii) x + 92
(c) 710654 (d) 710661 x x
48. 4 –4
If x + x = 47, (x > 0), then the value of (iii) x102 + x96 + x101 + x95 + x100 + x94
A

(2x – 3)2 is : 1
SSC CHSL 8/07/2019 (Shift-03) (iv) x6 – +2 (v) x18 + x12 + x6 + 1
x6
(a) 2 (b) 3 2
(c) 5 (d) 4  1
56. If  x   = 3, then what is the value of
49. 4 –4
If x + x = 194, (x > 0), then the value of x
(2x – 4)2 is : x6 + x–6?
SSC CHSL 09/07/2019 (Shift-01) SSC CGL MAINS (08/08/2022)
(a) 15 (b) 20 (a) 6 (b) 2
(c) 12 (d) 16 (c) –2 (d) –6
 65. If P = 7 + 43 and PQ = 1, then what is
1
57. If x + = 3 , then the value of
x  1 1 
the value of  2  2  = ?
x18 + x12 + x6 + 1 is : P Q 
SSC CHSL 21/10/2020 (Shift- 03) (a) 148 (b) 189
(a) 0 (b) 2 (c) 194 (d) 204
(c) 3 (d) 1
5– 4 5+ 4
66. If x = and y = then the
1 5+ 4 5– 4
58. If x + = – 3 then x67 + x53 + x43 + x29 +
x
x 2 – xy + y 2
x24 + x12 + x6 + 3 is value of =?
x 2 + xy + y 2
(a) 3 (b) 0 SSC CGL 08/12/2022 (Shift-02)
361 341

(c) 2 2 + 3  
(d) 2 2 – 3  (a)
363
(b)
343
384 321
1 (c) (d)
387 323

r
59. If x = 2 + 3 , find x + x
1 1

si
(a) 5 (b) 4 67. If m  = 5, then (m – 3)8 + =?
m–3 (m – 3)8
(c) –4 (d) 2 3

an by1
(a) 2
(c) –2
(b) 0
(d) 15

n
60. If x = 7 + 4 3 then x –
x 1
68. If x  = 11, then (x – 11) 12 +
x–9
(a) 8 3

ja (b) 4
R s
(c) –4 (d) –2 3 1
=?
a th

(x – 11)13
61. If x = 2 + 5 then the value of x3 – x–3 is :
(a) 0 (b) 2
SSC CHSL 08/07/2019 (Shift-03) (c) –2 (d) 1
ty a

(a) – 52 (b) 52
1 1
(c) 76 (d) – 76 69. If a  = 6, then (a – 3)7 + =?
a–4 (a – 7)3
di M

1
62. If x = 3 + 2 2 , then the value of x –
x 7 1
(a) 63 (b) 255
is: 8 8
SSC CHSL 26/10/2020 (Shift-03)
7
(a) 2 (b) 1 (c) 127 (d) 216
8
(c) 0 (d) 3
70. If x² –22x + 111 = 0, then what is the value
 1  1
63. If x = 1 + 2 , then the value of x +
 x  of (x – 8)² –
(x – 8)2
=?
is:
1
;fn x² –22x + 111 = 0 rks(x – 8)² – (x – 8)2 =\
A

SSC CHSL 17/03/2020 (Shift-03)

(a) 2.1014 (b) 2.1973
(a) 1210 (b) 85
(c) 1.9876 (d) 1.9996
(c) 83 (d) 18
64. If p = 7 + 43, then what is the value of
p6 + p4 + p2 + 1 1
? 71. If x² + 13x + 39 = 9, then (x + 8)5 –
p3 (x  8)5
SSC CGL 06/12/2022 (Shift-02)
=?
(a) 2617 (b) 2167 (a) 393 (b) 396
(c) 2716 (d) 2176 (c) 392 (d) 394
72. If x² – 12x + 33 = 0, then what is the value (a) 110 (b) 1
1 (c) 125 (d) 140
of (x – 4)4 + =? 1 1
(x – 4)4 5
74. If x –  7 , then find x 35 – =?
x5 x 35
1
;fn x² – 12x + 33 = 0 rks(x – 4)4 + (x – 4)4 = (a) 791 7 (b) 789 7
(c) 790 7 (d) 792 7
dk eku D;k gS\
1
(a) 227 (b) 326 75. If 3 x  = 3, x > 0 then find x²(18x² – 7)?
(c) 167 (d) 194 2 x
73. If (x – a)(x – b) = 1 and a – b + 5 = 0, then 1
;fn 3 x  = 3, x > 0 gS]rksx²(18x² – 7) Kkr
1 2 x
(x – a)3 –
(x – a)3
=? dhft,\
;fn (x – a)(x – b) = 1 vkSja – b + 5 = 0 rks 1 2
(a) – (b) –
1 36 63
(x – a)3 – =?
(x – a)3 1 8
(c) – (d) –
72 81

r
si
ANSWER KEY
1.(i)7 (ii)18 an by
(iii)47 (iv)123 (v)322 (vi)843

n
2.(i)18 (ii) 8 5
ja(iii)76 (iv)322 (v)1364 (vi)5778
R s
a th

3.(d) 4.(a) 5.(b) 6.(b) 7.(b) 8.(c) 9.(d) 10.(d) 11.(a)

12.(c) 13.(d) 14.(c) 15.(a) 16.(c) 17.(a) 18.(a) 19.(c) 20.(b)

ty a

21.(c) 22.(a) 23.(a) 24.(a) 25.(a) 26.(a) 27.(b) 28.(a) 29.(b)

di M

30.(c) 31.(a) 32.(c) 33.(d) 34.(b) 35.(a) 36.(c) 37.(c) 38.(a)

39.(a) 40.(a) 41.(d) 42.(a) 43.(i) 29 (ii) 5 44.(i) 7 (ii) 4 7

45.(d) 46.(a) 47.(b) 48.(c) 49.(c) 50.(a)

51.(i) 21 (ii) 2 (iii) 2 (iv) 2 (v) 0 52.(c)

A

53.(i) 0 (ii)101 (iii) 0 (iv) 2 (v) 0

54.(i) 4 (ii) 0 (iii) –14 (iv) 2
55.(i) –2 (ii) –1 (iii) 0 (iv) 2 (v) 0 56.(c) 57.(a) 58.(c) 59.(b)
60.(a) 61.(c) 62.(a) 63.(b) 64.(c) 65.(c) 66.(d) 67.(a) 68.(a)
69.(c) 70.(a) 71.(a) 72.(d) 73.(d) 74.(a) 75.(c)
 ALGEBRA / chtxf.kr
(CLASSROOM SHEET-02)
9. If a2 + b² + c² + 84 = 4 (a – 2b + 4c), then
Concept of Perfect Square ab – bc  ca is equal to:
SSC CPO 24/11/2020 (Shift-2)
1. If (a – 3)2 + (b – 4)2 + (c – 9)2 = 0,
(a) 4 10 (b) 10
then a + b + c = ?
(a) –4 (b) 4 (c) 5 10 (d) 2 10
(c) ±4 (d) ±2 QUESTIONS BASED ON (a + b + c)2
2. If (a – 1)2 + (b + 2)2 + (c + 1)2 = 0, then

r
find 2a – 3b + 7c = ? 10. If (a + b + c) = 16, and (a 2 + b 2 + c 2 )
= 90, find the value of (ab + bc + ca).

si
(a) 12 (b) –11
SSC CGL 14/07/2023 (Shift-01)
(c) 3 (d) 1
3.
an by
If (a – 4) + (b – 5)2 + (c – 3)2 = 0,
2 (a) 84 (b) 83

n
a +b (c) 82 (d) 81
then the value of is:
c 11. If a + b – c = 5 and ab – bc – ac = 10, then
(a) 0
ja (b) 3 find the value of a2 + b2 +c2.
R s
(c) 1 (d) –3 SSC CGL 18/04/2022 (Shift- 01)
a th

4. If (x + y – z)2 + (y + z – x)2 + (z + x – y)2 = 0,

(a) 15 (b) 45
then x + y + z = ?
(c) 5 (d) 40
(a) 3 (b) 3 3
ty a

(c) 3 (d) 0 12. If a² + b² + c² = 6.25 and (ab+bc+ca) = 0.52,

2 2 2 what is the value of (a+b+c), if (a+b+c)<0?
5. If a + b + c = 2 (a – b – c) – 3, then the
di M

value of 4a – 3b + 5c is – SSC CGL 11/04/2022 (Shift- 03)

(a) 2 (b) 3 (a) ± 2.7 (b) – 2.7
(c) 5 (d) 6 (c) – 2.8 (d) ± 2.8
6. If a² + b² + 49c² + 18 = 2 (b–28c–a) then
13. If x + y + z = 13, x² + y² + z² = 91 and
the value of (a+b – 7c) is:
xz = y², then the difference between z and x is:
SSC CGL 12/04/2022 (Shift- 2)
SSC CHSL 05/08/2021 (Shift- 01)
(a) 4 (b) 3
(c) 2 (d) 1 (a) 3 (b) 8
7. If a2 +b2 + 49c2 + 18 = 2 (b – 28c – a), then (c) 5 (d) 9
A

the value of (a – b – 7c) is:

1 1 1
SSC CGL 18/04/2022 (Shift- 2) 14. If a + b + c = 3, a² + b² + c² = 6 and + + = 1,
a b c
(a) 2 (b) 1
Here, a, b & c are non-negative, then abc = ?
(c) 3 (d) 4
8. If a² + b² + 49c² + 18 = 2 (b + 28c –a), then 2 3
the value of (2a – b + 7c) is: (a) (b)
3 2
SSC CGL 19/04/2022 (Shift- 2)
(a) – 3 (b) 1 1 1
(c) (d)
(c) – 4 (d) 5 2 3
 22. I f x + y + z = 2, x³ + y³ + z³ – 3xyz = 74,
QUESTIONS BASED ON
then (x² + y2 + z²) is equal to :
(a³ + b³ + c³–3abc) SSC CGL 23/08/2021 (Shift- 03)
15. If a + b + c = 6, a² + b² + c² = 14 and ab + (a) 22 (b) 29
bc + ca = 11, then what is the value of a³ + (c) 26 (d) 24
b³ + c³ –3abc ? 23. If x + y + z = 13, x2 + y2 + z2 = 133 and x3 +
SSC CGL 02/12/2022 (Shift-03) y3 + z3 = 847, then the value of 2 xyz is:
(a) 31 (b) 12 SSC CPO 24/11/2020 (Shift- 02)
(c) 18 (d) 42 (a) 8 (b) 7
16. If a + b + c = 5, a³ + b³ + c³ = 85 and abc (c) –9 (d) –6
= 25, then find the value of a² + b² + c² –
24. If a = 355, b = 356, c = 357 find a 3 +
ab – bc – ca.
b 3 + c 3 – 3abc = ?
(a) 2 (b) 4
SSC CHSL 15/10/2020 (Shift- 01)
(c) 6 (d) 8 (a) 3208 (b) 3202

r
17. Factorize the given algebraic expression. (c) 3206 (d) 3204

si
x³ + 27y³ + 64z³ – 36xyz 25. If x = 222, y = 223 and z = 224, then find
(a) (x + 3y + 4z) (x² + 9y² + 16z² + 3xy + the value of x³ + y³ + z³ – 3xyz.
12yz + 4xz)
an by
(b) (x + 3y + 4z) (x² + 9y² + 16z² – 12xy² – (a) 2007
SSC CGL 12/12/2022 (Shift-04)
(b) 2004

n
3yz – 4xz) (c) 2006 (d) 2005
(c) (x – 3y – 4z) (x² + 9y² + 16z² – 3xy – 26. Find a 3 + b 3 + c 3 – 3abc if a = 1001,

ja b = 1004 and c = 1007.

R s
12yz – 4xz)
(d) (x + 3y + 4z) (x² + 9y² + 16z² – 3xy – (a) 80235 (b) 81234
a th

12yz – 4xz) (c) 79356 (d) 81324

18. What is the value of 27. If x = z = 225, and y = 226 then x3 + y3 + z3
– 3xyz = ?
(a) 765 (b) 676
ty a

2.2473  1.7303  1.0233 

  (c) 576 (d) 674
 –3  2.247  1.730  1.023 
28. If a = 2022, b = 2021 and c = 2020, then
di M

2.2472  (1.730)2  1.0232 – 2.247  1.730

  value of a2 + b2 + c2 – ab – bc – ca is:
 –1.730  1.023 – 2.247  1.023  SSC CHSL 26/05/2022 (Shift- 03)
(a) 1.730 (b) 4 (a) 2 (b) 4
(c) 5 (d) 5.247 (c) 3 (d) 1
19. If a + b + c = 5 and ab + bc + ca = 7, then 29. If x = 2015, y = 2014 and z = 2013 then
the value of a³ + b³ + c³ – 3abc is: the value of x2 + y2 + z2 – xy – yz – zx
SSC CPO 05/10/2023 (Shift-01) (a) 3 (b) 4
(a) 20 (b) 25 (c) 5 (d) 6
(c) 15 (d) 30 30. If a = 101, b = 102 and c = 103, then a² +
20. If (a + b + c) = 14, and (a3 + b3 + c³ – 3abc) b² + c² – ab – bc – ca
A

= 98, find the value of (ab + bc + ca). SSC CGL 24/07/2023 (Shift-01)
SSC CGL 26/07/2023 (Shift-04) (a) 2 (b) 4
(a) 60 (b) 64 (c) 3 (d) 6
(c) 65 (d) 63 31. If x = 32, y = 33 and z = 35, then evaluate
21. If (a + b + c) = 12, and (a2 + b2 + c2) = 50, the expression x3 + y3 + z3 – 3xyz.
find the value of (a3 + b3 + c3 – 3abc).
SSC CHSL 01/06/2022 (Shift- 03)
SSC CGL 19/07/2023 (Shift-02)
(a) 1120 (b) 1000
(a) 36 (b) 24
(c) 42 (d) 48 (c) 900 (d) 700
32. If ab + bc + ca = 8 and a2 + b2 + c2 = 20, 40. Find the value of (1.6)3 – (0.9)3 – (0.7)3.
1 (a) 3.24 (b) –3.24
then possible value of (a + b + c)[(a – b)2 (c) 3.024 (d) –3.024
2
+ (b – c)2 + (c – a)2] is : 41. If (3x + 1)3 + (x – 3)3 + (4 – 2x)3 + 6 (3x + 1)
(x – 3)(x – 2) = 0, then what is the value
SSC CGL 06/06/2019 (Shift- 01)
of x?
(a) 72 (b) 56
SSC CHSL 11/07/2019 (Shift- 01)
(c) 84 (d) 80
33. If (a + b + c) = 19, and (a2 + b2 + c2) = 155,
1
(a) – 1 (b) –
find the value of (a – b)2 + (b – c)2 + (c – a)2. 2
SSC CGL 18/07/2023 (Shift-02) 1
(c) 1 (d)
(a) 104 (b) 108 2
(c) 100 (d) 98 42. If (4x – 5)³ + (x —2)³ + 27 (2x – 5)³ = 9
34. 3 3 3
If (a + b + c – 3abc) = 405, and (a + b + c) (4x –5)(x – 2) (2x – 5), then the value of
= 15, find the value of (a – b)2 + (b – c)2 +
 3
(c – a)2.  x  2  will be:

r
 
SSC CGL 20/07/2023 (Shift-01)

si
(a) 27 (b) 54 SSC CHSL 05/08/2021 (Shift- 02)
(c) 18 (d) 45 1 5
35. 3 3 3

an by
If (a + b + c – 3abc) = 405, and (a – b)2 +
(b – c)² + (c – a)² = 54, find the value of (a
(a)
2
3
(b)
2
7

n
(c) (d)
+ b + c). 2 2
SSC CGL 26/07/2023 (Shift-02) 43. If (5x + 1) 3 + (x – 3) 3 + 8(3x – 4) 3 =

ja
R s
(a) 15 (b) 45 6(5x + 1)(x – 3)(3x – 4), then x is equal to:
(c) 9 (d) 27 CGL Tier II 12/09/2019
a th

36. Simplify the following expression. 5 1

(a) (b)
6 3
(59  59  59)  (54  54  54)  (57  57  57) 2 3
ty a

–3(59) (54) (57) (c) (d)

3 4
(59  54  57) 44. Given that (5x – 3)3 + (2x + 5)3 + 27 ( 4 – 3x)3
di M

= 9(3 – 5x)(2x + 5)(3x – 4), then the value

SSC CHSL 27/05/2022 (Shift- 01) of (2x +1) is :
(a) 38 (b) 76 SSC CGL Tier II 13/09/2019
(c) 170 (d) 19 (a) – 13 (b) 15
37. If a = 25, b = 15, c = –10, then (c) – 15 (d) 13
a 3 + b 3 + c 3 – 3abc QUESTIONS BASED ON
2 2 2 =?
a – b  + b – c  + c – a  (a3 + b3) and (a3 – b3)
(a) 30 (b) –15 45. If a3 – b3 = 216 and a – b = 6, then
(c) –30 (d) 15
(a + b)2 – ab is equal to :
38. If a3 – b3 – c3 – 3abc = 0, then
A

(a) a = b = c (b) a + b + c = 0 SSC CPO 15 /03/2019 (Shift- 03)

(c) a + c = b (d) a = b + c (a) 38 (b) 42
39. What is the value of a3 + b3 + c3 if (a + b + c) (c) 52 (d) 36
= 0?
46. If (a – b) = 4 and ab = 2, then (a3 – b3)
SSC CGL 14/07/2023 (Shift-02) is equal to :
(a) a + b + c2 – 3abc
2 2
SSC CGL 13/06/2019 (Shift- 01)
(b) 0
(c) 3abc (a) 92 (b) 88
(d) a2 + b2 + c2 – ab – bc – ca (c) 84 (d) 80
47. If (a – b) = 9, and (a3 – b3) = 4410, find the
value of ab. 52. If 2 2x3 – 3 3y 3 =  
2x – 3y (Ax2 + By2+

SSC CHSL 02/08/2023 Shift-02 Cxy), then the value of A + B2 – C2 is :

2

(a) 190 (b) 112 CGL Tier II (11/09/2019)

(c) 162 (d) 136 (a) 11 (b) 7
48. If a + b = 10 and ab = 6, then the value
(c) 19 (d) 10
of a3 + b3 is: 53. If (8x + 27y ) ÷ (2x + 3y) = (Ax2 + Bxy +
3 3

SSC CPO 03/10/2023 (Shift-01) Cy2), then the value of (5A + 4B + 3C) is :
(a) 860 (b) 820 SSC CGL 7/06/2019 (Shift- 03)
(c) 800 (d) 840 (a) 26 (b) 23
(c) 24 (d) 27
49. If a3 + b3 = 432 and a + b = 12,
then (a + b)2 – 3ab is equal to : 54. If  
5 5x3  2 2y 3 = A x  2y (Bx 2 +2y 2 +
SSC CPO 16 /03/2019 (Shift- 02) Cxy), then the value of (A2 + B2 – C2) is
(a) 42 (b) 52 CGL Tier II (13/09/2019)

r
(c) 36 (d) 38 (a) 15 (b) 20

si
(c) 30 (d) 40
x y
If y  x = 1 and x + y = 2, then the value 55. If 27(x + y)3 – 8(x – y)3 = (x + 5y)(Ax2 +
50.
of x3 + y3 is: an by By2 + Cxy), then what is the value of (A

n
+ B – C)?
SSC CPO 03/10/2023 (Shift-01)
CGL 2019 Tier II (15/11/2020)
(a) 0 (b) 1

ja (a) 13 (b) 16
R s
(c) 3 (d) 2
(c) 18 (d) 11
51. If (8x3 – 27y3) ÷ (2x – 3y) = (Ax2 + Bxy +
a th

56. If 8(x + y) – (x – y)3 = (x + 3y)(Ax2 + Cy2

3
Cy2), then the value of (2A + B – C) is : + Bxy), then the value of (A – B – C) is :
SSC CGL 6 /06/2019 (Shift- 1) SSC CHSL 30/6/019 (Shift- 03)
ty a

(a) 4 (b) 6 (a) – 2 (b) – 6

(c) 5 (d) 3 (c) 10 (d) 14
di M

ANSWER KEY
1.(b) 2.(d) 3.(b) 4.(d) 5.(a) 6.(a) 7.(a) 8.(b) 9.(d) 10.(b)

11.(c) 12.(b) 13.(b) 14.(b) 15.(c) 16.(a) 17.(d) 18.(c) 19.(a) 20.(d)

21.(a) 22.(c) 23.(d) 24.(d) 25.(a) 26.(d) 27.(b) 28.(c) 29.(a) 30.(c)
A

31.(d) 32.(a) 33.(a) 34.(b) 35.(a) 36.(d) 37.(d) 38.(d) 39.(c) 40.(c)

41.(a) 42.(d) 43.(a) 44.(b) 45.(d) 46.(b) 47.(d) 48.(b) 49.(c) 50.(a)

51.(c) 52.(b) 53.(b) 54.(b) 55.(b) 56.(a)

 ALGEBRA / chtxf.kr
(CLASSROOM SHEET-03)

CONCEPT OF SYMMETRY 7. If
x – a2
2 2 +
x – b2
2 2 +
x – c2
= 3, find the
b c c a b2  a2
1 y2 value of x.
1. If xy + yz + zx = 1, then x  y y  z = ?
   (a) a2 + b2 – c2 (b) a2 + b2 + c2
(c) a2 – b2 – c2 (d) a2 + b2
(a) 0 (b) 1
8. If bc + ca + ab = abc then
(c) 2 (d) 3
If x2 + y2 + z2 = xy + yz + zx then the value b c c a a b

r
2.
 
4 4
3x  7y  5z 4 bc a – 1 ca b – 1 ab c – 1 = ?

si
of is
5x y 2  7y 2z 2  3z 2x 2
2
(a) 0 (b) 1

(a) 2
an by (b) 1
9.
(c) 2 (d) 3
If a = b + c, b = c + a, c2 = a + b, then
2 2

n
(c) 0 (d) –1
a c 1 1 1
If a2 + b2 + c2 = ab + bc + ca then + +
3.

ja b
is 1 a 1 b 1 c
R s
(a) 0 (b) 1
(a) 0 (b) 2
a th

(c) 2 (d) 3
(c) 1 (d) –1
1 1
4x – 3 4y – 3 4z – 3 10. If a, b, c are non zero, a  = 1 & b
4. If + + = 0 then b c
ty a

x y z
1
1 1 1 = 1 then (i) abc is (ii) c + is
di M

  is a
x y z
(a) –1, 1 (b) 3, –1
(a) 9 (b) 3 (c) –3, 1 (d) 1, 1
(c) 4 (d) 6 11. If a = (x + y + z) , a = (x + y + z) z and
x y y

2a 2b 2c az = (x + y + z)z, then x + y + z = ? (a  0)

5. If + + = 4 then the value
a b c (a) 0 (b) 1
3
(c) a (d) a
ab  bc  ca 
of is 12. If a + b + c = 0, then the value of
abc
a2  b2  c2
(a) 2 (b) 1
A

a 2 – bc
1 (a) 0 (b) 1
(c) 0 (d)
2
(c) 2 (d) 3
x  a 2  2c 2 x  b 2  2a 2 x  c 2  2b 2 a 2 – bc
6. If + + 13. If a + b + c = 0 then the value of
b c c a a b b 2 – ca
= 0, find x is :

(a) a2 + b2 + c2 (b) –(a2 + b2 + c2) (a) 0 (b) 1

(c) a2 + 2b2 + c2 (d) –(a2 + 2b2 + c2) (c) 2 (d) 3

14. If a + b + c = 0 then the value of 22. If x2 + 9y2 = 40 and xy = 4, where x > 0, y
> 0, then what is the value of (x3 + 27y3)?
a 2  b2  c2
is SSC PHASE IX 2022
ab  bc  ca
(a) 224 (b) 416
(a) 2 (b) –2
(c) 440 (d) 800
(c) 0 (d) 4
23. If xy = – 6 and x³ + y³ = 19 (x and y are
CONCEPT OF VALUE PUTTING integers), then what is the value of
1 1
15. If x + y = 1, then what is the value of x³ +  ?
x –1 y –1
3xy + y3?
SSC CGL MAINS (08/08/2022)
SSC CGL MAINS (08/08/2022)
(a) –2 (b) 2
(a) 0 (b) 2
(c) 1 (d) –1 (c) –1 (d) 1
16. If a + b = 6 and ab = 5, then what is the 24. If If a + b + c = 0, then
value of a3 + b3?

r
a b b c c a   a b c 
     b  c  c  a  a  b  is :
SSC CGL MAINS (08/08/2022)  c a b   

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(a) 136 (b) 126
SSC CGL 06/06/2019 (Shift- 01)

17.
(c) 116

an by (d) 106
If a3 + b3 = 217 and a + b = 7, then the
(a) 8
(c) – 3
(b) 9
(d) 0

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value of ab is:
25. If x – y + z = 0, then find the value of
SSC CPO 24/11/2020 (Shift- 02)
(a) –6
ja (b) –1 y² x² z²
R s
- - .
(c) 7 (d) 6 2xz 2yz 2xy
a th

SSC CGL 12/04/2022 (Shift- 03)

64 1
18. If r   16 , then the value of r 4  3 is.
r r 3 1
(a) (b)
2 2
ty a

SSC CHSL 25/05/2022 (Shift- 03)

1 3
(d) -
di M

(a) 4096 (b) 512 (c) – 6

2
512
26. If a + b + c = 0, then what is the value of
1
(c) 512 (d) 4096 (b  c)2 (c  a)2 (a  b)2
4096   ?
bc ca ab
12 16
19. If a –  1 , a > 0, find a2 + 2 ? SSC CGL 18/08/2021 (Shift- 03)
a a
(a) 15 (b) 19 (a) 1 (b) –3
(c) 17 (d) 11 (c) –1 (d) 3
20. If (x+6y) = 8, and xy = 2, where x > 0, what 27. If a + b – c = 0, then what is the value of
is the value of (x³ + 216 y³) ?
A

(b – c )2 (c – a )2 (a  b )2
SSC CGL 11/04/2022 (Shift- 01) ?
4bc 4ca 4ab
(a) 288 (b) 224
(c) 368 (d) 476 SSC CHSL 06/08/2021 (Shift- 01)

21. If a2 + b2 = 65 and ab = 8, a > b > 0, then 3 3

find the value of a2 – b2. (a) – (b)
4 4
SSC CGL 21/04/2022 (Shift- 02)
(a) 65 (b) 60 1 1
(c) (d) –
(c) 72 (d) 63 2 2
28. If x + y + z = 0, then what will be the value 35. If a + b – c = 5 and ab – bc – ac = 10,
2 2 2
x  y   z  then find the value of a2 + b2 +c2.
of      
 yz    zx    xy  ?
      SSC CGL 18/04/2022 (Shift- 01)
SSC CHSL 24/05/2022 (Shift- 01) (a) 15
3(x 2  y 2  z 2 ) (b) 45
(a) (b) x2 + y2 + z2
xyz (c) 5
(d) 40
x 2 y2z 2 36. If a + b + c = 11 and ab + bc + ca = 28,
(c) (d) 3
x then find the value of a³ + b³ + c³ – 3abc.
29. If If a + b + c = 0, then SSC CGL 19/04/2022 (Shift- 03)
a b b c c a   a b c 
(a) 407 (b) 1639
     b  c  c  a  a  b  is :
 c a b  
(c) 2255 (d) 1093
SSC CGL 06/06/2019 (Shift- 01)
(a) 8 (b) 9

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(c) – 3 (d) 0 MISCELLANEOUS QUESTIONS

si
a b b c c a 37. If p = 11, then the value of p(p2 + 3p + 3) is:
30. If x = + , y= + , z= + ,
b a c

an by
b a
then what is the value of xyz – x2 – y2 –
c

(a) 1629
SSC CHSL 25/05/2022 (Shift- 03)
(b) 1225

n
z2 = ?
(c) 1727 (d) 1111
(a) –4 (b) –2
(c) –1
ja
(d) –6 38. If p = 38, then the value of p(p2 + 3p + 3)
R s
31. If a3b = abc = 180, a, b, c are positive in- is _________.
a th

tegers, then the value of c is : SSC CHSL 06/06/2022 (Shift- 03)

(a) 110 (b) 1 (a) 39313 (b) 59319
(c) 4 (d) 25 (c) 39318 (d) 59318
39. If x = 9, then the value of x5 – 10x4 + 10x3
ty a

1 1
32. If x = a and y = a– then – 10x2 + 10x – 1 is:
a a SSC CHSL 30/05/2022 (Shift- 02)
di M

4 4 2 2
is equal to : (a) 10 (b) 8
x  y – 2x y
(c) 9 (d) 1
SSC CGL 06/06 2019 (Shift- 01) 40. 4 4
If a + b + a²b² = 273 and a² + b² – ab =
2
(a) 16a (b) 8 1 1 
21, then one of the values of 
   is:
8 a b 
(c) (d) 4 SSC CGL 24/08/2021 (Shift- 01)
a2
33. If x = 2 – p, then x3 + + 6xp + p3 is equal to : 9 3
(a) – (b) –
4 4
SSC CGL 07/06/2019 (Shift- 01)
9 3
(a) 12 (b) 6 (c) (d)
8 2
A

(c) 8 (d) 4
1
34. Find the product of 41. If x4 + y4 + x2y2 = 17 and x² – xy + y²
16
(a + b + 2c)((a2 + b2 + 4c2 – ab – 2bc – 2ca). 1
=5 , then one of the values of (x – y) is:
SSC CGL 07/03/2020 (Shift- 03) 4
SSC CGL MAINS 03/02/2022
(a) a + b + 8c3 – 6abc
3 3

5 3
(b) a3 + b3 + 6c3 – 6abc (a) (b)
2 4
(c) a3 + b3 + 8c3 – 2abc 5 3
(c) (d)
(d) a3 + b3 + 8c3 – abc 4 2
 49. ab(a – b) + bc(b – c) + ca(c – a) is equal to :
1 1 1
42. If 2x = 3y = 6z, then + – is equal to: CGL Tier-II 13 09/2019
x y z
(a) (a + b)(b – c)(c – a)
SSC Phase X 01/08/2022 (Shift- 03) (b) (a – b)(b + c)(c – a)
3 (c) (a – b)(b – c)(c – a)
(a) 1 (b) (d) (b – a)(b – c)(c – a)
2
–1 3
(c) 0 (d) 50. Let x = 6
27 – 6 and y =
2 4
43. If 3a = 27b = 81c and abc = 144, then the
45 + 605 + 245
1 1 1  , then the value of x2 +
value of 12  + +  is : 80 + 125
 a 2b 5c 
y2 is :
SSC CGL 06/03/2020 (Shift- 01) CGL Tier-II 13/09/2019
17 18 223 221
(a) (b)

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120 10 (a) (b)
36 36

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18 33 221 227
(c) (d) (c) (d)
120 10
44.
an by
If 2x + y – 2z = 88z – 5 – y ; 54y – 6z = 25y + z ; 34x – 51.
9 9
If x4 + x2y2 + y4 = 21 and x2 + xy + y2 = 7,

n
3z
= 9x + z then the value of 2x + 3y + 5z is :
 1 1 
CHSL 13/10/2020 (Shift- 01) then the value of  2  2  is :
(a) 56
ja (b) 44  x y 
R s
(c) 32 (d) 28 SSC CGL 03/03/2020 (Shift- 02)
a th

3x 3x 3x
45. If 5 +12 =13 , then the value of x is : 5 7
(a) (b)
SSC CHSL 05/07/2019 (Shift- 02) 2 4
ty a

(a) 2 (b) 8 5 7
(c) 1 (d) 4 (c) (d)
4 3
di M

4x 4x
46. If 6 +8
4
x
= 10 , then the value of x is : 52. If x4 + x2y2 + y4 = 273 and x2 – xy + y2 =
13, then the value of xy is :
SSC CHSL 08/07/2019 (Shift- 02)
SSC CGL 05/03/2020 (Shift- 02)
(a) 2 (b) 16
(a) 4 (b) 8
(c) 4 (d) 8
(c) 10 (d) 6
47. If 86 – 60 2 = a – b 2 , then what will be 53. If 16a4 + 36a2b2 + 81b4 = 91 and 4a2 + 9b2
– 6ab = 13, then what is the value of 3ab?
the value of a 2 + b 2 , correct to one
SSC CGL 04/03/2020 (Shift- 01)
decimal place?
CGL Tier-II 11/09/2019 3
(a) – 3 (b)
(a) 8.4 (b) 8.2 2
A

(c) 7.8 (d) 7.2

3
3 3 (c) 5 (d) –
48. If x = 1+ – 1– , then the value of 2
2 2
2
2–x 7(x 2  1) – 17x  x +1 
will be closest to : 54. If = 2, then  =?
2+x 7x  x 
CGL Tier-II 11/09/2019
(a) 6.45 (b) 6.53
(a) 0.17 (b) 0.12 (c) 6.86 (d) 6.31
(c) 1.4 (d) 1.2
 61. Simplify the following expression.
55. If

7 2 2–x  14
= 7 + 2 then
x x 5(a 6 – b6 )3  5(b6 – c6 )3  5(c6 – a 6 )3
2(a 3 – b3 )3  2(b3 – c3 )3  2(c3 – a 3 )3
23x 64
 42x 12  163
SSC CGL 24/08/2021 (Shift- 02)
(a) 12 (b) 12 2 5
(a) (a³ + b³)(b³ + c³)(c³ + a³)
(c) 12 3 (d) 24 2
6 3
56. If 8k + 15k – 2 = 0, then the positive 5
(b) (a³ + b³)(b³ – c³)(c³ – a³)
 1 2
value of  k +  is:
 k 5
(c) (a³ – b³)(b³ + c³)(c³ + a³)
SSC CGL 12/04/2022 (Shift- 03) 2
1 1 5
(a) 2 (b) 2 (d) (a³ – b³)(b³ – c³)(c³ + a³)
2 8 2
1 1 62. If x3 + 4x = 8, find x7 + 64 x2 ?
(c) 8 (d) 8

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2 8 (a) 32 (b) 64
3 3 (c) 128 (d) 256

si
57. If x = 1 – 1– then the value of 63. If x2 – 5x + 25 = 0 then x5 – 5x4 – 3100 ?
2 2
3–x
3x an by
(correct to one decimal place) is?
(a) 25
(c) 25 3
(b) 25 2
(d) None

n
SSC CGL MAINS 29/01/2022 9
(a) 0.25 (b) 0.17 64. If (x + 5) (x + 6) + = 0, Find 3x2

ja x (x – 1)
R s
(c) 0.19 (d) 0.27
+ 15x – 1 ?
58. If (ab + bc + ca) = 0, then what is the value
a th

(a) 4 (b) 6
 1 1 1 
of   2  2  ? (c) 8 (d) 10
 2
 a – bc b – ca c – ab  65. If ax + by = 8, ay + bx = 6, a2 + b2 + x2 +
SSC CHSL 09/06/2022 (Shift- 01) y2 = 29
ty a

(a) 2 (b) 0 a 2 + b2
(c) 1 (d) a + b + c Find ; a 2 + b 2 > x2 + y2 ?
(x 2 + y 2 )
di M

59. If px3 + x 2 + 3x + q is exactly divisible by

(x + 2) and (x – 2), then the values of p 25 27
(a) (b)
and q are: 4 4
SSC CHSL 08/06/2022 (Shift- 03)
23 29
3 (c) (d)
(a) p  – and q  4 4 4
4 66. a b c
If 3 = 27 = 81 and abc = 144, then the value
3
(b) p  and q  4 1 1 1 
4 of 12     is :
3 a 2b 5c 
(c) p  and q  – 4
4 33 18
(a) (b)
A

3
(d) p  – and q  – 4 10 10
4
60. What is the value of the following 17 18
expression? (c) (d)
120 120
 x a  (a  b)  b  (b  c)
2 x 
 c  (a  c)
–2  x 
22 
xb 
  3  xc   6  xa  a 2  b2  c2 – 1024
      67. If = – 2 and a + b = 5c,
ab – bc – ca
SSC CHSL 08/06/2022 (Shift- 02) where c > 0, then the value of c is _____.
(a) 1 (b) 0 (a) 8 (b) 4
(c) 4 (d) 9 (c) 12 (d) 5
68. If a3 – b3 = 602 and a – b = 2, then find the 70. Select the option that is true regarding the
value of (a2 + b2). following labelled Assertion (A) and Reason (R).
IB ACIO GRADE II 17/01/2024 (Shift-01) Assertion (A): The value of
(a) 156 (b) 240  3.93  9  1.3  4.29  11.7  1.12  1.331
 
(c) 202 (d) 260
1.23  1.23  0.77  0.77  0.6  4.1
2 1
69. If x  7  7 3  7 3 , then which of the is 31.25.
options below is correct. Reason (R): Using (a + b)3 = a3 + b3 + 3ab
IB ACIO GRADE II 17/01/2024 (Shift-02) (a + b) and (x + y)2 = x2 + y2 + 2xy

(a) x3 + 21x2 – 126x – 252 = 0 IB ACIO GRADE II 18/01/2024 (Shift-01)

(b) x3 + 21x2 – 126x + 252 = 0 (a) A is true and R is false.

(c) x3 – 21x2 – 126x + 252 = 0 (b) Both A and R are true but R is not a
correct explanation of A.
(d) x3 – 21x2 + 126x – 252 = 0
(c) Both A and R are true and R is a correct
explanation of A.

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(d) A is false and R is true.

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an by ANSWER KEY

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1.(b) 2.(b) 3.(b) 4.(c) 5.(d) 6.(b) 7.(b) 8.(b) 9.(b) 10.(a)

ja
R s
11.(d) 12.(c) 13.(b) 14.(b) 15.(c) 16.(b) 17.(d) 18.(a) 19.(c) 20.(b)
a th

21.(d) 22.(a) 23.(d) 24.(b) 25.(a) 26.(d) 27.(a) 28.(d) 29.(b) 30.(a)

31.(b) 32.(d) 33.(c) 34.(a) 35.(c) 36.(a) 37.(c) 38.(d) 39.(b) 40.(b)
ty a

41.(a) 42.(c) 43. (d) 44. (b) 45.(b) 46.(b) 47.(c) 48.(a) 49.(d) 50.(a)
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51.(c) 52.(a) 53.(d) 54.(a) 55.(a) 56.(a) 57.(d) 58.(b) 59.(d) 60.(a)

61.(a) 62.(c) 63.(a) 64.(c) 65.(a) 66.(a) 67.(a) 68.(c) 69.(d) 70.(c)
A
 (Types of Triangle)/f=kHkqt ds izdkj
(Practice Sheet With Solution)
LEVEL - 01 lef}ckgq f=kHkqt dh rhuksa Hkqtkvksa dk ;ksx
1. The measure of sides of a triangle is (x2 – 1), vkSj cjkcj yackbZ okyh Hkqtkvksa esa ls ,d Hkq
(x2 + 1) and 2x then find the triangle. dk vuqikr 3 % 4 gSA f=kHkqt dh Å¡pkbZ Kkr d
fdlh f=kHkqt dh Hkqtk
(x2 –1), (x2 + 1), 2x gS f=kHkqt gksxk& (a) 3 3 cm (b) 4 5 cm
(a) Equilateral (c) 3 5 cm (d) 2 5 cm
(b) Isosceles
7. The perimeter of an isosceles triangle is 91 cm
(c) Right angled triangle

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1
(d) Acute angled triangle and its base is 1 times each of its equal side
4

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2. The measures of three angles of a triangle are
What is the length (in cm) of its base?
in the ratio of 3 : 2 : 1 The triangle is a/an
fdlh lef}ckgq f=kHkqt dk ifjeki 91 lseh gSA vk

an by
,d f=kHkqt ds rhuksa dks.kksa dh eki dk3vuqikr
:2:1
1
gS f=kHkqt gksxkA vk/kj bldh izR;sd leku Hkqtk dk1 xquk gSA

n
4
(a) Equilateral triangle/leckgq f=kHkqt vk/kj dh yackbZ (lseh esa) Kkr djsaA
(b) Obtuse angled triangle/vf/ddks.k f=kHkqt

ja
(a) 20 (b) 28
R s
(c) Acute angled triangle/U;wudks.k f=kHkqt (c) 25 (d) 35
(d) Right angled triangle/ledks.k f=kHkqt 8. In ABC, find the measure of B if AB = A
a th

and BAC = 40°.

3. Which of the set of three sides can't form a
triangle? ABC esa] ;fn AB = AC rFkk BAC = 40° gS]
B dk eki Kkr dhft,A
rhu Hkqtkvksa dk dkSu lk lewg f=kHkqt ugha cuk ldrk gS\
ty a

(a) 50° (b) 60°

(a) 5 cm, 6 cm, 7 cm
(c) 40° (d) 70°
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(b) 5 cm, 8 cm, 15 cm

9. The length of each equal side of an isoscele
(c) 8 cm, 15 cm, 18 cm triangle is 15 cm and the included angl
(d) 6 cm, 7 cm, 11 cm between those two sides is 90°. Find the are
4. The sides of a triangle are in the ratio 3 : 4 : 6. of triangle.
The triangle is: lef}ckgq f=kHkqt ds izR;sd cjkcj Hkqtkvksa15
dhcm
y
,d f=kHkqt dh Hkqtk,¡ 3 % 4 % 6 ds vuqikr esa gSaA f=kHkqt gS% mu nksuksa Hkqtkvksa ds chp
gS] vkSj 90°dk
gSA
dks.f
(a) Acute-angle triangle/U;qu&dks.k f=kHkqt dk {ks=kiQy Kkr dhft;sA
(b) Right-angled triangle/ledks.k f=kHkqt 225
(a) cm2 (b) 225 cm2
(c) Obtuse-angled triangle/vf/d dks.k okyk f=kHkqt 2
(d) Either acute-angled or right-angle triangle 255 125
;k rks U;qu&dks.k f=kHkqt ;k le dks.k f=kHkqt (c) cm2 (d) cm2
A

2 2
5. In an isosceles ABC, the measure of A is 10. An acute-angle isosceles triangle has two of it
twice the measure of the base B. then what sides equal to 10 unit and 16 unit. Find th
will be the measure of C? area of this triangle.
,d lef}ckgq ABC esa] ;fnA dk eki] vk/kj B ds ,d U;wudks.k lef}ckgq f=kHkqt dh nks Hkqtk,¡
eki ls nksxquk gS]C
rksdk eki D;k gksXk
k\ 16 bdkbZ ds cjkcj gSaA bl f=kHkqt dk {ks=kiQ
(a) 30° (b) 33° (a) 48 sq.units (b) 1266 sq.units
(c) 45° (d) 90° (c) 24 sq.units (d) 5231 sq.units
6. The sum of three sides of an isosceles triangle 11. The perimeter of a right triangle is 60 cm
is 20 cm, and the ratio of an equal side to the and its hypotenuse is 26 cm. What is th
base is 3 : 4. The altitude of the triangle is : area (in cm2) of the triangle?

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs1

 fdlh ledks.k f=kHkqt dk ifjeki60 cm gS vkSj bldk
d.kZ26 cm gSA f=kHkqt dk {ks=kiQy
(cm2 esa
) Kkr djsaA the correct relation between the sum of th
squares of sides to the sum of the squares o
(a) 120 (b) 96 medians is.
(c) 90 (d) 60
;fn AD, BE, CF ,d ABC dh ekfè;dk,¡ gSa rks H
12. The sides of a right-angled triangle are in the
ratio x : (x –1) : (x –18). What is the perimeter
ds oxks± ds ;ksx dk ekfè;dk ds oxks± ds ;ksx ls lgh
of the triangle? (a) 2(AB² + BC² + AC²) = 3(AD² + BE² + CF²)
(b) 4(AB² + BC² + AC²) = 3(AD² + BE² + CF²)
,d ledks.k f=kHkqt dh Hkqtk,¡
x : (x –1) : (x –18) ds
(c) 3(AB² + BC² + AC²) = 4(AD² + BE² + CF²)
vuqikr esa gSaA f=kHkqt dk ifjeki D;k gS\
(d) None of the above
UPSC CDS-1 2021 18 . The angles of a triangle are in the ratio 1 : 1 : 2
(a) 28 units (b) 42 units What percentage of the total internal angl
(c) 56 units (d) 84 units is the greatest angle?
13. If each side of an equilateral triangle is tripled, ,d f=kHkqt ds dks.kksa
vuqikr
dk 1 % 1 % 2 gSA lcls cM
then what will be the area of the new dks.k] dqy vkarfjd dks.k
ksa
dk fdruk izfr'kr gS\

r
equilateral triangle?
SSC CPO 09/11/2022 (Shift-01
;fn fdlh leckgq f=kHkqt dh izR;sd Hkqtk dks rhu xquk dj (a) 50% (b) 65%

si
fn;k tkrk gS] rks u, leckgq f=kHkqt dk {ks=kiQy fdruk gksxk\
(c) 45% (d) 40%
(a) 12 times the initial area 19. In ABC, 2A = 3B = 6C. What is the valu

an by
izkjafHkd {ks=kiQy dk 12 xquk of the largest angle among these three angles
ABC esa]2A = 3B = 6C gSA bu rhuksa d

n
(b) 6 times the initial area
izkjafHkd {ks=kiQy dk 6 xquk ls lcls cM+s dks.k dk eku D;k gS\

ja
(c) 3 times the initial area SSC CPO 11/11/2022 (Shift-03
R s
izkjafHkd {ks=kiQy dk 3 xquk (a) 170° (b) 90°
(d) 9 times the inital area (c) 80° (d) 150°
a th

izkjafHkd {ks=kiQy dk 9 xquk 7

14. Let ABC be an equilateral triangle and AD 20. In ABC, a : b : c = 3 : 5 : , find b : s.
2
perpendicular to BC then AB2 + BC2 + CA2 = ?
7
ty a

eku yhft, ABC ,d leckgq f=kHkqt gS vkSj

AD] BC ij ABC esa]a : b : c = 3 : 5 :
, b : s Kkr dhft;sA
2
yacor gS rks
AB2 + BC2 + CA2 = ? (a) 12 : 23 (b) 20 : 23
di M

(a) 2AD2 (b) 3AD3 (c) 14 : 23 (d) 10 : 23

(c) 4AD2 (d) 5AD3 21. The length of the three sides of a right-angle
15. From the interior point of an equilateral triangle are (x – 2) cm, x cm and (x + 2) cm
triangle, perpendiculars are drawn on all three respectively. Then the value of x is:
sides. The sum of the lengths of the ,d ledks.k f=kHkqt dh rhu Hkqtkvksa dh yackbZ(x –Ø2
perpendiculars is 'S'. Then the area of the
triangle is. lseh] x lseh vkSj
(x + 2) lseh gSA rks
x dk eku gS%
,d leckgq f=kHkqt ds vkarfjd fcanq ls] rhuksa rjiQ yEc(a) 10 (b) 8
(c) 4 (d) 0
•haps tkrs gSaA yacksa dh yackb;ksa
'S' dk
gksrk
;ksxgSA rks
22. Find the area of triangle (in sq cm) in give
f=kHkqt dk {ks=kiQy gSA
figure:
(a)
S²
(b)
2S² nh xbZ vkd`fr esa f=kHkqt dk (in
{ks=kiQy
sq cm) Kkr dhft,
A

2 3 3 A
S² 3S²
(c) (d) 15 cm
3 2
16. Find the area of a triangle, whose sides are 8
cm, 7 cm and 5 cm.
ml f=kHkqt dk {ks=kiQy Kkr dhft,] ftldh Hkqtk,¡ 8 lseh] 30°
7 lseh vkSj 5 lseh gSaA B 16 cm
C
(a) 10 cm2 (b) 20 3 cm 2 (a) 60 3 (b) 60
2 2
(c) 10 3 cm (d) 15 3 cm (c) 75 (d) 75 3

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs2

 ,d ABC esa]A = 90°, ;fn BM vkSjCN nks ekfè;
The ratio of their corresponding sides is: 2 2
BM  CN
,d ABC ds dks.k 1 % 2 % 3 ds vuqikr esa gSaA mudhgSa] rks BC2 fdlds cjkcj gS\
laxr Hkqtkvksa dk vuqikr gS%
3 4
(a) 1 : 3 :2 (b) 1 : 2 : 3 (a) (b)
5 5
(c) 3:1:2 (d) None of these 5 3
(c) (d)
LEVEL - 02 4 4
30. The lengths of the sides of a right-angle
24. If the length of the sides of a triangle are in the
triangle are consecutive even integers (in cm
ratio 4 : 5 : 6 and the inradius of the triangle is
What is the product of these integers?
3 cm, then the altitude of the triangle
corresponding to the largest side as base is : ,d ledks.k f=kHkqt dh Hkqtkvksa dh yackbZ Øekx
(lseh esa) gksrh gSA bu iw.kkZadksa dk xq.kuiQ
;fn ,d f=kHkqt dh Hkqtkvksa dh yackbZ 4 % 5 % 6 ds vuqikr
esa gS vkSj f=kHkqt dh vkarfjd f=kT;k 3 lseh gS] rks vk/kj ds UPSC CDS-2 202
:i esa lcls cM+h Hkqtk ds vuq:i f=kHkqt dh ÅapkbZ gS% (a) 60 (b) 120

r
(a) 7.5 cm (b) 6 cm (c) 360 (d) 480

si
31. Triangle ABC is right angled at B and D is
(c) 10 cm (d) 8 cm
point on BC such that BD = 5 cm, AD = 1
25. If the ratio of the angles of a triangle is

an by
cm and AC = 37 cm, then find the length o
18 : 31 : 43 then type of triangle is: DC in cm.
;fn fdlh f=kHkqt ds dks.kksa dk vuqikr 18 % 31 % 43 gS] f=kHkqtABC esa, B ij ledks.k gS vkSjBC ij fcUnqD b

n
rks f=kHkqt dk izdkj gS% izdkj fLFkr gS fdBD = 5 cm, AD = 13 cm vk
(a) acute (b) right AC = 37 cm gSA DC dh yackbZ (lseh esa) Kkr djs

ja
R s
(c) obtuse (d) isosceles (a) 25 (b) 35
26. PQR is an isosceles triangle such that PQ = QR (c) 5 (d) 30
a th

= 10 cm and PQR = 90°. What is the length 32. In triangle ABC, a line is drawn from the verte
of the perpendicular drawn from Q on PR? A to a point D on BC. If BC = 9 cm and DC =
PQR ,d ,slk lef}ckgq f=kHkqt gS PQ fd = QR = 10 cm, then what is the ratio of the areas o
cm vkSjPQR = 90° gSA Hkqtk PR ij fcanqQ ls [khaps
ty a

triangle ABD and triangle ADC respectively?

tkus okys yac dh yackbZ crkb,A f=kHkqtABC esa]BC ij fcanqA ls fcanq
D rd js[kk [khaph
gSA ;fnBC = 9 cm vkSjDC = 3 cm gS] rks Øe'k%
di M

(a) 4 2 cm (b) 7 2 cm
ABD vkSj f=kHkqtADC ds {ks=kiQyksa dk vuqikr D;k
(c) 6 2 cm (d) 5 2 cm (a) 1 : 1 (b) 2 : 1
27. ABC is an isosceles triangle with AB = AC (c) 3 : 1 (d) 4 : 1
= 10 cm. AD = 8 cm is median on BC from A. 33. In the adjoining figure, if BC = a, AC = b, AB =
The length of BC is: and CAB = 120°, then the correct relation is:
ABC ,d lef}ckgq f=kHkqt gS ftlesa
AB = AC = 10 lseh gSA layXu vkÑfr esa] ;fnBC = a, AC = b, AB = c vk
AD = 8 lseh] A ls BC ij ekfè;dk gSA BC dh yackbZ%gS CAB = 120° gS] rks lgh laca/ gS%
(a) 8 cm (b) 12 cm C
(c) 10 cm (d) 6 cm
28. If the perimeter of an isosceles right angle
A

 
triangle is 16 2  16 cm, then the area of the
triangle is:

;fn ,d lef}ckgq ledks.kh; f=kHkqt dk ifjeki

16 2  16 
D A B
lseh gS] rks f=kHkqt
k {ks=kiQy
d fdruk gksxk\
(a) a² = b² + c² + 2bc (b) a² = b² + c² – 2bc
(a) 76 sq. cm (b) 64 sq. cm
(c) a² = b² + c² + bc (d) a² = b² + c² – bc
(c) 58 sq. cm (d) 66 sq. cm
34. In a triangle ABC, medians AD and BE ar
29. In an ABC, A = 90°, if BM and CN are two perpendicular to each other, and have length
2 2
medians, BM  2CN is equal to: 12 cm and 9 cm, respectively. Then, the are
BC of triangle ABC (in sq cm) is.

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs3

 ,d f=kHkqtABC esa] ekfè;dk,¡AD vkSj BE ,d&nwljs
ds yacor gSa vkSj budh yackbZ Øe'k% 12 lseh vkSj 9 lseh
of the angles A, B and C can be:
gSA rks (oxZ lseh esa) f=kHkqt
ABC dk {ks=kiQy gSA ;fn ABC ,d vf/ddks.k f=kHkqt gS] rks dks.k
A, 
(a) 80 (b) 68 vkSjC dk vuqikr gks ldrk gS%
(c) 72 (d) 78 (a) 4 : 9 : 3 (b) 11 : 14 : 21
35. The lengths of the sides of a triangle are 3x, (c) 1 : 2 : 3 (d) 5 : 8 : 12
4 y , 53 z , where 3x < 4 y < 53 z . If one of 41. In the given figure, if AB = 8cm, AC = 10 cm
ABD = 90° and AD = 17 cm, then the mea
the angles is 90°, then what are the minimum
sure of CD is:
integral values of x, y, z respectively?
nh xbZ vkd`fr esa] ;fn
AB = 8 cm, AC = 10 cm
,d f=kHkqt dh Hkqtkvksa dh3x,
yackbZ
4 y, 53 z , gS] tgk¡
ABD = 90° vkSj AD = 17 cm gS rks
CD dh yac
3x < 4 y < 53 z gSA ;fn dks.kksa esa 90°
ls ,dgS] rks fdruh gS\
Øe'k%x, y, z ds U;wure vfHkUu eku D;k
Sa\ g A
UPSC CDS-1 2020

r
(a) 1, 2, 3 (b) 2, 3, 4
(c) 1, 1, 1 (d) 3, 4, 5

si
36. Sides of a right angle triangle are 6, 10 and x.
For what value of x is the area of the  the

an by
maximum? B C D
,d ledks.k f=kHkqt dh Hkqtk,¡ 6] 10 vkSj
x gSa]x ds (a) 10 cm (b) 11 cm

n
fdl eku ds fy,  dk {ks=kiQy vf/dre gS\ (c) 9 cm (d) 8 cm

ja
(a) 136 (b) 9 LEVEL - 03
R s
(c) 115 (d) 10
42. In ABC, AB = AC = 17 cm and D is point o
37. In the ABC we have AB = 5 cm, AC = 6 cm,
a th
BC. If CD = 4 cm and AD = 15 cm, then th
A = 60°. Find the length of the side BC. length of BD is:
ABC esa gekjs iklAB = 5 cm, AC = 6 cm, A = 60°
ABC esa]AB = AC = 17 lseh vkSjD, BC ij ,d fcan
gSA Hkqtk
BC dh yackbZ Kkr dhft,A
gSA ;fnCD = 4 lseh vkSjAD = 15 lsehgS]rks BD d
ty a

(a) 31 (b) 2 3 yackbZ%gS

(a) 16 cm (b) 20 cm
di M

(c) 2 2 (d) 2
38. ABC and DEF are two triangles such that (c) 14 cm (d) None of these
ABC  FDE. If AB = 5 cm, B = 40° and 43. In ABC, AB = AC and D is a point on BC.
A = 80°, then which of the following BD = 5 cm, AB = 12 cm and AD = 8 cm, the
option is true? the length of CD is:
ABC vkSjDEF nks ,sls f=kHkqt gSa fd  FDE
ABC ABC esa]AB = AC vkSjD, BC ij fLFkr ,d fcanq
gSA ;fnAB = 5 cm, B = 40° vkSjA = 80° gS] rks ;fn BD = 5 lseh] AB = 12 lseh vkSjAD = 8 lseh
fuEufyf[kr esa ls dkSu lk fodYi lR; gS\ rksCD dh yackbZ gS%
SSC CPO 09/11/2022 (Shift-01) (a) 16 cm (b) 16.2 cm
(a) DE = 5 cm, E = 60° (c) 14.8 cm (d) 14 cm
(b) DE = 5 cm, F = 60° 44. In ABC, C = 90° and D is a point on CB suc
A

(c) DE = 5 cm, D = 60° that AD is the bisector of A. If AC = 5 cm an

(d) DF = 5 cm, E = 60° BC = 12 cm, then what is the length of AD?

39. For a triangle, base is 6 3 cm and two base

ABC esa]C = 90° gS vkSj
D, CB ij ,d ,slk fcanq g
angles are 30° and 60°. Then the height of the
ftlls fd AD, A dk lef}Hkktd gSA ;fnAC = 5 ls
triangle is: vkSjBC = 12 lseh gS] rksAD dh yEckbZ fdruh gS\
,d f=kHkqt ds fy,] vk/kj 6 3 lseh gS vkSj nks vk/kj 10 5 3
(a) cm (b) cm
dks.k30° vkSj60° gSaA rks f=kHkqt dh ÅapkbZ gS% 3 6
(a) 3 3 cm (b) 4.5 cm
5 13 20
(c) cm (d) cm
(c) 4 3 cm (d) 2 3 cm 3 3

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs4

 trisected at D such that BC = 3 BD. What is the internal bisector of C. If DC is produced t
ratio of AD2 to AB2? E such that AC = CE, then CAE is equal to:
ABC ,d leckgq f=kHkqt gS- Hkqtk
BC dks D ij bl çdkj ABC ,d leckgq f=kHkqt gS vkSj
CD] C dk vkarfj
foHkkftr fd;k x;k gS fdBC = 3 BD gSA
AD2 ls AB2 dk lef}Hkktd gSA ;fnDC dks E rd bl çdkj c<+k;k tk,
vuqikr D;k gS\ fd AC = CE gS] rks
CAE cjkcj g S%
UPSC CDS-2 2020 (a) 45° (b) 75°
(a) 7 : 9 (b) 1 : 3 (c) 30° (d) 15°
(c) 5 : 7 (d) 1 : 2
49. In an ABC, 2ABC = 9ACB and 2BA
46. The sides of a scalene acute angle triangle are = 7ACB. If AB = 8 cm, AC = 17 cm, the
14,22 and x. x is an integer and 22 is the
the length of BC is:
largest side of triangle, what will be the average
of minimum and maximum value of x? ,d ABC esa] ABC
2 = 9ACB vkSj 2BA
,d U;wu dks.k fo"keckgq f=kHkqt dh Hkqtk,¡ 14] = 7ACB gSA ;fnAB = 8 lseh]AC = 17 lseh] rksB
x 22 vkSj
gaSA Hkqtk dh yEckbZ gS%
x ,d iw.kkZd gS tcfd f=kHkqt dh lcls cM+h22

r
gSAx ds U;qure vkSj egÙke eku dk vkSlr D;k gksxk\ (a) 8 cm (b) 25 cm

si
(a) 18 (b) 19 (c) 15 cm (d) 9 cm
(c) 21 (d) 20.5 50. In a triangle ABC, AC = 20 cm, BC = 10 cm

an by
47. If two sides of an obtuse angle triangle are 8 and area of triangle is 80 sq cm then find th
cm and 15 cm respectively. How many triangles length of AB.

n
are possible given the length of the third side
is an integer (in cms)? ,d f=kHkqt
ABC esa]AC = 20 lseh] BC = 10 lseh v
f=kHkqt dk {ks=kiQy 80 oxZ lseh ABgS]
dh rks
yackbZ
Kk

ja
;fn ,d vf/d dks.k f=kHkqt dh nks Hkqtk,¡ Øe'k% 8 lseh
R s
dhft,
vkSj 15 lseh gSaA ;fn rhljh Hkqtk dh yackbZ ,d iw.kkZad A
(lseh esa) gS rks fdrus f=kHkqt laHko gSa\
a th

(a) 3 39 cm (b) 2 78 cm
(a) 6 (b) 10
(c) 9 (d) None (c) 2 52 cm (d) 2 65 cm
ty a

Answer Key
di M

1.(c) 2.(d) 3.(b) 4.(c) 5.(c) 6.(d) 7.(d) 8.(d) 9.(a) 10.(d)

11.(a) 12.(c) 13.(d) 14.(c) 15.(c) 16.(c) 17.(c) 18.(a) 19.(b) 20.(b)

21.(b) 22.(b) 23.(a) 24.(a) 25.(a) 26.(d) 27.(b) 28.(b) 29.(c) 30.(d)

31.(d) 32.(b) 33.(c) 34.(c) 35.(c) 36.(a) 37.(a) 38.(d) 39.(b) 40.(a)

41.(c) 42.(a) 43.(a) 44.(c) 45.(a) 46.(b) 47.(b) 48.(d) 49.(c) 50.(d)
A

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs5

 SOLUTIONS
1. (c) Sides are (x² – 1), (x² + 1), 2x A
8. (d)
Let, x = 2
So, sides are 3, 4, 5
 It is right angled triangle 40°
2. (d) Ratio  3 : 2 : 1
Here, 3 = 2 + 1
So, triangle is right angle triangle x
x
3. (b) In the given options, the set, 5cm, 8cm,
15cm can't form a triangle (  5 + 8 < 15)
B C
4. (c) Here, 3² + 4² < 6²  40° + x + x = 180°
So, triangle is obtuse angled triangle. 2x = 140°

r
5. (c) x = 70°

si
A 9. (a) A

2

an by
n
15

ja
 
R s
B C
a th

 2 +  +  = 180° B 15 C
180°
 θ = = 45° 1 225
4 Area =  15  15 = cm²
2 2
ty a

C =  = 45°
10. (d) For acute angle triangle of side a, b, c
A a2 + b2 > c2
di M

6. (d) (10)2 + (10)2 > 162

3x 3x 200 > 256 (not correct)
(16)2 + (16)2 > 102
256 × 2 > 100 (correct)
A
B D C
4x
 10x = 20 cm
x = 2 cm
16 16
In ADC,
A

AD = AC2 – DC2 = 36 – 16 = 2 5 cm
7. (d) Given,
5 B 10 D C
Base = equal side
4 AD = Altitude of triangle
Perimeter = 91 cm BD = DC = 5 cm
 (4x + 4x + 5x) = 91 cm
 13x = 91 cm AD = 16² – 5² = 231
 x = 7cm 1
 Area = ×10 × 231 = 5 231 sq.units
 Length of base = 5x = 5 × 7 = 35 cm 2

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs6

 11. (a) A
2
a S
3
26cm 3 4 S2
So, area =   S2 
4 3 3
8 7 5
16. (c) Semi perimeter =  10 cm
2
B C  Area = 10 10 – 810 – 7 10 – 5
AB + BC = 60 – 26 = 34 cm  10  2  3  5  10 3 cm2
So, AB = 10 cm
BC = 24 cm A
1 17. (c)
 Area =  10  24 = 120 cm²
2

r
12. (c) Sides ratio = x : (x – 1) : (x – 18)
F E
Let, sides are ax, a(x – 1), a(x – 18)

si
 (ax)2 = [a(x – 1)]2 + [a(x – 18)]2
 x2 = x2 + 1 – 2x + x2 + 324 – 36x

an by
x2 – 38x + 325 = 0  x = 25, 13
So, x = 25 D

n
B C
Perimeter = x + x – 1 + x – 18 = (75 – 19) then 3(AB2 + BC2 + AC2) = 4(AD2 + BE2 + CF2)
= 56 units 18. (a) Angles ratio  1 : 1 : 2

ja
13. (d) If side of equilateral triangle is tripled then
R s
area becomes 9 times the initial area because  x + x + 2x = 180°
 4x = 180°
a th
3
area of equilateral triangle = (side)2 x = 45°
4
Largest angle = 45° × 2 = 90°
A
14. (c) 90º
 Required % = 100 = 50%
ty a

180º
19. (b)
di M

Let, 2A = 3B = 6C = 6

A : B : C = 3 : 2 : 1
(6) unit  180°
B D C (1) unit  30°
AB AC BC  Largest angle = 30° × 3 = 90°
BD = DC =  
2 2 2 7
Using Pythagoras theorem, 20. (b) a : b : c = 3 : 5 : = 6 : 10 : 7
2
3 3 3
AD  AB  BC  CA  b : s = 10 = 10 × 2 = 20
2 2 2 23 23 23
2 2
AB = BC = AC = AD 21. (b) Length of sides = (x – 2) , x, (x + 2) cm
3
A

4 4 4  So, x = 8 from option satisfies

AB2 + BC2 + AC2 = AD2     = 4AD2  6, 8, 10
3 3 3 

15. (c) A A
22. (b)

F E 15

30º
B D C B 16 C

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs7

 1 28. (b) Perimeter = (16 2  16) cm
Area of triangle = 16  15  sin 30º
2 So,
1
= 8 × 15 × = 60 cm2
2
23. (a) Angles  1 : 2 : 3
 (6) unit  180° 16
8 2
(1) unit  30°
Angles are 30°, 60°, 90°
So, side ratio = 1 : 3 : 2
24. (a) Sides = 4x : 5x : 6x
8 2
Area
 r = 1
Semi perimeter Area =  8 2  8 2 = 64 cm²
2

r
1 29. (c)
 6x  h B
 3 = 2

si
15x
2

an by
N
3  15

n
 h= = 7.5 cm
6
25. (a) Ratio of angles = 18 : 31 : 43

ja
R s
Here, 18 + 31 > 43 A M C
So, triangle is acute angled triangle We know that, 4(BM + CN2) = 5 BC2
2
a th

26. (d) Q 2 2
So, BM  CN  5
2
BC 4
ty a

10 10 30. (d)
di M

10
6
P X R
PR = 100  100 = 10 2 cm
10  10 10 2 8
Then, QX = =  = 5 2 cm
10 2 2 2  Product = 480

A 31. (d) A
27. (b)
37cm
10 10 13
A

B D C
B 5cm D C
AD² = AB² – (BD)²
8² = 10² – BD²  AB = 132 – 52 = 12 cm
BD² = 100 – 64
BD² = 36 So, BC = 372 – 122
BD = 6 BC = 35 cm
BC = 2BD = 12 cm CD = (35 – 5) = 30 cm

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs8

 32. (b) A BG 2
 
GE 1
(3) unit  9 cm
(1) unit  3 cm
BG = 6 cm
 Area of ABC = 3 × Area of BGA
1 
= 3 ×   6  8  = 72 cm2
2 
B 6cm D 3cm C
9cm 35. (c) Given, length of sides = 3x, 4 y , 53 z
BD 6 2
 
DC 3 1
Area (ABD) 2 53 z
 Area  ADC   1 3x

r
33. (c)

si
C
4 y

an by
(3, 4, 5) is a triplet.
30° So, minimum value of x, y, z = (1, 1,1)

n
36. (a)

ja
R s
x
60° 120° 6
a th

D A B
Angle  30 60 90 10
Opposite Sides  1 : 3 : 2
ty a

x= 62  102  136 unit

Since, A is obtuse angle
A
di M

BC2 = AB2 + AC2 + 2 AB.AD 37. (a)

1  1  60°
= AB2+ AC2 + 2AB × AC  AD  AC
2  2  5 6
2 2
= AB + AC + AB × AC
 a2 = c2 + b2 + cb
C B C
 BC = AB + AC – 2AB × AC cos60°
2 2 2
34. (c)
1
E D BC2 = 52 + 62 – 2 × 5 × 6 × = 25 + 36 – 30
2
G BC2 = 31
A

90°  BC  31 cm
38. (d) Given,
A B
 AD is median A  80º , B  40º

AG 2 and ABC  FDE

 
GD 1  A  F, B  D and C  E
(3) unit  12 cm
  C  180 –   A   B 
(1) unit 4 cm
AG = 8 cm  C  180 – 120
 BE is median.  C  60

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs9

  C  E  60
43. (a) A
and AB = FD = 5 cm
Hence, Option (d) is correct.
39. (b)
12 12
8

B 5 D C
 AD² = AB² – BD × CD
6 3 (8)² = (12)² – 5 × CD
Side ratio when angles are 30°, 60°, 90° 5 × CD = 144 – 64
=1: 3 :2 80
CD = = 16 cm
(2) unit  6 3 cm 5
A

r
(1) unit  3 3 cm 44. (c)
So,

si
AB  3 3  3  9 cm

an by
AC  1  3 3  3 3 cm 5cm

1 1

n
 Area  =  3 3  3  3   6 3  AD
2 2
AD = 4.5 cm

ja
40. (a) If triangle is obtuse then [greatest angle >
R s
C D B
sum of other two angles]
12cm
a th
So, option (a) satisfies this AB = 13 cm
9 > 4 + 3  AD is angular bisector
4 : 9 : 3
AC CD 5x
A   
ty a

41. (c) AB DB 13x

18x = 12 cm
di M

17cm 2
 x = cm
10

8cm 3
cm

2 10
CD = 5x = 5 ×  cm
3 3
B 6cm C D 100 225  100
 AD² = AC² + CD² = 25 + =
15cm 9 9
 CD = (15 – 6) = 9 cm 325
 AD² =
9
42. (a) A
5 13
 AD = cm
3
45. (a)
A

17 17 Let, side of equilateral triangle be 6 unit.

15 A

B D 4cm C 6
 AD² = AB² – BD × CD
(15)² = (17)² – BD × 4
BD × 4 = 289 – 225
BD × 4 = 64 B 2 D1E 3 C
BD = 16 cm  Height of equilateral triangle, AE

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs1

 3 3
= a  6  3 3 unit
2 2
 ABC  9
In AED, ACB 2
(AD)2 = DE2 + AE2 = 1 + 27 = 28 unit2 2BAC = 7ACB (Given)
AD2 : AB2 = 28 : 36 = 7 : 9
46. (b) Sides = 14, 22, x  BAC  7
 Triangle is acute. ACB 2
x2 + 142 > 222 ABC : BAC : ACB = 9 : 7 : 2
 x2 > 288 (18) unit 180°
289 = 172 (1) unit  10°
 xmin = 17 ABC = 90°
xmax = 21
A
17  21 38
 Average =   19

r
2 2
47. (b)

si
17
Let, third side of triangle be 'x'. 8
Sides = 8, 15, x

an by
According to property of triangle,
15 – 8 < x < 15 + 8

n
 7 < x < 23
B C
If a, b, c are the sides of obtuse angle triangle

ja
then,  BC  172 – 82 = 15 cm
R s
a2 + b2 < c2 (where, c = longest side) 50. (d)
a th

If 15 is longest side then, B

82 + x2 < 152
x2 <161
 x is integer
ty a

10cm
 x = 12, 11, 10, 9, 8
If 'x' is longest side then,
di M

82 + 152 < x2

289 < x2 17 < x
A 20cm C
 x = (18, 19 20, 21, 22)
 10 triangles are possible. Area of triangle = 80 cm2
E 1
  20  10  sin θ  80
48. (d) 2

C 4
 sin θ 
° 5
30
A

3
 cos θ 
60° 5
A D B then, AB2 = AC2 + BC2 – 2AC × BC × cos
ACE = 180°– 30° = 150° (Linear pair)
AC = EC (Given) 3
AB2 = (20)2 + (10)2 – 2 × 20 × 10 ×
 CAE = CEA 5
In CAE, AB2 = 500 – 240 = 260
150° + 2CAE = 180°
AB = 2 65 cm
CAE = 15°

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs1

 ALGEBRA/chtxf.kr
(Practice Sheet With Solution)
Level - 01  3  1 
;fn  3y – y  = 5 gS] rks y dk eku Kkr djasA
2

1 1 y 2 
2
1. If x   7 , then the value of x  2 is:
x x SSC CGL 19/07/2023 (Shift-02)

1 1 47 49
;fn x   7 , rks x 2  2 dk eku gS% (a)
9
(b)
9
x x

r
SSC CHSL 02/08/2023 Shift-01 41 43
(c) (d)
(a) 49 (b) 51 9 9

si
(c) 47 (d) 5 6. If p + q = 12 and pq = 14, then find the

2.
 1 an by  2 1 
If  y   = 8, find the value of  y  2  .
 y  y 
value of p2 – pq + q2
;fn p + q = 12 vkSjpq = 14, rksp2 – pq + q2 dk

n
eku Kkr dhft,A
 2 1  SSC CHSL 14/08/2023 (Shift-1)
;fn

ja
1
 y  y  = 8 gS] rks
 y  y 2  dk eku Kkr dhft,A
R s
(a) 192 (b) 181
(c) 102 (d) 144
a th

SSC CHSL 03/08/2023 (Shift-03)

(a) 64 (b) 66 1
7. If x2 + x = 19, find (x + 5)2 + (x + 5)2 .
(c) 62 (d) 60
ty a

 1 1
3. If  z    4 , then what will be the value ;fn x2 + x = 19 rc (x + 5)2 + (x + 5)2 dk
z
di M

eku Kkr dhft,A

1 2 1  (a) 77 (b) 79
of  z  2  ?
2 z (c) 81 (d) 83
8. If 2a + 3b = 14 and 2a – 3b = 10, then find
 1 1 2 1 
;fn  z    4 ] rks  z  2 
z 2 z
dk ewY; D;k gksxk the value of ‘ab’.
;fn 2a + 3b = 14 vkSj2a – 3b = 10 gS] rks
‘ab’
\
dk eku Kkr dhft,A
SSC CHSL 03/08/2023 (Shift-02)
SSC CHSL 03/08/2023 Shift-04
(a) 14 (b) 16
(a) 5 (b) 3
(c) 7 (d) 8
(c) 4 (d) 2
A

4. If x + y = 7 and xy = 19, then calculate the

value of x2 + y2 9. Simplify the given expression:
(5p + 3q) (5p – 3q)
;fn x + y = 7 vkSj xy = 19 gS] rksx2 + y2 dk
eku Kkr djsaA fn, x, O;atd dks ljy dhft,A
SSC CGL 21/07/2023 (Shift-01) (5p + 3q) (5p – 3q)
(a) 17 (b) 12 SSC CHSL, 17/08/2023 (Shift-2)
(c) 11 (d) 19 (a) 25p – 9q2 + 30 pq
2

(b) 25p2 – 9q2

 3  2 1 
5. If  3y –  = 5, find the value of  y  2  . (c) 25p2 + 9q2
 y  y 
(d) 25p2 + 9q2 – 30pq
 1 1   1 3
 2 1
10. If x = ,  x  0  , then the value of x  is : ;fn  a  a   6 , rks 4  a  a2  dk ekuD;k gS \
x –3 x

1 1 SSC CHSL 02/08/2023 Shift-02

;fn x = ,  x  0  rksx  dk eku D;k gS\ (a) 22.5 (b) 34
x –3 x
(c) 25.5 (d) 36
SSC CGL 26/07/2023 (Shift-01)
 2 1 
(a) 11 (b) 17 16. If  y  y 2  = 167 and y > 0, find the value
(c) 15 (d) 13
 1
1 1 of  y   .
11. If a – = 4, then the value of a + is:  y
a a
 1   1
;fn  y  y2  = 167 vkSjy > 0, rks  y  y  dk
2
1 1
;fn a – =4 gS] rks
a+ dk eku gS%
a a

r
SSC CPO 05/10/2023 (Shift-2)
eku Kkr dhft,A

si
SSC CHSL 14/08/2023 (Shift-1)
(a) 5 5 (b) 4 5
(a) 13 (b) – 165
(c) 2 5
an by (d) 3 5
(c) 165 (d) –13

n
1  1
12. If a = (a > 0), then the value of  a   is:  3  2 1 
a 6 a
17. If  3y   = 8, find the value of  y  y 2  .
 y
ja
R s
1  1
;fn a = (a > 0)] rks  a  
a
dk eku gS%  3  1 
a 6 ;fn  3y  y  = 8 rks  y  y2  dk eku Kkr dhft,A
2
a th

SSC CHSL 02/08/2023 Shift-03

SSC CHSL 11/08/2023 (Shift-1)
(a) 6 (b) 10
1 5
ty a

(c) 15 (d) 7 (a) 5 (b) 4

9 6
1 1
di M

13. If x  (x  0) , then the value of x  is: 1 1

x5 x (c) 7 (d) 9
9 9
1 1 18. If x2 – 3.2 x + 1 = 0 and x > 1, the value of
;fn x  (x  0) , rks x  dk eku gS%
x5 x  2 1 
 x – 2  is:
SSC CHSL 02/08/2023 Shift-03 x
(a) 41 (b) 29  2 1 
;fn x2 – 3.2 x + 1 = 0 vkSjx > 1, rks  x – 
x2 
(c) (d) 43
23 dk eku D;k gS\
2
14. If x – 8x – 1 = 0, what is the value of SSC CHSL 11/08/2023 (Shift-3)
1 (a) 16.8 0.39 (b) 12.8 0.39
x2 + 2 .
x
A

(c) 16.8 0.32 (d) 12.8 0.32

1
;fn x – 8x – 1 = 0 gS] rksx + 2
2
x
2
dk eku D;k gS\ 19. 2
If x – 7x + 1 = 0, and 0 < x < 1, what is the
2 1
SSC CGL 26/07/2023 (Shift-01) value of x – 2 ?
x
(a) 68 (b) 62 ;fn x2 – 7x + 1 = 0 vkSj 0 < x < 1 gS] rks
(c) 64 (d) 66 1
x 2 – 2 dk eku D;k gksxk\
 1 x
15. If  a    6 , then what is the value of
a SSC CGL 19/07/2023 (Shift-03)

3 2 1  (a) 21 5 (b) –21 5

 a  2  ?
4 a (c) 28 5 (d) –28 5
 (a) 27 (b) 81
 2 1 
20. If  x  2   7 , and 0 < x < 1, find the value (c) 48 (d) 47
x
1
25. If x > 0, and x4 + = 254, What is the value of
1
2 x4
of x – 2 .
x 1
x5 + ?
x5
 2 1
;fn  x  x 2   7 ] vkSj 0 < x < 1 gS] rks 1 1
;fn x > 0 vkSjx4 + = 254 gS] rksx5+ dk
x4 x5
x2 –
1
dk eku Kkr djsaA eku D;k gksxk\
x2 SSC CGL 14/07/2023 (Shift-04)
SSC CGL 17/07/2023 (Shift-01) (a) 717 2 (b) 723 2
(a) 3 5 (b) 4 3
(c) 720 2 (d) 726 2
(c) 4 3 (d) 3 5

r
1 1
26. If x + = 7, then the value of x6 + 6 is:

si
21. If x2 – 5 5x  1  0 , and x > 0, what is the x x

 3 1  1 1
;fn x + = 7 gS] rks dk eku Kkr dhft,A
an by
value of  x  3 
x x
x6 + 6
x

n
SSC CGL 14/07/2023 (Shift-03)
 3 1  (a) 113682 (b) 103682
;fn x – 5 5x  1  0 vkSjx > 0, rks x  3 
2
x
ja (c) 103882 (d) 103862
R s
dk eku D;k gS 1 x 2  7x  1
27. If x  = 1, then the value of x 2  11x  1  ?
a th

SSC CHSL 08/08/2023 Shift-02 x

(a) 1331 (b) 1364
1 x 2  7x  1
(c) 1296 (d) 1244 ;fn x = 1 gS] rks 2
x  11x  1
dk eku D;k gS\
x
ty a

 1  4 1  SSC CGL 26/07/2023 (Shift-02)

22. If  x   = 10, what is the value of  x  4  ?
x x
3 2
di M

(a) (b)
 1  1  4 3
;fn gS rks x  dk eku D;k gS
\
4
 x   = 10, 
x x4 
1 1
(c) (d)
SSC CGL 17/07/2023 (Shift-04) 3 4
(a) 9604 (b) 9602
(c) 9600 (d) 9606 28. If P = 7 + 4 3 and PQ = 1, then what is
4 1 1 1
23. If x  4  194, x  0 , then find the value 
x the value of ?
P 2 Q2
3 1 1
of x  3  x  .
x x 1 1
1 1 1 ;fn P = 7 + 4 3 vkSjPQ = 1 gSa] rks 
P 2 Q2
;fn x  4  194, x  0 gS] rks
4
x3  3  x 
A

x x x dk eku D;k gS\

dk eku Kkr dhft,A
(a) 196 (b) 194
SSC CHSL 08/08/2023 Shift-03
(c) 206 (d) 182
(a) 76 (b) 66
(c) 56 (d) 46 29. If x = 2  3 , y = 2  3 , then what will be
the value of x2 + y2.
1 1
24. If a +
a
= 3, then the value of a4 + 4 is:
a ;fn x = 2  3 , y = 2  3 rks x2 + y2 dk eku
D;k gksxkA
1 1
;fn a + = 3 gS] rks
a4 + 4 dk eku D;k gksxk\ UPSI 13/12/2017 (Shift-01)
a a
(a) 14 (b) 3
SSC CPO 04/10/2023 (Shift-01) (c) 8 (d) 4
30. If x = 1  2 and y = 1  2 then what is the ;fn a = 17, b = 13 gks] rks O;atd
(a3 – b3 – 3a2b +
value of x2 + y2? 3ab2) dk eku Kkr dhft,A

;fn x = 1  2 vkSjy = 1– 2 rksx2 + y2 dk eku SSC CGL 14/07/2023 (Shift-04)

D;k gksxk\ (a) – 64 (b) –2700

UPSI 13/12/2017 (Shift-01) (c) 2700 (d) 64
(a) 8 (b) 4 37. If a + b + c = 1 and a + b3 + c3 = 4, then
3

(c) 6 (d) 2 1 1 1
+ + =?
31. If a + b = 10 and ab = 6, then the value a+bc b+ac c+ab
of a3 + b3 is:
;fn a + b + c = 1 vkSja3 + b3 + c3 = 4 rks
;fn a + b = 10 vkSj ab = 6 rks a3 + b3 dk 1 1 1
eku D;k gksxk\ + +
a+bc b+ac c+ab
=?
SSC CPO 03/10/2023 (Shift-01) (a) 1 (b) –1
(a) 860 (b) 820

r
(c) –2 (d) 3
(c) 800 (d) 840

si
38. If (a + b + c) = 19, and (a2 + b2 + c2) = 155,
x y
32. If  = 1 and x + y = 2, then the value find the value of (a – b)2 + (b – c)2 + (c – a)2.
y x
an by
of x3 + y3 is: ;fn (a + b + c) = 19 vkSj (a 2 + b 2 + c2)
= 155 gS] rks(a – b) 2 + (b – c) 2 + (c – a) 2 dk

n
x y
;fn y  x = 1 vkSj x + y = 2 gS] rksx3 + y3 eku Kkr djsaA
dk eku D;k gksxk\
ja SSC CGL 18/07/2023 (Shift-02)
R s
SSC CPO 03/10/2023 (Shift-01) (a) 104 (b) 108
(a) 0 (b) 1
a th

(c) 100 (d) 98

(c) 3 (d) 2
3 3 39. If a = 101, b = 102 and c = 103, then
33. If x = 270 + y and x = (6 + y) then what is the
a² + b² + c² – ab – bc – ca
value of (x + y)? (given that x > 0 and y > 0)
ty a

;fn x3 = 270 + y3 vkSjx = (6 + y) rks(x + y) dk eku ;fn a = 101, b = 102 vkSjc = 103 gS] rks
D;k gS\ (;g ns•rs gq, fd x > 0 and y > 0) a² + b² + c² – ab – bc – ca dk eku D;k gksxk\
di M

SSC CHSL 03/08/2023 Shift-04 SSC CGL 24/07/2023 (Shift-01)

(a) 2 3 (b) (a) 2 (b) 4

3
(c) 3 (d) 6
(c) 3 3 (d) 4 3
40. If (a + b + c) = 20 and a2 + b2 + c2 = 262, then
34. If a – b = 5 and ab = 24, find the value of a3 –
find the value of ab + bc + ca.
b3.
;fn a – b = 5 vkSjab = 24 gS] rksa3 – b3 dk eku ;fn (a + b + c) = 20 vkSja2 + b2 + c2 = 262, rks
Kkr dhft,A ab + bc + ca dk eku Kkr dhft,A
SSC CHSL 14/08/2023 (Shift-3) SSC CHSL 10/08/2023 (Shift-2)
(a) 455 (b) 485 (a) 48 (b) 84
A

(c) 385 (d) 360 (c) 72 (d) 69

35. If x + y = 25 and xy = 20, then find the 41. If (a + b + c) = 17, and (a2 + b2 + c2) = 115,
value of x3 + y3. find the value of (a + b)2 + (b + c)2 + (c + a)2.
;fn x + y = 25 vkSjxy = 20 gS] rksx3 + y3. dk ;fn (a + b + c) = 17, vkSj(a2 + b2 + c2) = 115, rks
eku Kkr dhft,A (a + b)2 + (b + c)2 + (c + a)2 dk eku Kkr dhft,A
SSC CHSL 11/08/2023 (Shift-2)
SSC CGL 27/07/2023 (Shift-01)
(a) 402 (b) 408
(a) 13152 (b) 13125
(c) 394 (d) 404
(c) 14125 (d) 14152
42. If x + y + z = 22 and xy + yz + zx = 35
36. If a = 17, b = 13, then find the value of the then what is the value (x – y) 2 + (y – z) 2
expression (a3 – b3 – 3a2b + 3ab2) + (z – x) 2
 ;fn x + y + z = 22 rFkkxy + yz + zx = 35 gS] ;fn (2a – 3)2 + (3b + 4)2 + (6c + 1)2 = 0 gS
rks(x – y)2 + (y – z)2 + (z – x)2 dk eku D;k gksxk\ rks
a 3 +b3 +c3 –3abc
+ 3 dk eku gS%
(a) 793 (b) 681 a 2 – b2 +c2
(c) 758 (d) 715 (a) abc + 3 (b) 6
43. a, b, c are three positive number such that (c) 0 (d) 3
a + b + c = 20, a2 + b2 + c2 = 152 the value 49. If a + b + c = 5 and a + b2 + c2 = 15, then
2

of ab + bc + ca is equal to. find the value of a3 + b3 + c3 – 3abc – 27.

a, b, c rhu /ukRed la[;k,¡ bl izdkj gSa fd ;fn a + b + c = 5 vkSja2 + b2 + c2 = 15, rksa3
a + b + c = 20, a2 + b2 + c2 = 152 gSaA
ab + bc + + b3 + c3 – 3abc – 27. dk eku Kkr djsa
ca dk eku D;k gksxkA SSC CHSL 08/08/2023 Shift-03
(a) 124 (b) 110 (a) 23 (b) 27
(c) 112 (d) 102 (c) 25 (d) 21
44. If a + b + c = 6 and ab + bc + ca = 1 then 50. Simplify the given expression.
fn, x, O;atd dks ljy dhft,A

r
find the value bc (b + c) + ca(c + a) +
ab(a + b) + 3abc.

si
x 3  y 3  z 3  3xyz
;fn a + b + c = 6 rFkkab + bc + ca = 1 gS rks
(x  y)2  (y  z)2  (z  x )2
bc(b + c) + ca(c + a) + ab(a + b) + 3abc dk
an by
eku D;k gksxkA
SSC CHSL 07/08/2023 Shift-04

n
(a) 3 (b) 6 1
(a) (x  y  z) (b) (x  y  z)
(c) 5 (d) 2 3
45.
ja
If (a+b+c) = 17 and (a + b2 + c2) = 117, what
2
1 1
R s
is the value of (a – b)2 + (b – c)2 + (c – a)2 (c) (x  y  z) (d) (x  y  z)
4 2
;fn (a + b + c) = 17 vkSj(a2 + b2 + c2) = 117, rks 51.
a th

x2 + y2 – 10x + 12y + 61 = 0, then 2x + 3y = ?

(a – b)2 + (b – c)2 + (c – a)2 dk eku D;k gksxk\ x2 + y2 – 10x + 12y + 61 = 0, rks2x + 3y = ?
(a) 57 (b) 72 (a) – 1 (b) – 8
(c) 42 (d) 62
ty a

(c) 1 (d) 8
46. If ab + bc + ca = 119, a 2 + b 2 + c 2 = 162 52. (a + 2)2 + (b – 3)2 + (c – 9)2 = 0. Find the
and a, b & c are positive values then
di M

value of a + b + c = ?
what is the value of a 2(b + c) + b 2(c + a)
(a + 2)2 + (b – 3)2 + (c – 9)2 = 0. a + b + c
+ c 2(a + b) + 3abc ?
dk eku Kkr dhft;s\
;fn ab + bc + ca = 119, a2 + b2 + c2 = 162
(a) 10 (b) 14
vkSja, b vkSjc /ukRed gSa] rks
a2(b + c) + b2(c + a)
(c) 12 (d) 13
+ c (a + b) + 3abc dk eku D;k gS\
2
53. If (a – 1)2 + (b + 2)2 + (c +1)2 = 0, then 2a –
(a) 2380 (b) 2499 3b + 7c = ?
(c) 2450 (d) 1760 ;fn (a – 1)2 + (b + 2)2 + (c +1)2 = 0 rks 2a –
47. If (a + b + c) 0, then ( a + b + c) ( a 2 + b 2 + 3b + 7c = ?
c 2 – ab – bc – ca) is equal to: (a) 1 (b) 0
;fn (a + b + c)  0] rks( a + b + c) ( a2 + b2 + c2 – ab (c) 2 (d) 3
A

– bc – ca) cjkcj gS%  1 1

If  x   = 2, then x  117 = ___.
2
54.
SSC CHSL 03/08/2023 Shift-04 x x
3 3 3
(a) a  b  c  3abc  1  1
; fn  x  x  = 2 gS] rksx 2  117 dk eku D;k gksxk\
(b) a3  b3  c3  3abc x
SSC CGL 19/07/2023 (Shift-04)
(c) a3  b3  c3  3abc
3 3 3
(a) 1 (b) 2
(d) a  b  c  3abc (c) 4 (d) 3
48. If (2a – 3)2 + (3b + 4)2 + (6c + 1)2 = 0, then 55. If x + y = 41, then find (x –20)2021 + (y – 21)2021?

the value of
3 3 3
a +b +c –3abc
+ 3 is:
;fn x + y = 41 gS] rks(x –20)2021 + (y – 21)2021
a 2 – b2 +c2 Kkr dhft;s\
 (a) 0 (b) 1 61. What is the value of (27x3 + 58x2y + 31xy2
(c) 2 (d) 3 + 8y3), when x = 5 and y = –7 ?
56. If a2 + b2 + c2 = ab + bc + ac, then the value of tc x = 5 vkSjy = –7 gS] rc (27x3 + 58x2y + 31xy2
11a 4 +13b4 +17c 4 + 8y3) dk eku D;k gksxk\
is:
17a 2 b2 +9b2 c2 +15c2 a 2 SSC CGL 21/07/2023 (Shift-02)
;fn a2 + b2 + c2 = ab + bc + ac gS] rks (a) 1924 (b) –1926
11a 4 +13b4 +17c 4 (c) –1924 (d) 1926
17a 2 b2 +9b2 c2 +15c2 a 2 dk eku Kkr dhft,A 62. If x + 3y = 6, what is the value of x3 + 27y3 + 54
xy?
SSC CGL 18/07/2023 (Shift-03)
;fn x + 3y = 6 rksx3 + 27y3 + 54 xy dk eku D;k gS\
(a) 1 (b) 2
SSC CHSL 14/08/2023 (Shift-4)
(c) 11 (d) 4
(a) 264 (b) 258
57. If x2 + y2 + z2 = xy + yz + zx, then the value
(c) 216 (d) 220
3x 4 +7y 4 +5z 4 63. If 2x + 3y = 15, then find the maximum

r
of is value of x2y3?
5x y 2 +7y 2 z 2 +3z 2 x 2
2

; fn r ks ;fn 2x + 3y = 15 gks] rks

x2y3 dk vf/dre

si
x 2 + y 2 + z 2 = xy + yz + zx
3x 4 +7y 4 +5z 4
eku Kkr dhft;s\
dk eku D;k gksxk\
(a) 0
an by
5x 2 y 2 +7y 2 z 2 +3z 2 x 2
(b) 2
(a) 243
(c) 27
(b) 81
(d) 15

n
64. If a + b = 3, then find the maximum value
(c) 1 (d) – 1 of a2b4?
;fn a + b = 3 gks] rks
a2b4 dk vf/dre eku
ja 3y 2  x 2  z 2
R s
58. If x + y + z = 0 then Kkr dhft;s\
2y– xz
a th

(a) 1 (b) 2
3y 2  x 2  z 2 (c) 4 (d) 16
;fn x + y + z = 0 gS rks 2y– xz dk eku
65. If 2x + y = 10, then find the maximum
value of x2y3?
D;k gksxkA
ty a

(a) 2 (b) 1
;fn 2x + y = 10 gks] rks
x2y3 dk vf/dre eku
Kkr dhft;s\
di M

3 5 (a) 684 (b) 864

(c) (d)
2 3 (c) 846 (d) 468
66. The value of/dk eku Kkr dhft,A
3y  x  z  2 2 2

59. If x + y + z = 0, then
2y – xz  = ?
2 2 2
2
p2 –  q – r  q2 –  p – r  r 2 – p – q
 p + r 2 – q2 +  p + q2 – r 2 +  q + r 2 – p2 is:
3y 2 2
 x z 2

;fn x + y + z = 0 gS] rks 2y 2 – xz  = ? SSC CGL 24/07/2023 (Shift-01)
(a) 1 (b) 2
(a) 2 (b) 1 (c) 0 (d) 3
3 5 2ab
67. If ax + by = 1 and bx + ay = , then
A

(c) (d) a  b2 2
2 3
the value of x (in terms of a and b) is:
2–x 1 2y  1 2ab
60. If y = , then  2 =? ;fn ax + by = 1 vkSjbx + ay = , rksx dk
1+ x y  1 y –1 a 2  b2
eku ( a vkSjb ds lanHkZ esa) Kkr dhft,A
2–x 1 2y  1
;fn y= rc 
y  1 y2 – 1
dk eku Kkr dhft,A SSC CHSL 11/08/2023 (Shift-4)
1+ x
2b a
(1 – x )(2 – x ) (1 – x )(2 – x ) (a) (b)
(a) (b) a 2  b2 a 2  b2
2x – 1 x –1
(1 – x )(2 – x ) (1 – x )(1 – 2x ) b 2a
(c) (d) (c) (d)
1 – 2x 2–x a – b2
2
a – b2
2
 (a) 2 (b) 1
1
68. If x + = 1 , then x53 + x50 = ? (c) 0 (d) – 1
x
1 1
1 74. If x = a  , y = a – then find the value
;fn x + =1 gks] rks
x53 + x50 ? a a
x
x4 + y4 – 2x2y2
(a) 1 (b) 2
1 1
(c) 0 (d) 5 ;fn x = a  , y = a – rc x4 + y4 – 2x2y2
a a
1
= 1 , then x28 + x25 + x21 + x18 + x12
dk eku Kkr dhft,A
69. If x +
x (a) 15 (b) 21
+ x9 + x 6 + x3 = ?
(c) 16 (d) 32
1
;fn x + =1, gks] rks
x28 + x25 + x21 + x18 +  4 
x 75. If  5a  – 2  = 13 and a > 0, what is the
a
x12 + x9 + x6 + x3 = ?

r
(a) 0 (b) 2  2 16 
(c) 1 (d) 5 value of  25a  2  ?

si
a
1
70. If x  , find x103 + x100 + x53 + x50 + x23 +  4 
x
an by
x20 + x18 + x15 + x12 + x9 + x6 + x3 + 1
;fn  5a  – 2  = 13
a
vkSj a > 0 gS] rks

n
1  16 
dk eku Kkr dhft,A
2
;fn x  rc x103 + x100 + x53 + x50 +  25a  2 
x a

ja
x23 + x20 + x18 + x15 + x12 + x9 + x6 + SSC CHSL 10/08/2023 (Shift-4)
R s
x3 + 1 dk eku Kkr dhft,A (a) 158 (b) 157
a th

(a) 0 (b) 1 (c) 185 (d) 175

(c) 2 (d) 4
 2 5 1
X X  5Y 7 76. If x  5 –   , then the value of x2 + 2 is:
71. What is the value of If  x x x
Y X  5Y 13
ty a

X  5Y 7 X  2 5 1
;fn  gks] rks dk eku Kkr dhft,A ;fn x  5 – x   x gS] rks
x2 + 2 dk eku gS%
di M

X  5Y 13 Y x
SSC CGL 17/07/2023 (Shift-04) SSC CPO 03/10/2023 (Shift-3)
27 24
(a) (b) 54 53
7 9 (a) (b)
23 28
50 100
(c) (d)
3 7 53 54
72. Find the minimum value of 4x2 – 2x + 5 ? (c) (d)
27 25
4x2 – 2x + 5 dk U;wuÙke eku Kkr dhft;s\
19 4 1
(a) (b) 77. If 7b – = 7, then what is the value of
4 19 4b
19 4 1
(c) – (d) –
A

4 19 16b2 + ?
49b2
Level-02
1 1
58 – 252
2
262 – 152 ;fn 7b – 4b = 7 gS] rks16b 2 + 49b2 dk eku
73. If A = 2 2 , B = , then the
46 – 37 562 – 152
1 20 Kkr dhft,A
value of – is:
B A SSC CGL 14/07/2023 (Shift-01)
582 – 252 262 – 152
; fn A = 2 2
,B = g S ] r ks (a)
80
(b)
104
46 – 37 562 – 152
49 7
1 20
– dk eku D;k gksxk\ 120 7
B A
(c) (d)
SSC CGL MAINS 26/10/2023 7 2
  2 1  (a) – 7776 (b) – 6726
78. If  x  2   4 6 , and x > 1, what is the (c) – 6730 (d) – 6732
x
1  83. If x2– 3 7x + 13 = 0, then
 3
value of  x  3  ?
x 6 1
x – 7  – 6

;fn
 2 1 
 x  2   4 6 vkSj x > 1 gS] rks
x – 7 
x
;fn x2 – 3 7x + 13 = 0 rc
 3 1 
 x  3  dk eku D;k gksxk\ 6 1
x x – 7  – 6
dk eku Kkr dhft,A
SSC CGL 21/07/2023 (Shift-03) x – 7 
(a) 20 2 (b) 24 2 (a) 80 77 (b) –80 77
(c) 18 2 (d) 22 2 (c) 77 (d) none
79. If x2 – 9.76 x + 1 = 0 and x > 1, the value

r
 1
84. If  x    5 2 , and x > 1, what is the
1 x

si
3
of x – is :
x3  6 1 
value of  x  6  ?
;fn x2 – an by
9.76 x + 1 = 0 vkSj x > 1, rks x

n
1  1  6  1
x3 – dk eku D;k gS\ ;fn  x  x   5 2 , vkSjx > 1, rks  x  x 6 
x3
dk eku D;k gS\
ja SSC CHSL 10/08/2023 (Shift-3)
R s
(a) 21.042 (b) 24.024 SSC CHSL 02/08/2023 Shift-02
a th

(c) 21.024 (d) 24.042 (a) 22970 23 (b) 23030 23

7 1 (c) 23060 23 (d) 22960 23
80. If 7a   4  0 , then find a 3  3  1 .
a a
 1
ty a

7 1 85. If  x    2 2 , and x > 1, what is the

;fn 7a   4  0 gS] rksa  3  1
3
dk eku x
a a
di M

 6 1 
Kkr dhft,A value of  x – 6  ?
x
SSC CGL 25/07/2023 (Shift-02)
 1   6  1
(a)
995
(b)
875 ;fn  x  x   2 2 vkSjx > 1, gS] rks x – x 6 
343 248
dk eku D;k gksxk\
694 765
(c) (d) SSC CGL 19/07/2023 (Shift-01)
315 262
(a) 140 2 (b) 116 2
1 1
81. If a   7 , then a 5  5 is equal to:
a a (c) 144 2 (d) 128 2
1 1
;fn a   7 gS] rksa 5  fuEu esa ls fdlds 86. If x > 0 and x4 + 14 = 142, what is the
a5
A

a x
cjkcj gS\ 1
SSC CGL 17/07/2023 (Shift-02) value of x7 + ?
x7
(a) 15127 (b) 13127
(c) 14527 (d) 11512 1 1
;fn x > 0 vkSjx4 + = 142 gS] rksx7 +
x4 x7
1 1
82. If x + = – 6 , what will be the value of x 5 + 5 dk eku D;k gksxk\
x x
SSC CGL 20/07/2023 (Shift-02)
1 1
;fn x + = – 6 gS] rks
x5  dk eku D;k gksxk\ (a) 1561 14 (b) 1563 14
x x5
SSC CGL 17/07/2023 (Shift-03) (c) 1560 14 (d) 1562 14
  1 (a) 0 (b) 18
87. If  x    6 , and x > 1, what is the (c) 21 (d) 20
x
 8 1  3 2 3– 2 a 2 + b2 + ab
92. If a = ,b  then =?
value of  x  8  ? 3– 2 3 2 a 2 + b2 – ab
x
3 2 a 2 + b2 + ab
3– 2
 1  8 1  ;fn a = ,b  gS] rks2
=?
;fn  x    6
x
vkSjx > 1 gS] rks x  x 8  dk 3– 2 3 2 a + b2 – ab
97 99
eku D;k gksxk\ (a) (b)
99 98
SSC CGL 17/07/2023 (Shift-02) 98 99
(c) (d)
(a) 1024 15 (b) 992 15 99 97
93. If y = 3  2 2 , then find the value
(c) 998 15 (d) 1012 15 2 –2
(approximately) of 2 (y – y )
 1
88. If  x   = 5, and x >1, what is the value vxj y = 3  2 2 , rks 2 (y2 – y–2) dk eku yxHkx

r
x
Kkr djsaA
 8 1 

si
of  x – 8  ? UPSI 12/12/2017 (Shift-02)
x
(a) 48 (b) 96
;fn

an by
1
vkSj x >1 gS]
 x   = 5
x
 8 1 
rks x – x 8  (c) 24 (d) 28

n
dk eku D;k gksxk\ 94. If
a 8
 
b
and (a+b) = 30, then what is
SSC CGL 18/07/2023 (Shift-03) b 3 a

ja the value of ab?

R s
(a) 60605 21 (b) 60615 21
a 8 b
;fn   vkSj(a+b) = 30 gS] rksab dk
a th

(c) 60705 21 (d) 60725 21

b 3 a
 1 eku D;k gS\
89. If  x    6 and x > 1, what is the value
x SSC CHSL 03/08/2023 (Shift-01)
ty a

8 1 (a) 64 (b) 28
of x – 8 ? (c) 81 (d) 36
x
di M

 1 1 3+ 2 3– 2
;fn  x  x   6 vkSjx > 1 gS] rksx 8 – 8 dk 95. a=
3– 2
,b=
3+ 2
a3 + b 3 = ?
x
eku D;k gksxk\ (a) 970 (b) 1000
SSC CGL 26/07/2023 (Shift-03) (c) 1030 (d) 90

(a) 120 3 (b) 128 3 2 1 2 1

96. If p = and q = then find the
2 1 2 1
(c) 112 3 (d) 108 3
p2 q 2
1 4 3 2
x  3 x  5x  3 x  1 value of  .
90. If x + = 5, then =? q p
x x4  1
2 1 2 1 p2 q2
1
;fn x + = 5 rks
x 4  3 x 3  5x 2  3 x  1
=? ;fn p = vkSjq = gS] rks q + p
2 1 2 1
A

x x4  1
12 43
dk eku Kkr dhft,A
(a) (b) SSC CGL 18/07/2023 (Shift-04)
31 23
(a) 200 (b) 196
15 31
(c) (d) (c) 198 (d) 188
26 52
97. If x = 3  2 , then the value of x3 + x +
1
1 x4+ 2 1 1
91. If x + = 3, then x =?  is "
x 2
x –2x +1 x x3
1 1 1
1 x4+ 2 ;fn x = 3  2 , rksx3 + x +  3 dk eku D;k gS\
;fn x + = 3] rks x =\ x x
x 2
x –2x +1 UPSC CDS 2015 (2)
 (a) 10 3 (b) 20 3 103. If (Ax + By – Cz) 2 = 4x 2 + 3y 2 + 2z 2 +
4 3 xy – 2 6 yz – 4 2 xz, then find the
(c) 10 2 (d) 20 2
value of A2 + B2C2
98. If 343 (x + y)3 + 216 (x – y) 3 = (13x + y)
(Ax2 + By2 – 2 Cxy), find A + B + C ? ;fn (Ax + By – Cz)2 = 4x2 + 3y2 + 2z2 +
;fn 343 (x + y) 3 + 216 (x – y) 3 = (13x + 4 3 xy – 2 6 yz – 4 2 xz gS] rks
A2 + B2C2
y) (Ax 2 + By 2 – 2 Cxy) rc A + B + C dk dk eku Kkr djas
eku Kkr dhft,A
(a) 16 (b) 8
(a) 157 (b) 147
(c) 12 (d) 10
(c) 137 (d) 167
104. If a + b + c = 7, ab + bc + ca = 11 and abc
x 3 –y3 x+y 3xy 2 –3x 2 y = – 1, then a 3 + b 3 + c 3 is equal to:
99. If A = x 2 –y 2 , B = x 2 + y 2 + xy and C = x 2 –y2 ,
;fn a + b + c = 7, ab + bc + ca = 11 vkSjabc
then find the value of AB (A + C)
=–1 gS] rksa3 + b3 + c3 fdlds cjkcj gksxk\
x 3 –y 3 x +y 3xy 2 –3x 2 y
;fn A = x vkSjC = SSC CGL 20/07/2023 (Shift-03)

r
2
–y 2 , B = x 2 + y 2 + xy x 2 –y 2
(a) 101 (b) 107

si
gS]rksAB(A + C) dk eku Kkr djsa (c) 109 (d) 111
(x – y) (x – y) 105. If (a + b + c) = 12, and (a2 + b2 + c2) = 50,
(a) (x + y) an by (b) (x + y)2 find the value of (a3 + b3 + c3 – 3abc).
;fn (a + b + c) = 12 vkSj(a2 + b2 + c2) = 50 gS]

n
(x – y)2 (x – y)2
(c) (x + y) (d) (x + y)2 rks(a3 + b3 + c3 – 3abc) dk eku Kkr dhft,A

ja SSC CGL 19/07/2023 (Shift-02)

R s
2x 2 + x – 6 x 2 + 3x + 9
100. If P = 2(x + 2)(x – 2) , Q = and R (a) 36 (b) 24
(2x – 3)(x 2 + 4)
a th

(c) 42 (d) 48
x 3 – 27 106. If x = (b – c)(a – d), y = (c – a)(b – d), z
= , then find the value of P × Q  R. = (a – b)(c – d), then
x 4 – 16
(x  y)(y  z)(z  x )(x 3  y 3  z 3 )
ty a

2x 2 + x – 6 x 2 + 3x + 9 =?
;fn P =
2(x + 2)(x – 2)
, Q =
(2x – 3)(x 2 + 4) 9x 2 y 2 z 2
;fn x = (b – c) (a – d), y = (c – a) (b – d), z = (a – b)
di M

x 3 – 27
vkSjR = gS] rks
P × Q  R dk eku Kkr djsa
A (x  y)(y  z)(z  x )(x 3  y 3  z 3 )
x 4 – 16 (c – d) rc 9x 2 y 2 z 2
(x  2) (x + 2)
(a) (b) dk eku Kkr dhft,A
[2(x – 3)] (x – 3)
1
2(x + 2) 4(x + 2) (a) (b) 1
xyz
(c) (d)
(x – 3) (x – 3)
1 –1
101. Find the value of/Ekku Kkr djsa% (c)
3
(d)
3
x 4 – 81 2x + 3 [(x – 3)2 + 3x ](x + 5) 107. If ( a + b – c) = 20, and a2 + b2 + c2 = 152, find
 3
× the value of a3 + b3 – c3 + 3abc.
2x 2 +13x +15 x + 27 (x + 3)2 – 6x
;fn ( a + b – c) = 20, vkSja2 + b2 + c2 = 152 gS]
A

(a) (x2 – 9) (b) (x – 3)

rksa3 + b3 – c3 + 3abc. dk eku Kkr dhft,A
(c) (x + 5) (d) (2x + 3)
SSC CGL 25/07/2023 (Shift-04)
102. If a = 97.5, b = 100, c = 102.5 then find
a2 + b2 + c2 – ab – bc – ca. (a) 480 (b) 720
(c) 640 (d) 560
;fn a = 97.5, b = 100, c = 102.5 gSa rks108. If (a3 + b3 + c3 – 3abc) = 405, and (a – b)2
a2 + b2 + c2 – ab – bc – ca dk eku D;k gksxkA + (b – c)² + (c – a)² = 54, find the value
81 70 of (a + b + c).
(a)
4
(b)
4 ;fn (a3 + b3 + c3 – 3abc) = 405 vkSj(a – b)2 + (b – c)²
75 15 + (c – a)² = 54 gS] rks(a + b + c) dk eku Kkr dhft,A
(c) (d) SSC CGL 26/07/2023 (Shift-02)
4 4
 (a) 15 (b) 45 115.If a 2 + b 2 + c 2 + 96 = 8 (a + b – 2c),
(c) 9 (d) 27 then ab – bc + ca is equal to:
109. Find the value of (a + b3 + c3 – 3abc), where
3
;fn a2 + b2 + c2 + 96 = 8 (a + b – 2c), rks
a = 335, b = 215 and c = 180.
ab – bc + ca fuEu esa ls fdlds cjkcj gS\
(a3+b3+c3 – 3abc) dk ekuKkr dhft,]
tgk¡ a = 335, b = 215 vkSjc = 180. (a) 6 (b) 2 2

SSC CHSL 04/08/2023 Shift-03 (c) 4 (d) 2 3

2 2 2
(a) 15452630 (b) 14502230 116. If a + b + 64c + 16c + 3 – 2(a + b), then
(c) 14472250 (d) 15421320 the value of 4a7 + b7 + 8c2 is:
110. If x = 999, y = 1000, z = 1001, then the ;fn a2 + b2 + 64c2 + 16c + 3 – 2(a + b) gS]
x 3 + y 3 + z 3 – 3xyz rks4a7 + b7 + 8c2 dk eku Kkr dhft;s\
value of is
x – y+z
7 7
;fn x = 999, y = 1000, z = 1001 rks (a) 3
8
(b) 4
8

r
x 3 + y 3 + z 3 – 3xyz
dk eku Kkr djks\

si
x – y+z 1 1
(c) 4 (d) 5
8 8
(a) 1000 (b) 6
(c) 1 an by (d) 9
111. If a = 25, b = 15, c = – 10, then find the
117. If ab + bc + ca = abc find
 (b + c)   (a + c)   (a + b) 

n
a 3 +b3 +c3 –3abc  bc(a – 1)  +  ac(b – 1)  +  ab(c – 1) 
value of ?      
a– b2 + b–c2 +c–a2
ja  (b + c)   (a + c) 
R s
; fn a = 25, b = 15, c = – 10 g S, r ks ;fn ab + bc + ca = abc rc   + 
 bc(a – 1)   ac(b – 1) 
a 3 +b3 +c3 –3abc
a th

 a– b2 + b– c2 +c– a 2 dk eku D;k gksxk\  (a + b) 

+  dk eku Kkr dhft,A
(a) 30 (b) – 15  ab(c – 1) 
(c) – 30 (d) 15
ty a

(a) 0 (b) 4
112. If a + b + c = 5 and ab + bc + ca = 7, then (c) 1 (d) –2
the value of a³ + b³ + c³ – 3abc is:
di M

118. If x1 x2 x3 = 4(4 + x1 + x2 + x3), then the

;fn a + b + c = 5 vkSjab + bc + ca = 7 gS] rks what is the value of
a³ + b³ + c³ – 3abc dk eku D;k gksxk\
 1   1   1 
SSC CPO 05/10/2023 (Shift-01)  + + 
 (2 + x1 )   (2 + x 2 )   (2 + x 3 ) 
(a) 20 (b) 25
;fn x1 x2 x3 = 4(4 + x1 + x2 + x3) rc dk eku
(c) 15 (d) 30
3 3 3 Kkr dhft,A
113. If (a + b + c – 3abc) = 405, and (a + b + c)
= 15, find the value of (a – b)2 + (b – c)2 +  1   1   1 
(c – a)2.  + + 
 (2 + x1 )   (2 + x 2 )   (2 + x 3 ) 
;fn (a3 + b3 + c3 – 3abc) = 405 vkSj(a + b + c)
1
= 15 gS] rks(a – b)2 + (b – c)2 + (c – a)2 dk eku (a) 1 (b)
A

2
Kkr dhft,A
1
SSC CGL 20/07/2023 (Shift-01) (c) 2 (d)
3
(a) 27 (b) 54
119. What is the value of (3x + 5x2y + 12xy2 + 7y3),
3
(c) 18 (d) 45 when x = – 4 and y = – 1 ?
114. If 8x2 + y2 – 12x – 4xy + 9 = 0, then value
of (14x – 5y) is: x=–4 vkSjy = – 1 gksus ij(3x3 + 5x2y + 12xy2
;fn 8x2 + y2 – 12x – 4xy + 9 = 0 gks] rks
(14x – + 7y3) dk eku Kkr dhft,A
5y) dk eku Kkr dhft;s\ SSC CGL 14/07/2023 (Shift-01)
(a) 9 (b) 6 (a) –329 (b) –359
(c) 5 (d) 3 (c) –361 (d) –327
120. What is the value of 64x3 + 38x2y + 20xy2 + 1 7 1 13 1 9
y3, when x = 3 and y = – 4? 125. a + = ,b+ = ,c+ = Find abc ?
b 3 c 4 a 2
x = 3 vkSjy = – 4 gksus ij64x3 + 38x2y + 20xy2
+ y3 dk eku Kkr dhft,A 1 7 1 13 1 9
;fn a + = ,b+ = ,c+ = rc
SSC CGL 14/07/2023 (Shift-02) b 3 c 4 a 2
(a) 1236 (b) 488 abc dk eku Kkr dhft,A
(c) 536 (d) 1256 (a) 21 (b) 22
121. What is the value of (27x3 – 58x2y + 31xy2 – (c) 23 (d) 24
8y³), when x = – 5 and y = –7? x y z
126. If = = = K ,the possible
x = – 5 vkSjy = – 7 gksu ij (27x3 – 58x2y + (y + z) (x + z) (x + y)
31xy2 – 8y³), dk eku Kkrdhft,\ value of K is.....
SSC CGL 17/07/2023 (Shift-04) x y z
= = = K ] K dk laHkkfor
(a) 1924 (b) –1924 (y + z) (x + z) (x + y)
(c) –1926 (d) 1928 eku gS-----A

r
122. If p + q + r = 0, then what is the simplified
UPSI 15/12/2017 (Shift-03)
value of the expression

si
 p2 q2 r2  1
(a) – 2 (b) or –1
 p2 - qr + q2 - pr + r 2 - pq 
an by  p2 q2 r2
;fn p + q + r = 0 gS] rks p2 - qr + q2 - pr + r 2 - pq 

(c) –
1
or 1 (d) – 1
2

n
2
O;atd dk ljyhÑr eku D;k gS\ 1 1 1
ja SSC CHSL 03/08/2023 (Shift-02) 127. If a + b + c = 20 and + + = 30 then
R s
a b c
(a) 0 (b) 2 a b b c c a
find the value of + + + + + .
a th

(c) 1 (d) – 1 b a c b a c
123. If abc = 1 then find the value of
1 1 1
1 1 1 ;fn a + b + c = 20 rFkk + + = 30 gS rks
  a b c
1  b  c  1  a  b  1  c  a 1 
1 1
ty a

a b b c c a
1 1 + + + + + dk eku D;k gksxkA
b a c b a c
;fn abc = 1] rks 1  b  c 1  +  +
di M

1  a  b 1  (a) 597
1 (b) 600
1  c  a 1  dk eku Kkr djsA (c) 599
UPSI 20/12/2017 (Shift-02)
(d) Can't be determind
(a) a + b + c (b) 1
a 1 l 2 – m2
128. Simplify the expression
(c)
b
(d)
a+b+c  l  m 2 ,
provided (l + m)  0.
1 1 1
124. a – = m, b – = n, c – = p l 2 – m2
b c a O;atd  l  m2 dk eku Kkr dhft,] ;fn (l + m)
1  0 gksA
A

What is abc – ?
abc SSC CGL 20/07/2023 (Shift-01)
1 1 1 lm l–m
;fn a – = m, b – = n, c – = p rc (a) (b)
b c a lm lm
1 (c) 0 (d) l
abc – dk eku Kkr dhft,A
abc
1 1 1
1 3 129. What is the value of     if
–n –n a b c
(a) (b)
2 2
2a  5 4b  5 6c  5
3 4    0?
(c) –m (d) p a b c
2 5
 134. What is added in x (x + 3) (x + 6)(x + 9)
1 1 1
   
a b c
dk eku D;k gS] ;fn = 0 to make a perfect square.
x (x + 3) (x + 6)(x + 9) = 0eas D;k tksM+k t
2a  5 4b  5 6c  5
  0 fd ;s ,d iw.kZ oxZ gks tk;sA
a b c
(a) 3 4 (b) 35
SSC CGL 21/07/2023 (Shift-02) (c) 36 (d) 37
4 8 135. Find the value of expression
(a) (b) –
5 5 O;atd dk eku Kkr djsaA
2 12 300 × 301 × 302 × 303 +1 = ?
(c) (d) –
5 5 (a) 90101 (b) 900901
130. If x2 – 15x + 1= 0, what is the value of (c) 90091 (d) 90901
x4 – 223x2 + 6?

r
136.Find the value of the expression
;fn x2 – 15x + 1= 0 gS] rks
x4 – 223x2 + 6? dk
O;atd dk eku Kkr djsaA

si
eku D;k gksxk \
600×601×602×603+1 = ?
(a) 9
an bySSC CGL 25/07/2023 (Shift-01)
(b) 5 (a) 361801 (b) 360801

n
(c) 6 (d) 0 (c) 360180 (d) 36001
131. If x2 – 11x + 1 = 0, what is the value of 4x  x 2 2
ja
x8 – 14159x4 + 11? 137. If  ,then what is the value
R s
2
x – 3x  4 3
;fn x2 – 11x + 1 = 0, rksx8 – 14159x4 + 11 dk
a th

8
eku D;k gS\ of x –
x
SSC CHSL 08/08/2023 Shift-01
4x  x 2 2 8
(a) 9 (b) 10 ;fn 2  gS] rks
x– dk eku fdruk
ty a

x – 3x  4 3 x
(c) 12 (d) 11 gksxk\
di M

132. Simplify the following expression (a) –24 (b) –18

fuEufyf•r O;atd dks ljy dhft;sA (c) –36 (d) –42
[(x – 5)(x – 1)] – [(9x – 5)(9x –1)] ÷16x 1
138. If (ab) 3 = 5 then which of the following is
SSC CHSL 07/08/2023 Shift-01
not (a + b)?
(a) 2x (5x – 3) (b) – (5x – 3)
1
(c) x (5x – 3) (d) 6x(5x – 3) ;fn (ab) 3 = 5 rks fuEu esa ls dkSu&lk
(a + b) ugha S\
g
133. Simplify the expression
UPSI 13/12/2017 (Shift-01)
O;atd dks ljy djsA
(a) 30 (b) 126
(u – v)3 + (v – w)3 + (w – u)3
(c) 47 (d) All of the above
A

(u 2 – v 2 )3 + (v 2 – w 2 )3 + (w2 – u 2 )3
a a–2
SSC CHSL 04/08/2023 Shift-01 139. If – = 1 then find 'a'.
3 5
1 a a–2
(a)
(u + v)(v + w)(w + u) ;fn – =1 rks'a' Kkr djsA
3 5
UPSI 14/12/2017 (Shift-01)
(b) 1
8
3 (a) (b) 3.5
(c) 3
(u + v)(v + w)(w + u)
16
(c) 4.5 (d)
(d) 0 3
140. Simplify the following.
A M B N
144. If + = 1 and + = 1, then the value
fuEufyf[kr dks ljy dhft,A L B M C
3a + b a – 3b L C
– + 2b of + is :
2 3 A N
SSC CGL 19/07/2023 (Shift-01)
A M B N L C
5(a  3b ) 7a  3b
;fn  =1 vkSj  =1 gS] rks 
L B M C A N
(a) (b)
6 6 dk eku D;k gksxk\
7(a  3b ) a  3b SSC CPO 03/10/2023 (Shift-02)
(c) (d)
6 6
B
(a) 1 (b)
Level -03 M
141. Simplify the expression:

r
M
(c) (d) 0
1 1 1 2 4  B

si
– – 2 – 4
8  b – 1 b  1 b  1 b  1 

1 1 1
  = 1, then find
an by
fuEufyf[kr O;atd dks ljy dhft,A
145. If p + q + r = pqr =

p3 + q3 + r3.
p q r

n
SSC CHSL, 14/08/2023 (Shift-3)

1 8 1 1 1
;fn p + q + r = pqr = p  q  r = 1 gS] rks
(a) 8
b –1 ja (b) 8
b 1
p3 + q3
R s
+ r3 dk eku Kkr dhft,A
a th

8 1
(c) 8 (d) 8 SSC CGL 25/07/2023 (Shift-02)
b –1 b 1
1
(a) 1 (b) –1
142. If x   
6 1 3 , then the value of (c) 5 (d) –5
ty a

3
 1  1 1 1 1 13
 x    3  x   is: + + =
146. If 25 (3a – 2b) = 5(b – a) = 52 and ,
di M

x x a b c 35
3 then find the value of ab – c
1  1  1
;fn x   6  1 gS] 3 rks x  x   3  x  x  1 1 1 13
;fn 25 (3a – 2b) = 5(b – a) = 52 vkSj + + =
dk eku gS% a b c 35

SSC CHSL 09/08/2023 Shift-01

gS] rks
ab – c dk eku Kkr djsa
(a) 1 (b) 0
2 66 4 66 (c) –1 (d) –2
(a) (b)
5 5 (4b – 3a) (2b + c) (c – a)
147. If 4 =2 =8 and a + b + c = 11,
4 66 4 36  1 1 1 
(c) (d)
A

3 5 then find the value of 4  + + 

ab bc ca 
143. If abc=1, then find the value of
;fn 4(4b – 3a) = 2(2b + c) = 8(c – a) vkSja + b + c = 11
123 123 123
+ + ?  1 1 1 
1 + a + ab 1 + b + bc 1 + c + ca gS] rks
4  + +  dk eku Kkr djsa
ab bc ca 
123 123 123
;fn abc = 1 gS] rks1 + a + ab + 1 + b + bc + 1 + c + ca
11 9
(a) (b)
dk eku Kkr dhft;s\ 9 11

(a) 3 (b) 41 6 5
(c) (d)
(c) 123 (d) 369 5 6
 a4 + 2 1 1 1
148. If a + a2 + a3 + a4 – 2 = 0, Find ? 150. If a – = b, b – = c, c – = a, find
a a b c

a4 + 2  1 1 1 
;fn a + a2 + a3 + a4 – 2 = 0 rc dk  + + .
a ab bc ca 
eku Kkr dhft,A
1 1 1
(a) 1 (b) 2 ;fn a – = b, b – = c, c – = a rc
a b c
(c) 3 (d) 4
149. 25x4 – 9x2y2 + 49y4 = 114, 5x2 + 3xy + 7y2  1 1 1 
 + +
ab bc ca 
 dk eku Kkr dhft,A
x y
= 19 find 5 +7
y x –1  1 1 1
(a)    
;fn 25x – 9x y + 49y = 114, 5x + 3xy
4 2 2 4 2 2  a 2 b2 c 2 

r
x y 1 1 1 1
+ 7y2 = 19 rc 5 y + 7 dk eku Kkr dhft,A (b)  2  2  2 
2 a b c

si
x
1 1 1 1
25 75  2  2  2 
(a)
13 an by (b)
13
(c)
3 a b c

n
–1  1 1 1
150 175 (d)  2  2  2 
(c) (d) 3 a b c
13 13
ja
R s
Answer Key
a th

1.(c) 2.(c) 3.(c) 4.(c) 5.(d) 6.(c) 7.(b) 8.(c) 9.(b) 10.(d)
ty a

11.(c) 12.(b) 13.(b) 14.(d) 15.(c) 16.(a) 17.(a) 18.(b) 19.(a) 20.(d)

21.(b) 22.(b) 23.(c) 24.(d) 25.(a) 26.(b) 27.(b) 28.(b) 29.(a) 30.(c)
di M

31.(b) 32.(a) 33.(d) 34.(b) 35.(c) 36.(d) 37.(c) 38.(a) 39.(c) 40.(d)

41.(d) 42.(c) 43.(a) 44.(b) 45.(d) 46.(a) 47.(c) 48.(d) 49.(a) 50.(d)

51.(b) 52.(a) 53.(a) 54.(b) 55.(a) 56.(a) 57.(c) 58.(a) 59.(a) 60.(c)

61.(c) 62.(c) 63.(a) 64.(d) 65.(b) 66.(a) 67.(b) 68.(c) 69.(a) 70.(b)

71.(c) 72.(a) 73.(b) 74.(c) 75.(c) 76.(d) 77.(c) 78.(d) 79.(c) 80.(a)

81.(a) 82.(b) 83.(a) 84.(b) 85.(a) 86.(a) 87.(b) 88.(a) 89.(c) 90.(b)
A

91.(b) 92.(d) 93.(a) 94.(c) 95.(a) 96.(c) 97.(b) 98.(a) 99.(c) 100.(a)

101.(b) 102.(c) 103.(d) 104.(c) 105.(a) 106.(d) 107.(d) 108.(a) 109.(c) 110.(d)

111.(d) 112.(a) 113.(b) 114.(b) 115.(c) 116.(d) 117.(c) 118.(b) 119.(d) 120.(d)

121.(a) 122.(b) 123.(b) 124.(a) 125.(d) 126.(b) 127.(a) 128.(b) 129.(a) 130.(b)

131.(b) 132.(b) 133.(a) 134.(a) 135.(d) 136.(a) 137.(b) 138.(c) 139.(c) 140.(c)

141.(a) 142.(b) 143.(c) 144.(a) 145.(a) 146.(b) 147.(a) 148.(c) 149.(b) 150.(a)
 SOLUTIONS
SOLUTION
1. (c) 6. (c)
p + q = 12, pq = 14,
1
x + =7 p2 – pq + q2 = ?
x
p2 + q2 – pq = (p + q)2 – 3pq
1 = 144 – 42 = 102
x2  = 72 – 2 = 47
x2
7. (b)
2. (c) x2 + x = 19
1 1
y 8 (x + 5)2 +
y (x + 5)2

r
Put x + 5 = y

si
1
y2   82  2 = 62
y2 x = (y – 5)

3. (c) an by (y – 5)2 + (y – 5) = 19
y2 + 25 – 10y + y – 5 = 19

n
1  y2 – 9y + 1 = 0
z+ =4
z
ja
R s
1
y 9
1 y
z 2 + 2 = 14
a th

z
1
Now, y2 
y 2 = 81 – 2 = 79
1 2 1  1
ty a

 z + 2  = ×14 = 7
2 z 2 (x  5)2 
1
(x  5)2 = 79
di M

4. (c)
Given, 8. (c)
x + y = 7, xy = 19 We know that,
According to question, (2a + 3b)2 – (2a – 3b)2 = 24ab
 (x + y)2 = x2 + y2 + 2xy (14)2 – (10)2 = 24 ab
 49 = x2 + y2 + 2 × 19 96 = 24 ab
 x2 + y2 = (49 – 38) = 11 ab = 4
5. (d) 9. (b)
(5p + 3q) (5p –3q)
3 1
A

3y –  5 , y 2  2  ? = (25p2 – 9q2)
y y
10. (d)
2 2
 1  5 1
  y       x  3 
 y 3 x

1 25 1
2
 y  – 2  x – 3
y2 9 x

1 25 43 1
2
 y   2 =  x   32  4 = 13
y 2
9 9 x
11. (c) 16. (a)
1 1
a– =4 y2  = 167, y > 0,
a y2

1  1
2 1
y ?
a+ = a –   4 y
a a
2
= 16  4  20 = 2 5  1 2 1
 y  y   y  y 2  2
12. (b)
= 167 + 2 = 169
1
a=
a 6  1
 y  y   13
1
a – 6
a  y>0

r
1 1
 a  6 y  13

si
a y
17. (a)
 a
1
a
an by  6
2
4= 10
3y 
3
 8, y2 
1
y2
?

n
13. (b) y

1 1 8
x=
ja y 
y 3
R s
x–5
1 2 2
a th

 x – 5  1  8
x  y  y    3 
1
 x– 5 1 64
x y2   2
ty a

y2 9
1
 x+ = 52 + 4 29 46 1
di M

=
x = 5
9 9
14. (d) 18. (b)
Given,
x 2 – 8x – 1 = 0 2 1
x2 – 3.2x + 1 = 0, x > 1, x – =?
x2
1
 x – =8 1 16
x x+ = 3.2 =
x 5
2 1
 x  = 82 + 2
x2 1  1
2

x  x    4
= 66 x  x
15. (c)
A

256 156 2 39
1 = 4 = =
a 6 25 25 5
a
1  1 1
1 2
 x – x   x  
2 =
2
 a   62  2 = 34 x  x   x
a2

3 2 1  2 16 32 39
= 39  =
  a  2  5 5 25
4 a

3 51 128 39
=  34   25.5 = = 12.8 0.39
4 2 100
19. (a) 23. (c)
2
x – 7x + 1 = 0, 0 < x < 1,
1
x4 + =194
1 x4
x 7
x
1
 0<x<1 x2 + =14
x2
1
(x –
x
)= 72 – 4 = 45 = 3 5 1
x+ =4
x
2 1  1 1
 x – 2 = 
x   x   1
x  x   x x3 + =52
x3
 7  3 5 = 21 5

r
1 1
20. (d) x3 + 3
+x+
x x

si
Given,
= 52 + 4 = 56
 2 1 
an by
 x  2   7
x
24. (d)

n
1
a 3
1 a
 x  3
x
ja
R s
1
a2  7
a2
a th

1
 x    5 [0< x <1]
x
1 2
  7 – 2  47
4
a  4
1  1  1 a
ty a

2
 x  2 = x   x  
x  x  x 
25. (a)
di M

= 3  ( 5) = (3 5) 1
x4+ =254
21. (b) x4

1 1
x 5 5  x²   254  2
x x²

1 1
x – =11 x2+ =16
x x2

1 1
 x3 – x + = 18
x3 x
A

= (11)3 + 3 × 11 = 1364
1  2 1  3 1 
22. (b)  x5+ = x + 2  x + 3 
x5  x  x 
1
x = 10
x = 16 ×15 18 – 18

2 1 = 240 18 – 18
 x + = 102 – 2 = 98
x2
 239 18
14 2
 x + 4 = 98 – 2 = 9602 = 717 2
x
26. (b) 31. (b)

1 a + b = 10, ab = 6
x+ =7
x a3 + b3 = (a + b) (a2 + b2 – ab)
a2 + b2 = (a + b)2 – 2ab
1
 x + 3 = 73 – 3 × 7
3
= 100 – 12 = 88
x
 a3 + b3 = 10(88 – 6) = 820
= 343 – 21 = 322
32. (a)
1 2
x + 6 = 322 – 2
6
x y
x + =1, x + y = 2
y x
= 103684 – 2 = 103682
x2 + y2 = xy
27. (b)
Given,  x3 + y3 = (x + y) (x2 + y2 – xy)
= 2(0) = 0

r
1
x + =1 33. (d)
x

si
According to question, A.T.Q.

x 2 +7x +1
= 2
an by x3 – y3 = 270
x–y=6

n
x +11x +1
x3 – y3 = (x –y) [(x –y)2 + 3xy]
Divided by x in nomi. and denomi.

ja 270 = 6[36 + 3xy)

R s
1
x + +7 45 = 36 + 3xy
x 8 2
a th

= =  3xy = 9
1 12 3
x + +11
x xy = 3
28. (b)  (x + y)2 = (x – y)2 + 4xy
ty a

P = 7 + 4 3 , Q = 7 –4 3 = 36 + 12
di M

Then, (x + y) = 48
The value of
(x + y) = 4 3
P +Q2
2  49+48
2
= = 194 34. (b)
P 2 × Q2 1
a – b = 5, ab = 24, a3 – b3 = ?
29. (a)
By value putting
x=2+ 3
a=8
y=2– 3 b=3
x + y = 4 a3 – b3 = (83 – 33)
xy = 4 – 3 = 1
A

= (512 – 27) = 485

We know
35. (c)
x2 + y2 = (x + y)2 – 2xy
Given,
x2 + y2 = 42 – 2 × 1
= 16 – 2 = 14 x + y = 25, xy = 20,
30. (c)  x3 + y3 = (x + y)[(x + y)2 – 3xy]

x=1+ 2 x = 3 + 2 2
2 According to question,
x3 + y3 = 25[252 – 3 × 20]
y = 1 – 2 y2 = 3 – 2 2
= 25[625 – 60]

x + y = 32 2 + 3–2 2 = 6
2 2
   = 25 × 565 = 14125
36. (d) 42. (c)
According to question, Let, z = 0
(a3 – b3 – 3a2b + 3ab2) x + y = 22, xy = 35
3
= (a – b) Then,
3
= (17 – 13) = 64 (x – y)2 = (484 – 140) = 344
37. (c) &
Let, x2 + y2 = 484 – 70 = 414
c=0 The value of (x – y)2 + x2 + y2 = (344 + 414)
a+b=1 = 758
a3 + b3 = 4 43. (a)
Then, Let,
4 = 1 (1 – 3ab) c=0

r
ab = – 1 a + b = 20, a2 + b2 = 152
The value of Then,

si
1 1 1 a + b +1 400 –152
+ +

38. (a)
a b ab
Or
an by
ab
= –2

44.
ab =

(b)
2
=124

n
(a + b + c)² = a² + b² + c² + 2 (ab + bc + ca) Let, c = 0

ja
361 = 155 + 2 (ab + bc + ca) a + b = 6, ab = 1
R s
 206 = 2 (ab + bc + ca) Then,
a th

 103 = (ab + bc + ca) The value of

 (a – b)² + (b – c)² + (c – a)² = 2 (155 – 103) ab (a + b) = 1× 6 = 6
= 2 × 52 = 104 45. (d)
ty a

39. (c) Let, c = 0

Given, a + b = 17
di M

a = 101, b = 102, c = 103, a2 + b2 = 117

Then,
Diff. = 102 – 101 = 1
2ab = 289 – 117
then, a2 + b2 + c2 – ab – bc – ca = 3d2
2ab = 172
=3×1=3
(a – b)2 = 289 – 344 = –55
40. (d) The value of
a + b + c = 20, (a – b)2 + a2 + b2 = – 55 + 117 = 62
a2 + b2 + c2 = 262, 46. (a)
ab + bc + ca = ? Let,
400 – 262 = 2(ab + bc + ca) c=0
A

ab + bc + ca = 69 Given that,
a2 + b2 = 162, ab = 119
41. (d)
We know,
Let,
(a + b)2 = a2 + b2 +2ab
c=0
= 162 + 238
a + b = 17 a + b = 20
a2 + b2 = 115 Then, the value of
then, a2 (b + c) + b2 (c + a) + c2 (a + b) – 3abc
(a+b)2 + a2 + b2 Or
= (289 + 115) = 404 ab (a+b) = 119 (20) = 2380
47. (c) 54. (b)
Here, (a + b + c)  0
1
(a + b + c) (a2+b2+c2 – ab – bc – ca) x 2
x
= a3 + b3 + c3 – 3abc
48. (d) 1
 x + 2
Given that, x
3 –4 –1  
a= ,b= ,c= (1) (1)
2 3 6
ATQ,
x=1
a+b+c=0
Then, 1
x7 
3 3 3
a + b + c – 3abc = 0 x 117

r
a3 +b3 +c3 –3abc 1
Then, the value of 3  (1)7   11  2

si
a2 –b2 +c2 (1)117
=0+3=3
49. (a) an by
(a + b + c)2 = (5)2
55. (a)
Let,

n
a2 + b2 + c2 + 2(ab + bc + ca) = 25 x = 20, y = 21
Then,
ja
15 + 2(ab + bc + ca) = 25
R s
ab + bc + ca = 5 The value of
a th

Now, (x – 20)2021 + (y – 21)2021

a3 + b3 + c3 – 3abc – 27 = (20 – 20)2021 + (21 – 21)2021 = 0
= 5 (15 – 5) – 27 = 23 56. (a)
ty a

50. (d) a² + b² + c² = ab + bc + ac
1 Let, a = b = c = 1
di M

x + y + z  x – y 2 +  y – z 2 + z – x 2 
= 2
 x – y  2 +  y – z  2 +  z – x 2 11a 4 +13b4 +17c 4
=
1 17a²b² + 9b²c² + 15c²a²
= x + y + z 
2
11 +13 +17 41
51. (b) = = =1
17 + 9 +15 41
x2 + y2 – 10x + 12y + 61 = 0
57. (c)
Or
Let,
(x – 5)2 + (y + 6)2 = 0
x=y=z=1
Then,
x = 5, y = –6 Then,
A

2x + 3y = (10 – 18) = – 8 3x 4 +7y 4 +5z 4 3 + 7 + 5 15

= = =1
52. (a) 5x y 2 +7y 2 z 2 +3z 2 x 2 5 + 7 + 3 15
2

(a + 2)2 = 0, (b – 3)2 = 0 , (c – 9)2 = 0

58. (a)
a = –2, b = 3, c = 9
If x + y + z = 0
a + b + c = (– 2 + 3 + 9) = 10
let x = 1, y = 1, z = –2
53. (a)
Then the value of
(a – 1)2 = 0, (b + 2)2 = 0, (c +1)2 = 0
a = 1, b = –2, c = –1 3y 2 + x 2 + z 2 3 +1 + 4
= =2
2a – 3b + 7c = (2 + 6 – 7 ) = 1 2y 2 – xz 2+2
59. (a) 64. (d)
Let, Given that
x = 1, y = 1, z = –2 a+b=3
Then, Let,
a&b=1&2
3y 2 + x 2 + z 2
The value of Then the value of a2b4 = 1 × 16 = 16
2y 2 – xz
65. (b)
3 +1 + 4 Given that,
= =2
2+2 2x + y = 10
60. (c) Let,
x = 2, y = 6
2x
If y  Then,

r
1 x
The value of
Find,

si
x2y3 = 4 × 216 = 864
1 2y  1 66. (a)
 2
an by
y  1 (y  1) Let, p = q = r = 1

n
Put x=0 1 1 1 3
=    1
Then, 3 3 3 3
ja
R s
2 67. (b)
y 2
1
a th

1
By option (c) Let x = y = 1 & a = b =
2
(1  0)(2  0) Value of x = 1
ty a

= 2 (Verify)
(1  0) by option
Option (ii)
di M

61. (c)
(3x + 2y)3 +xy (4x – 5y) 1 1
a 2
1 + (5 × –7) (55) = = 2 =1
a 2 + b2 1 1 1
+
1 + (–35 × 55) 4 4 2
1 – 1925 68. (c)
= – 1924 1
62. (c) If x + =1
x
x + 3y = 6, Then, x3 + 1 = 0
3 3
x + 27y + 54xy = ? The value of
A

Let, x = 3, y = 1 x53 + x50 = x50 (x3 + 1) = 0

3 3
x + 27y + 54xy 69. (a)
= 27 + 27 + 54 × 3 = 216
1
63. (a) If x + = 1, then x3 + 1 = 0
x
2x + 3y = 15 Then the value of
Let, x28 + x25 + x21 + x18 + x12 + x9 + x6 + x3
The value of x & y = 3, 3 Or
Then, x25(x3 +1) + x18 (x3+1) + x9 (x3 +1) + x3 (x3 +1)
The value of x2y3 = 9 × 27 = 243 =0
70. (b) 75. (c)
We know,
4
5a   2  13 , a > 0,
1 a
If x  = 1 then x3 + 1 = 0
x
16
ATQ, 25a 2  ?
a2
= x100 (x3 + 1) + x50 (x3 + 1) + x20 (x3 +1) +
x15 (x3 + 1) + x9 (x3 + 1) + x3 (x3+ 1) + 1 4
5a   15
= x100 (0) + x50 (0) + x20 (0) + x15 (0) + x9 (0) + x3 a
(0) + 1 = 1
2
71. (c)  4
 5a    15
a
X – 5Y 7
=
X + 5Y 13 16 4
2

25a2   2  5a  = 225

r
2
a a
X + 5Y 13
=

si
X – 5Y 7 16
25a 2  = 225 – 40 = 185
using comp. and divid. a2

X 20
an by 76. (d)

n
=
5Y 6 2 5

x 5 –  
 x x
X 50
ja
R s
=
Y 3
5
a th

72. (a) 5x – 2 =

x
Given that,
4x2 – 2x + 5 5
5x – =2
x
ty a

For minimum value of the expression

4ac – b2 4×4×5 – 4 76 19 1 2
di M

= = = x– =
4a 4×4 16 4 x 5
73. (b) 2
1  1
2 2  x2   x –   2
58 – 25 83  33 11 x2  x
A=  
462 – 372 83  9 3
4 54
26 2 – 152 41  11 11 = 2
B=   25 25
56 2 – 152 71  41 71
77. (c)
1 20 71 20  3
 –  – 1 1
B A 11 11 7b – 7
4b
74. (c)
A

ATQ, 4
Multiply by both sides
x4 + y4 – 2x2y2 7
(x2)2 + (y2)2 – 2x2y2
1
(x2 – y2)2  4b  4
7b
2 2 2
 1  1 
 a   –  a –   2 1 1
 16b   42  2  4 
  a a  
49b2 7
2
 1 8 120
 4  a   = 1  16  
a 7 7
78. (d) 81. (a)

1 1
x² – 4 6 a 7
x² a

 1  1 1
 x    x –   2 3  2 2  a² + = 47
x x a²
  3
1  1  1
12 2 2 8  a³ + =  a+  – 3  a+ 
a³  a  a
3
1  1  1 = (343 – 21) = 322
 x³ –  x –   3 x – 
x³  x  x 1  2 1  3 1   1
5
 a  =  a  2   a  3    a  
a5 a a a
 ( 8)³  3  8  8 8  3 8

r
= (47 × 322) – 7 = 15127
 11 8  22 2 82. (b)

si
79. (c)
1
x –6

1
an by
x2 – 9.76x  1  0 , x > 1, x

n
x3- =? 1  2 1  3 1   1
x5+ = x + 2  x + 3  – x + 
x3 x5  x  x   x
x2 + 1 =
ja9.76x = 34 × (–198) + 6
R s
= – 6726
1
a th

x+ = 9.76 83. (a)

x

2 x 2  3 7x  13  0
1  1
x– =  x +  -4
ty a

x  x Put, x  7  y
di M

= 9.76  4 x  (y  7)
= 5.76 = 2.4 (y  7)2  3 7(y  7)  13  0
3
1  1  1
x3  3
 x    3 x    y2  7  2 7y  3 7y  21  13  0
x  x   x 
 y2 – 7y –1= 0
= 2.4 3  3  2.4
= 13.824 + 7.2 = 21.024 1
y– = 7
80. (a) y
7
 7a – =–4 1
A

a y3  = 7 7  3 7 = 10 7
y3
1 4
 a – =–
a 7 1
y+ = ( 7)2  4 = 11
1 –64 12 – 652 y
3
 a – = – =
a 3 343 7 343
1
y3 + = 11 11  3 11 = 8 11
3 1 y3
 a – 3 –1
a
6 1
– 652 – 995 y6 
1 x – 7  – 6
= 80 77

343
–1 =
343 y6
Or
x – 7 
84. (b) 86. (a)

1 1
x+ =5 2 If x > 0, x4 + = 142,
x x4

7 1  4 1  3 1   1
3 1  x + = x + 4  x + 3  – x + 
 x + = 250 2 –15 2 x7  x  x   x
x3
4 1
= 235 2  x + =142
x4
1 1
x– = 50–4 = 46 2
 x + =12
x x2

1 1
x3 – = 49 46  x + = 12+2
x3 x

r
1
x+ = 14

si
1  1  1 
x – 6 = x3+ 3  x3 – 3 
6
x
x  x   x 

an by
= 49 × 2 × 23 × 235 2
3
x +
1
x3
=14 14 –3 14=11 14

n
= 23030 23 7 1  4 1  3 1   1
x + = x + 4  x + 3  – x + 

x 7
x  x   x
85. (a)
ja
R s
= 142  11 14 – 14
a th

 1
 x    2 2 , x > 1, = 1562 14 – 14
x
= 1561 14
1  3 1  3 1 
ty a

6
 x – 6 =  x – 3   x + 3  87. (b)
x x x
 1
di M

 x –  = 6
1 3 x
3
 x +
x3
= 2 2  – 3× 2 2
8 1  4 1  4 1 
 x – = x + 4  x – 4 
= 10 2 x8  x  x 

1  4 1  2 1  2 1 
x+ =2 2 =  x + 4   x + 2   x – 2 
x x x x

 4 1  2 1  1  1
1 1
2 =  x + 4   x + 2   x +   x – 
 x x x x
x– = x +  – 4
x  x
1
A

x+ = 6 + 4 = 10
x
2
= 2 2  –4 =2 1
x2+ =8
x2
1
x3 – = 23 + 3 × 2 = 14 1
x3 x4+ = 62
x4
1  3 1  3 1  1
x6 – = x – 3  x + 3  8
 x –
x6  x  x   62  8  10  6
x8

= 14 × 10 2 = 140 2 = 99 2 15
88. (a) 91. (b)
1 1
x+ =5 x+ =3
x x
Then, the value of
 1   1  1   1  1
  x 8 – 8  =  x 4 + 4   x 2 + 2   x+   x –  1 1
 x   x  x   x x x4+ x3+ 3
x 2 or x = 27 – 9 =18
1 x 2 – 2x +1 1 1
x– = 21 x+ –2
x x
92. (d)
1
x2 +  23
x2 a = 5 +2 6 , b = 5 –2 6

1 Then, the value of

x4 + = 527
x2 2 2
5+2 6  + 5 – 2 6  +1 99

r
2 2
a +b +ab
= 2 =2
8 1 a +b –ab  5+2 6  +  5 – 2 6  –1 97
2 2

si
 x + = 527 × 23 × 5 × 21
x8
93. (a)
= 60605 21
89. (c)
an by y=3+2 2

n
1 1 1 3–2 2
x+ = 6  = =
x y 3+2 2 1
ja
R s
2 1 1
 x + =4  y– 4 2
a th

x2 y

4 1 1
 x + =14 y+ =6
x4 y
ty a

1  1  1
 x– = 2 y2 – y–2 =  y –   y  
 y  y
di M

8 1  4 1   4 1  = 24 2
 x + =  x + 4  ×  x – 4 
x8  x x
Now, 2  y 2 – y –2   48
 4 1   2 1  1  1 94. (c)
=  x + 4  ×  x + 2   x +   x – 
x x x x
A.T.Q.
= 14 × 4 × 6 × 2
a b 8
 
= 112 3 b a 3
90. (b) a b 64
  = 2
A

1 b a 9
x+ =5
x
a 2  b2 82
Then, the value of =
ab 9
x 4 +3x 3 +5x 2 +3x +1
We know
x 4 +1
(a + b)2 = a2 + b2 + 2ab
On dividing numerator and denominator by x2
1 1
900 = 100x
2 
x + 2 + 3 x +  + 5
x  x 23 + 3 × 5 + 5 43 x=9
= =
2 1 23 23 then,
x + 2
x
ab = 9x = 81
95. (a) 36y2 – 72xy – 42x2 + 42y2)]
a = 5 +2 6 , b = 5 – 2 6 [(13x + y)] [(43x2 + 127y2 + 2 × 13xy)]
A = 43, B = 127, C = – 13
Then,
then,
The value of
A + B + C = (170 – 13) = 157
a3 + b3 = (5 + 2 6 )3 + (5 – 2 6 )3 99. (c)
= 970 x 3 –y 3 xy
96. (c) A= ,B= 2 , and C
x 2 –y 2 x  y 2  xy
2 1 2 –1
p= ,q  3xy 2 – 3 x 2 y
2 –1 2 1 =
x 2 – y2

2 1 2 1 2 then,
p=
2 –1 2 1
  2 1   32 2 AB (A + C)

r
= 32 2
 x 3 – y 3 3xy  y – x  

si
x 3 – y3 xy
 2 
1 = 2 ×
x – y 2 x 2  y 2  xy  x – y
2
 x 2 – y2  
 p 
q
an by  x – y   x 2  y2  xy 
 2
xy

n
 q  3 – 2 2 =
x  y x – y x  y 2  xy

p2 q2 p³ + q³
 + =
ja  x 3 – y 3 – 3 xy  x – y  
R s
q p pq × x 2 – y2 

a th

(p + q)(p² + q² – pq) 6 × 33
=
pq
=
1
 x – y 2
=
x  y
= 198
ty a

97. (b) 100. (a)

ATQ,
1
di M

x= 3 2, = P×Q÷R=
x 3– 2
2

then, 2x 2  x – 6

x2  3x  9

 x 2  – 4  2

=
2( x  2)( x – 2) (2 x – 3)( x 2  4) x 3 – 27
1 1
x3 + +x+
x3 x
3 3
 2x – 3  x  2

x 2  3x  9

 x 2  4   x  2  x – 2
=  3 2   3– 2   3 2 3– 2 2( x  2)( x – 2) 2
(2 x – 3)( x  4)  x – 3  x 2  3 x  9


= 2 3 33 3 2 2 3  
x2
2( x – 3)
= 20 3
101. (b)
A

98. (a)
2
[7 (x + y)]3 + [6(x – y)]3 x 4 – 81 2x  3 [(x – 3)  3x ](x  5)
 3 ×
Let, P = 7 (x + y)
2
2x  13x  15 x  27 (x  3)2 – 6 x
Q = 6 (x – y)
We know, (x 2 – 9)(x 2  9) 2x  3
2
[ x  9 – 6x  3x ](x  5)
= (x  5)(2x  3) 
(x  3)(x 2  9 – 3x )
× (x 2  9  6 x – 6 x )
P3 + Q3 = (P + Q) (P2 + Q2 – PQ)
then,
 [7 (x + y) + 6(x – y)] [7(x + y)]2 + (x – 3)(x  3)(x 2  9) 1 ( x 2  9 – 3x )
=  2
(x 2  9)
[6 ( x – y ) ] 2 – 4 2 ( x 2 – y 2 ) 1 (x  3)(x  9 – 3x )

[(13x + y)] [(49x2 + 49y2 + 98xy + 36x2 + = (x – 3)

102. (c) 106. (d)
1 x = (b – c) (a – d) = ab – bd – ca + cd
a2 + b2 + c2 – ab – bc – ca = (a – b)2 + (b – c)2 +
2 y = (c – a) (b – d) = cb – cd – ab + ad
(c – a)2 z = (a – b) (c – d) = ac – ad – bc + bd
1 x+y+z=0
= (97.5 – 100)2 + (100 – 102.5)2 + (102.5 –97.5)2
2
Then,
1
= (– 2.5)2 + (– 2.5)2 + (5)2 x3 + y3 + z3 = 3xyz
2
Required,
1
= (6.25 + 6.25 + 25)
2 (x  y)(y  z)(z  x )(x 3  y 3  z 3 )
75 9x 2 y 2 z 2
=
4
 z  ( x )  ( y)  3xyz 1

r
103.(d)
= 9 x 2 y 2z 2 =
We know that, 3

si
(a + b – c)² = a² + b² + c² + 2ab – 2bc – 2ca
107. (d)
Given that
an by
(Ax + By – Cz)2 = 4x2 + 3y2 + 2z2 + 4 3 xy –
a3 + b3 – c3 + 3abc

 a + b – c

n
2 6 yz – 4 2 xz = [3(a2 + b2 + c2) – (a + b – c)2]
2
Then,

ja
R s
(Ax + By – Cz)2 = (2x + 3 y – 2 z)
2
20
=
2
3  152  400
A = 2, B = 3 , C = 2
a th

the value of A2 + B2 × C2 = 10 × (456 – 400)

= 4 + 3 × 2 = 10 = 560
104. (c) 108. (a)
ty a

Given,
We know that
a + b + c = 7,
di M

ab + bc + ca = 11, (a3 + b3 + c3 – 3abc)

abc = – 1,
 a + b + c  2 2 2
according to question, = 
a – b +  b – c + c – a  
2
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
49 = a2 + b2 + c2 + 2 ×11 a + b + c
a2 + b2 + c2 = 27  405 =
2
54
a3 + b3 + c3 – 3abc = (a + b + c)
 [a2 + b2 + c2 –(ab + bc + ca)] 405
 (a + b + c) = = 15
a3 + b3 + c3 – 3 × –1 = 7(27 – 11) 27
a3 + b3 + c3 = 112 – 3 = 109
109. (c)
105. (a)
A

a + b + c = 12, a = 335, b = 215, c = 180

a2 + b2 + c2 = 50, a3 + b3 + c3 – 3abc
(a + b + c)2
a  b  b   2 2 2

= a2 + b2 + c2 + 2 (ab + bc + ca) =
2  a  b   b  c   c  a  
144 = 50 + 2(ab + bc + ca)
ab + bc + ca = 47 730  2 2 2
=   120  35  155 
a3 + b3 + c3 – 3abc 2

= (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)] = 365[14400 + 1225 + 24025]

= 12[50 – 47] = 365 × [39650]
= 12 × 3 = 36 = 14472250
110. (d) 114. (b)
x = 999, y = 1000, z = 1001 8x2 + y2 – 12x – 4xy + 9 = 0
We know, 4x2 + y2 – 4xy + 4x2 – 12x + 9 = 0

1 (2x – y)2 + (2x – 3)2 = 0

x3 + y3 + z3 – 3xyz = [(x + y + z)(x – y)2 + (y – z)2
2 2x = y, 2x = 3
+ (z – x)2]
3
Then, the value of x= ,y=3
2
x 3 + y3 + z 3 – 3xyz Then,
=
x +z – y 14x – 5y = (21 – 15) = 6
115. (c)
1 2 2
 x + y + z   x – y   y – z  + z – x 2  a2 + b2 + c2 + 96 = 2 (4a + 4b – 8c)
2
= x +z – y

r
a2 + b2 + c2 – 2 (4a + 4b – 8c) + 96
(a – 4)2 + (b – 4)2 + (c + 8)2 = 0

si
3000  3  1
= 9 a = 4, b = 4, c = –8
999  1001 – 1000
111. (d) an by ab – bc + ca = 16 + 32 – 32 = 4

n
116. (d)
1
a + b + c  a – b2 +  b – c2 + c – a 2  a2 + b2 + 64c2 + 16c + 3 – 2 (a + b) = 0
2
=
ja (a – 1)2 + (b – 1)2 + (8c + 1)2 = 0
R s
 a – b  2 +  b – c 2 +  c – a  2
–1
a th

25 +15 – 10 a = 1, b = 1, c =
= = 15 8
2
Then,
112. (a)
ty a

1 1
a+b+c=5 4a7 + b7 + 8c2 = 4 + 1 + =5
8 8
di M

ab + bc + ca = 7
117. (c)
a2 + b2 + c2 = (a + b + c)2 – 2(ab + bc + ca)
If, ab + bc + ca = abc
= 25 – 14 = 11
(Symmetry case)
a3 + b3 + c3 – 3abc = (a + b + c)
Put, a = b = c = k (Let)
[a2 + b2 + c2 – ab – bc – ca]
K3 = 3K2
= 5[11 – 7] = 20
113. (b) k3
a3 + b3 + c3 – 3abc = 405,
(b + c)
a + b + c = 15 Now, Required, ×3
bc(a – 1)
a3 + b3 + c3 – 3abc
A

1 6
= 3= 1
= (a + b + c) × [(a – b)2 + (b – c)2 + (c – a)2] 92
2
118. (b)
15 2 2 2
405 = ×   a - b +  b - c  +  c - a   Let,
2 
x1 = x2 = x3 = 4
810 2 2 2 then,
 = a - b +  b - c + c - a 
15
1 1 1 1 1 1 3 1
      
2 2 2
  a - b +  b - c  +  c - a  = 54 2  x1 2  x 2 2  x 3 6 6 6 6 2
119. (d) 125. (d)
(3x³ + 5x2y + 12xy2 + 7y3)
x = – 4 1 7
a+ =
y = –1 b 3
= 3(–4)3 + 5(–4)2 (–1) + 12(–4) (–1)2 + 7(–1)3
= – 192 – 80 – 48 – 7 = – 327 1 13
120. (d) b+ =
c 4
64x3 + 38x2y + 20xy2 + y3
 x = 3, y = – 4 1 9
64(3)3 + 38(3)2 (–4) + 20(3) (–4)2 + (–4)3 c+ =
a 2
= 1728 – 1368 + 960 – 64
= 1256 a = 2, b = 3, c = 4
121. (a) abc = 24
27x³ – 58x²y + 31xy² – 8y³
126. (b)
 x = – 5, y = – 7
= 27 (– 5)³ – 58 (– 5)² (– 7) + 31 (– 5) (– 7)² – 8 x y z
(– 7)³ = = =k
y+z x +z x +y

r
= – 3375 + 10,150 – 7595 + 2744 = 1924
122. (b) Therefore

si
(p + q + r) = 0
x = k(y + z) ___(1)
Put P = 1, q = 1, r = –2
y = k(x + z) ____(2)
  2 an by
 p2   q 2   r 2 
+ +
 p - qr   q2 - pr   r 2 - pq 
z = k (x + y) ____(3)
therefore

n
1 1 4 x + y + z = k[y + z + x + z + x + y]
  
3 3 3 x + y + z = 2k[x + y + z]
ja
R s
6 1
= =2 k =
3
a th

2
123. (b)
Given that, From eq.(1)
x = ky + kz
1 1 1
+ + kz = x – ky ___ (4)
ty a

1 1 1
1+ b + 1+ a + 1+ c + From eq.(2)
c b a
y = kx + kz
di M

Let, a = b = c = 1
1 1 1 kz = y – kx ___ (5)
   1 From eq.(4) and (5)
3 3 3
124. (a) x – ky = y – kx
1 x – y = ky – kx
a– =m x – y = – k(x – y)
b
k = – 1
1
b– =n 127. (a)
c
1 1 1 1
c– =p (a+b+c)  + +  = 30 × 20
a a b c
Put,
A

a = 2, b = 1, c = –1 1 1 1 1 1 1 1 1 1
m=1 a + +  +b + +  +c + + 
a b c a b c a b c
n=2
= 600
1 3
p = 1  = a a b b c c
2 2 1+ + + +1 + + + +1 = 600
b c a c a b
1 1
abc – = 2×1×(–1) –
abc 2  1  (1) a a b b c c
3+ + + + + + = 600
1 3 b c a c a b
= 2  =
2 2 a b a c b c
+ + + + + = 597
Option (A) is correct b a c a c b
128. (b) 132.(b)
2
l -m 2 [(x – 5) (x – 1)] – [(9x – 5) (9x – 1)] ÷ 16x
=? If l + m  0 [(x2 – 6x + 5) – (81x2 – 54x + 5)] ÷ 16x
l + m2
(– 5x2 + 3x) ÷ 16x
l 2  m2 l  m l  m = – (5x – 3)
2

l  m l  m2 133. (a)
A.T.Q.
l  m 
= lm 3 3 3
  u – v  +  v – w  + w – u
3 3 3
129. (a) u 2
- v2  + v 2
- w2  + w 2
- u2 
ATQ,
Here,
2a – 5 4b – 5 6c + 5
– + =0 u–v+v –w+w–u=0
a b c
So, (u – v)3 + (v – w)3 + (w – u)3

r
 5  5  5 = 3( u – v) (v – w) (w – u)
Or  2 –  –  4 +  +  6 +  = 0
a b c

si
3 u – v  v – w  w – u 
–5 5 5  3 u2 – v 2  v 2 – w2  w2 – u2 
= + + =–2+4–6
a b c an by
1 1 1 1

n
= –5  – – =  u + v  v + w  w + u 
 a b c  = – 4

1 1 1 4
ja 134. (a)
R s
= – – =
a b c 5 Given that,
130. (b) x(x + 3)(x + 6)(x + 9) = 0
a th

x² – 15x + 1 = 0 Or
1 x(x + 9)(x + 3)(x + 6) = 0
x+ = 15
x (x2 + 9x)(x2 + 9x + 18) = 0
ty a

1 Let,
 x² + = 225 – 2 = 223
x² y =x2 + 9x
di M

 x4 + 1 = 223x2 y(y + 18) = 0

 x4 – 223x2 + 1 = 0 y2 + 18y = 0
 x4 – 223x2 = – 1 Add 81 both sides
ATQ, y2 + 18y + 81 = 81
4 Required answer = 81 = 34
The value of expression x – 223 x 2 + 6

135. (d)
=–1+6=5 Let, 300 = a
131. (b) Then,
x2 – 11x + 1 = 0
300 × 301× 302 × 303 +1 = a(a + 1)(a + 2)(a + 3) +1
A

1
x + = 11
x (a2 + 3a)(a 2 + 3a + 2) +1

1 Let,
x2 + = 119
x2 a2 + 3a = b
1
x4 +
x4
= 14159 b  b + 2 +1 = b 2

+ 2b +1 = b +1 ........(i)

x8 + 1 = 14159x4 Put the value of b in equ.1

x8 – 14159x4 = – 1 a2 + 3a + 1
Now, Put the value of a (a = 300)
8 4
x – 14159x + 11 = –1 + 11 = 10 (90000 + 900 + 1) = 90901
136. (a) 141. (a)
Given that,
1 1 1 2 4 
 – – 2 – 4  =?
600 × 601× 602 × 60 +1 8   b – 1  b +1  b +1  b +1 
ATQ,
Value of the expression 1  b +1 – b +1 2 4 
 8  b2 – 1 – b2 +1 – b 4 +1
2
(600) + 3(600) + 1 = (360000 + 1800 + 1)  

= 361801 1   b2 +1 – b2 +1  4 
137. (b)  8 2  4
b –1  –
4 
 b +1 


4x + x 2 2 1 4 4   1 
=  – 
2
x – 3x + 4 3 8  b4 – 1 b4 +1  b8 -1 
12x + 3x2 = 2x2 – 6x + 8 142. (b)

r
2
x = 8 –18x 1

devided by x both the sides

x=  6 -1 3 

si
8  1
3
 1 1
x=
x an by
– 18 3
 x –  + 3  x –  = x – x 3
x x

n
8 x3 = 6 –1
x– = – 18
x
1 1 6 +1
ja = =
R s
138. (c) x 3
6 –1 5
1
a th

ab 3  5 1  6 +1 
x3 – = 6 –1–
ab = 5 = 125 3 x3 5
Possible value of a and b
5 6 – 5 – 6 –1
ty a

=
a b a+b 5
di M

5 25 30 4 6–6
1 125 126 =
5

47 is not possible 143. (c)

139. (c) Given that,
abc = 1
a a–2
– =1 Then,
3 5
123 123 123
 5a – 3a + 6 = 15 + +
1 + a + ab 1 + b + bc 1 + c + ac
2a = 9
a = 4.5 123 123 123
= + +
1 + a + ab 1 1
A

140. (c) 1+ b + 1+ c +
a b
3a + b a – 3b
– + 2b
2 3 123 123 × a 123
= + +
1 + a + ab a + ab +1 1 1
9a + 3b – 2a + 6b 1+ +
= + 2b ab b
6

7a + 9b + 12b 7a + 21b 123 123 × a 123 × ab

= = = + +
6 6 1 + a + ab a + ab +1 a + ab +1

7 a  3b 1 + a + ab
= = 123 × = 123
6 1 + a + ab
144. (a) 1 13 1 1 13 – 7 – 5
= – – =
A Μ c 35 5 7 35
+ =1
L B c = 35
Let, Value of ab – c
A B C = 5 × 7 – 35
=x, = y, = z
L M N = 35 – 35 = 0
Then, 147. (a)
1 1 4(4b – 3a) = 2(2b + c) = 8(c – a)
x+ = 1, y + = 1 22(4b – 3a) = 2(2b + c) = 23(c – a)
y z
2(8b – 6a) = 2(2b + c) = 2(3c – 3a)
1
Let, x = , z=–1 2(8b – 6a) = 2(3c – 3a)
2
8b – 6a = 3c – 3a

r
y=2
8b = 3a + 3c ......... (i)

si
L C 1 2 (2b + c)
=2 (3c – 3a)
 + = +z = 2 –1=1
A N x
2b + c = 3c – 3a
145. (a) an by 3a + 2b = 2c .......... (ii)

n
1 1 1 2 (8b – 6a)
=2 (2b + c)
p + q + r = pqr + + =1
p q r 8b – 6a = 2b + c
ja
R s
pq + qr + rp = pqr 6b = 6a + c .......... (iii)
(p + q + r)² = p² + q² + r² + 2(pq + qr + rp) From (i) and (ii)
a th

 1 = p² + q² + r² + 2 × 1 8b = 2c – 2b + 3c
 p² + q² + r² = –1  10b = 5c
Now, p³ + q³ + r³ – 3pqr = (p + q + r) c = 2b
ty a

[(p² + q² + r² – (pq + qr + rp)]

c
p³ + q³ + r³ – 3 × 1 = 1 [– 1 – 1] b=
2
di M

p³ + q³ + r³ – 3 = – 2
From (i) and (iii)
= p³ + q³ + r³ = + 1
8b = 3a + 3 (6b – 6a)  15a = 10b  3a
146. (b)
25(3a – 2b) = 5(b – a) = 52 2b c
= 2b  a = =
3 3
52 (3a – 2b)
= 5(b – a) = 52
5 (6a – 4b)
= 5(b – a) = 52 a + b + c = 11
5(6a – 4b) = 52 c c
  +   + c = 11
6a – 4b = 2 3 2
3a – 2b = 1 ........(i) c = 6, b = 3 and a = 2
A

(b – a) 2
5 =5
1 1 1
b–a=2 Value of 4  ab + bc + ac 
 
2b – 2a = 4 ...... (ii)
1 1 1
From (i) and (ii), we get = 4  ab + bc + ac 
 
a=5
b=7 1 1 1
= 4  6 + 18 + 12 
Now,  

1 1 1 13 1 1 1 13  6  2  3  11
+ + = , + + = =4
a b c 35 5 7 c 35  36  = 9
148. (c) 150. (a)
Given,
1
a + a2 + a3 + a4 – 2 = 0 a– = b .............(i),
a
a + a2 + a3 + a4 = 2
a(a 4 – 1) 1
=2 (G.P. series) b– = c .............(ii),
a –1 b
5
a – a = 2a – 2
1
2 c– = a ............(iii)
a4  =3 c
a
149. (b) Add all
If 25x4 – 9 x2 y2 + 49y4 = 114 ......(i) 1 1 1
& 5x2 + 3x y + 7y2 = 19 ...........(ii) + + = (a – b) + (b–c) + (c–a)
a b c
By (i) & (ii)

r
114 1 1 1
5x2 + 7y2 – 3xy = = 6 ........(iii) + + =0
19 a b c

si
Then, from (ii) & (iii) 2
1 1 1
   
an by
5x2 + 7y2 =
25
2
a b c
1 1 1  1 1 1

n
& 6xy = 13 = a 2 + b2 + c2 + 2  ab + bc + ca 
13
xy =
6
ja  1 1 1 
R s
  
x y ab bc ca 
Now, 5 y + 7 x
a th

–1  1 1 1
2 2 = 2  a 2 + b2 + c2 
5x + 7y 75
= xy =
13
ty a
di M
A
 ALGEBRA
Concept of Quadratic Equation
1. If  and  are the roots of 4x2 + 3x + 7 = 0, 6. Find the value of k so that the sum of the
roots of equation 3x2 + (2k + 1) x – k – 5 =
then the value of 1 + 1 is: 0 is equal to the product of the roots :
 
k dk og eku Kkr dhft, ftlds fy, lehdj.k
;fn  rFkk lehdj.k 4x2 + 3x + 7 = 0 ds ewy 3x2 + (2k + 1) x – k – 5 = 0 ds ewyksa dk ;ksxiQ

gks rks +
1 1
dk eku Kkr dhft,A muds xq.kuiQy ds cjkcj gksA
  (a) 4 (b) –4
(c) 2 (d) 8
4 –3
7. If sum of the roots of a quadratic equation

r
(a) (b)
7 7 is 1 and product of the roots is -20. find

si
3 –3 the quadratic equations
(c) (d) ;fn f}?kkr lehdj.k ds ewyksa dk ;ksx 1 gS vkSj
7 4
2.
an by
If  and  are roots of the equation x2 – x + 1
= 0, then write the value of 2 +  2.
dk xq.kuiQy &20 gSA f}?kkr lehdj.k Kkr dhft
(a) x² – x – 20 = 0 (b) x² + x + 20 = 0

n
;fn  rFkk lehdj.k x2 – x + 1 = 0 ds ewy gks rks (c) x² + x – 20 = 0 (d) x² – x + 20 = 0
2 +  2 dk eku Kkr dhft,A 8. Which of the following quadratic equation

ja
R s
(a) 1 (b) –1 has roots –3 and –5.
(c) 0 (d) None of these fuEUk esa ls fdlh f}?kkr lehdj.k ds ewy
–3 rFkk
a th

3. If the equation 2x2 – 7x + 12 = 0 has two

–5 gSA
  (a) x² – 8x + 15 = 0
roots a & b, then the value of    . (b) x² – 8x – 15 = 0
ty a

(c) x² + 8x + 15 = 0
;fn  rFkk lehdj.k 2x2 – 7x + 12 = 0 ds ewy
(d) x² + 8x – 15 = 0
  9. If ,  are roots of the equations x2 – 5x + 6
di M

gks rks
  dk eku Kkr dhft,A

= 0 then find the quadratic equation whose

97 7 roots are 1 , 1
(a) (b)  
24 2
1 7
;fn  vkSj lehdj.k x² – 5x + 6 = 0 ds ewy gSa
(c) (d)
24 24 1 1
4. One root of quadratic equation x² – kx +
rks f}?kkr lehdj.k ftldk ewy vkSj gSA
 
27 = 0 is 3, then find the value of 'k'. 2
(a) 6x – 5x + 1 = 0
f}?kkr lehdj.kx² – kx + 27 = 0 dk ,d ewy 3 gS] (b) 6x2 + 5x + 1 = 0
rksk dk eku Kkr dhft,A (c) 6x2 – 5x – 1 = 0
A

(a) 10 (b) 12 (d) 6x2 + 5x – 1 = 0

(c) –12 (d) 16 10. If  and  are the roots of equation x² – × +
5. Find the value of k if one root of the 1 = 0, then which equation will have roots
equation: x2 – 9x + k = 0 is twice the other ³ and ³
root. ;fn  vkSj lehdj.k x² – × + 1 = 0 ds ewy gSa
k dk og eku Kkr dhft, ftlds fy, lehdj.k x2 rks f}?kkr lehdj.k ftldk ewy³ vkSj³ gSA
– 9x + k = 0 ds ,d ewy dk eku nwljs ewy ds nks (a) x² + 2× + 1 = 0
xqus cjkcj gksA (b) x² – 2× + 1 = 0
(a) 18 (b) 16 (c) x² + 3× + 1 = 0
(c) 12 (d) 9 (d) x² – 3x + 1 = 0
11. If  and  are the roots of the equation 17. If px3 + x 2 + 3x + q is exactly divisible by
x² – 3x + 2 = 0, then the quadratic equation (x + 2) and (x – 2), then the values of p and
whose roots ar ( + 1) and ( + 1) is. q are:
;fn  vkSj lehdj.k x² – 3x + 2 = 0 ds ewy gSa] rks ;fn px 3 + x 2 + 3x + q] (x + 2) vkSj(x – 2) ls
f}?kkr lehdj.k ftldk ewy ( + 1) vkSj( + 1) gSA iw.kZr% foHkkT; pgS vkSj
rksq ds eku gS%
(a) x² – 5x + 6 = 0 (b) x² + 5x – 6 = 0 SSC CHSL 08/06/2022 (Shift- 03)
(c) x² + 5x + 6 = 0 (d) x² – 5x – 6 = 0
12.  and  are the roots of quadratic equation. 3
(a) p  – and q  4
If  +  = 8 and  –  = 25, then which of 4
the of the following equation will have roots
3
 and  ? (b) p  and q  4
4
 rFkk f}?kkrh; lehdj.k ds ewy gSA ;fn +=
8 rFkk –  = 25 gSa] rks
 rFkk  fuEufyf[kr esa 3
(c) p  and q  – 4
ls fdl lehdj.k ds ewy gSa\ 4
(a) x² – 1522x + 14641 = 0
3
(b) x² + 1921x + 14641 = 0 (d) p  – and q  – 4

r
4
(c) x² – 1764x + 14641 = 0
18. Let f(x) = x3 – 6x2 + 11x – 6, then which

si
(d) x² + 2520x + 14641 = 0
13. When (x4 – 3x3 + 2x2 – 5x + 7) is divided by one of the following is not a factor of f(x) ?
;fn f(x) = x3 – 6x2 + 11x – 6 gS] rks fuEu esa
an by
(x – 2), the remainder is –
tc (x4 – 3x3 + 2x2 – 5x + 7) dks (x – 2) ls dkSu&lk f(x) dk xq.ku[kaM ugha gS\

n
foHkkftr fd;k tkrk gS] rks -----------
'ks"kiQy izkIr gksrk gSA (a) (x – 1) (b) (x – 2)
(a) 3 (b) –3 (c) (x + 3) (d) (x – 3)
(c) 2

ja
(d) 0 19. If (x + 2) and (x – 3) are the factors of
R s
14. When f(x) = 15x3 – 14x2 - 4x + 10 is divided
x² + k1x + k2 , then:
by (3x + 2), then the remainder is:
a th

tc f(x) = 15x3 – 14x 2 - 4x + 10 dks (3x + 2) ;fn (x + 2) vkSj(x – 3), x² + k1x + k2 ds xq.kt
gS] rks%
ls foHkkftr fd;k tkrk gS] rks ----------------- 'ks"kiQy
izkIr gksrk gSA SSC CHSL 09/06/2022 (Shift- 02)
ty a

SSC CHSL 27/05/2022 (Shift- 2) (a) k1 = 1 and k2 = – 6

(a) – 1 (b) 1 (b) k1 = –1 and k2 = – 6
di M

(c) – 2 (d) 2 (c) k1 = –1 and k2 = 6

15. If 5x3 + 5x2 – 6x + 9 is divided by (x + 3), (d) k1 = 1 and k2 = 6
the remainder is : 20. For what value(s) of k will the expression
;fn 5x3 + 5x2 – 6x + 9 dks (x + 3) ls foHkkftr 1
fd;k tkrk gS rks izkIr 'ks"kiQy gSA p
9
p  k 2 be a perfect square ?
(a) 135 (b) 63
(c) –135 (d) –63 K ds fdl eku@fdu ekuksa ds fy, O;atd
16. If x3 + 2x2 – ax – b is exactly divisible by (x2 1
– 1), then the values of a and b are : p p  k2 ,d iw.kZ oxZ gksxk\
9
;fn x3 + 2x2 – ax – b, (x² – 1) ls iw.kZr% foHkkftr
SSC CHSL 10/06/2022 (Shift- 02)
gks tkrk gks rks
a rFkkb dk eku Kkr dhft,A
A

CHSL 2019 21/10/2020 (Shift- 02) 1 1

(a) k   (b) k  
(a) a = – 1 and b = 2 8 9
(b) a = 1 and b = – 2
(c) a = 1 and b = 2 1 1
(c) k   (d) k  
(d) a = 2 and b = 2 21 18
Answer Key
1.(b) 2.(b) 3.(c) 4.(b) 5.(a) 6.(a) 7.(a) 8.(c) 9.(a) 10.(a)
11.(a) 12.(a) 13.(b) 14.(d) 15.(d) 16.(c) 17.(d) 18.(c) 19.(b) 20.(d)
 Quadratic Equation
(Practice Sheet With Solution)
1. If  and  are the roots of the equation x2 + x ;fn a rFkkb lehdj.k Px2 – Qx + R = 0 ds ewy
– 1 = 0, what is the equation whose roots are
5 and  5 ?  1   1  a  b
rks a 2    b2    b    a  ? dk eku D;k gS\
;fn  rFkk lehdj.k x2 + x – 1 = 0 ds ewy gSa] rks        
og lehdj.k D;k gS ftlds ewy  rFkk  gSa\ (R  P)(Q2  2RP)
(a)
2
(a) x + 7x – 1 = 0 2
(b) x – 7x – 1 = 0 PR 2
(c) x2 – 11x – 1 = 0 (d) x2 + 11x – 1 = 0

r
(R  P)(Q2  2RP)
2. If the roots of the equation a(b – c)x2 +b(c – (b)
PR 2

si
a)x+ c(a – b) = 0 are equal, then which of the
follow- ing is true? (R  P)(Q2  2RP)

an by
(c)
;fn lehdj.k a(b – c)x +b(c – a)x + c(a – b) = 0
2
PR 2
ds eqy cjkcj gSa] rks fuEufyf•r esa ls dkSu lk lgh gS\

n
(R  P)(P 2  2RQ)
(d)
 a  c 2 1 1 PR 2
(a) b    

ja
(b)
ac R s b  a c 6.  and  are the roots of the quadratic equatio
x2 – x – 1 = 0. What is the value of  +  8?
1 1 rFkk f}?kkr lehdj.k x2 – x – 1 = 0 ds ewy gS
a th
(c) 2b       (d) abc = ab + bc + ca 
a c +  8 dk eku D;k gS\
3. If the difference between the roots of the equa- (a) 47 (b) 54
tion Ax2 – Bx + C = 0 is 4, then which of the
ty a

(c) 59 (d) 68
following is TRUE? 7. Sum of the roots of a quadratic equation is
;fn lehdj.k Ax2 – Bx + C = 0 ds ewyksa dk varj 4
di M

less than the product of the roots. If one roo

gS] rks fuEufyf•r esa ls dkSu&lk lR; gS \ is 1 more than the other root, find the produc
(a) B2 – 16A2 = 4AC + 4B2 of the roots?
(b) B2 – 10A2 = 4AC + 6A2 ,d f}?kkr lehdj.k ds ewyksa dk ;ksx ewyksa ds x
(c) B2 – 8A2 = 4AC + 10A2 5 de gSA ;fn ,d ewy nwljs ewy ls 1 vf/d gS] rks
(d) B2 – 16A2 = 4AC + 8B2 dk xq.kuiQy Kkr dhft,\
4.  and  are the roots of quadratic equation. If (a) 6 or 3 (b) 12 or 2
 +  = 8 and –  = 2 5 , then which of the (c) 8 or 4 (d) 12 or 4
following equation will have roots 4 and  4? 8. If x, y, z are three factor of a3 – 7a – 6 the
value of x + y + z will be
 rFkk f}?kkr lehdj.k ds ewy gSaA ;fn$  ¾ 8
;fn x, y, z, a3 – 7a – 6 ds rhu xq.ku•aM gSax rks
+
rFkk–  ¾ 2 5 gSa] rks 4 rFkk 4 fuEufyf•r esa ls
A

+ z dk eku gksxk
fdl lehdj.k ds ewy gSa \ (a) 3a (b) 3b
(a) x2 – 1522x + 14641 = 0 (c) 6a (d) 9b
(b) x2 + 1921x + 14641 = 0 9. The value of a for which one root of th
(c) x2 – 1764x + 14641 = 0 quadratic equation (a² – 5a+3) x² + (3a – 1)x
(d) x2 + 2520x + 14641 = 0 2 = 0 is twice as large as the other is
5. If a and b are the roots of the equation Px2 – a dk eku ftlds fy, f}?kkr lehdj.k (a² – 5a+3) x²
Qx + R = 0, then what is the value of (3a – 1)x + 2 = 0 dk ,d ewy nwljs ls nksxquk cM+
 1   1  a  b (a) –2/3 (b) 1/3
 2  2   ?
a   b   b  a (c) –1/3 (d) 2/3

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs1

 px + (1 – p) = 0 then its roots are (c) 1 (d) 2
;fn (1 – p) f}?kkr lehdj.k x² + px + (1 – p) = 0 18. What should be subtracted from x4 + x3 – 2x
dk ewy gS rks blds ewy gSa + x + 1 such that it is divisible by x – 1?
(a) 0,–1 (b) –1,1 x4 + x3 – 2x² + x + 1 esa ls D;k ?kVk;k tkuk
(c) 0,1 (d) –1,2 rkfd ;g x – 1 ls foHkkT; gks tk,\
11. A complete factorisation of x4 + 64 is (a) 3 (b) 2
x4 + 64 dk iw.kZ xq.ku•aMu gS\ (c) 1
(a) (x² + 8)² (d) More than one of the above
(b) (x² + 8) (x² – 8) 19. One of the linear factors of 3x² + 8x + 5 is
(c) (x² – 4x + 8) (x² – 4x +8)
3x² + 8x + 5 ds jSf•d xq.ku•aMksa esa ls ,d gSA
(d) (x² + 4x + 8) (x² – 4x + 8)
(a) (x + 1) (b) (x – 4)
12. (x + 2) is a factor of 2x3 + 5x² – x – k. The value
k is: (c) (x – 2) (d) (x + 2)
20. For a polynomial p(x), p(–1) and p(2) are bot

r
(x + 2), 2x3 + 5x² – x – k dk ,d xq.ku•aM gSA k dk
equal to zero .So, we can conclude that,
eku gS%

si
,d cgqin p(x) ds fy,] p(–1) vkSjp(2) nksuksa 'k
(a) –24 (b) 6
cjkcj gSaA blfy,] ge ;g fu"d"kZ fudky ldrs gSa f

an by
(c) –6 (d) 24
13. The polynomial 4x² - kx + 7 leaves a remainder (a) (x² + 2x-1) is a factor
(b) (x² – 2x + 1) is a factor

n
of –2 when divided by x – 3. Find the value of
k. (c) (x2 – x – 2) is a factor
cgqin 4x² - kx + 7 dks x – 3 ls foHkkftr djus ij (d) (x² – x + 2) is a factor

ja
'ks"kiQy
(a) 17
R s
–2 cprk gSAk dk eku Kkr dhft,A
(b) 19
21. Find roots of the equation 4x2 – 41x + 37
lehdj.k 4x2 – 41x + 37 ds ewy Kkr dhft,A
a th
(c) 15 (d) 23 37 27
14. If two polynomials 2x3 + kx2 + 4x – 12 and x3 (a) 1, (b) 2,
4 4
+ x² – 2x + k leave the same remainder when
ty a

divided by (x - 3), find the value of k and the 47 37

(c) 5, (d) 6,
remainder. 4 4
;fn nks cgqin2x3 + kx2 + 4x – 12 vkSj x3 + x² –
di M

22. x – 2 is the HCF of the equation 4x3 + 3x2

2x + k]
ls (x - 3) ls foHkkftr djus ij leku 'ks"kiQy 8x – p = 0 then find the value of P
NksM+rs gSa]
k dkrks
eku vkSj 'ks"kiQy Kkr dhft,A lehdj.k 4x3 + 3x2 – 8x – p = 0 dk ek-lk x – 2 g
(a) (–5, 29) (b) (7, 28) rksP dk eku Kkr djsA
(c) (–3,–27) (d) (– 3 ,27) (a) 24 (b) 17
15. Find the value of m, if x = 1/2 is one of the (c) 28 (d) 31
zeroes of the polynomial p(x) = 4x4 – 4x³ – mx² 23. What is the condition that the roots of the equatio
+ 12x – 3. ax 2 + bx + c = 0 are in the ratio c : 1?
m dk eku Kkr dhft, ;fn x = 1/2 cgqin p(x) = og D;k izfrca/ gS fd lehdj.k ax2 + bx + c = 0 ds ewyc
4x – 4x³ – mx² + 12x – 3 ds 'kwU;dksa esa ls ,d gS
4
1 ds vuqikr esa gksa\
(a) 9 (b) 11 (a) b2 = a(c + 1)2 (b) a2 = b(c + 1)2
A

2 2
(c) 7 (d) 13 (c) b = a(c – 1) (d) b2 = a(c + 2)2
16. Find the value of k if p(x) = (3x – 2)(x – k) – 8 24. (x + 2) is a factor of 2x + 5x2 – x – k. The value k is
3

is divided by (x – 2) leaving the remainder 4. (x + 2), 2x3 + 5x2 – x – k dk ,d xq.ku[kaM gSA

k dk eku g
k dk eku Kkr dhft, ;fn p(x) = (3x – 2)(x – k) – 8 (a) –24 (b) 6
dks(x – 2) ls foHkkftr djus ij 'ks"kiQy
4 cprk gSA (c) –6 (d) 24
(a) –1 (b) –2 25. If x – y = 1 and x² + y² = 41 where x, y  0, the
(c) +1 (d) +2 the value of x + y will be:
17. If (x – 8) is one of the factors of mx3 – 24x² + ;fn x – y = 1 vkSjx² + y² = 41 gS] tgk¡x, y  0 gS]
192x – 512, find the value of m. x + y dk eku gksxkA
;fn (x – 8), mx3 – 24x² + 192x – 512 ds xq.ku•aMksa (a) 9 (b) 8
esa ls ,d gS] rks
m dk eku Kkr dhft,A (c) 6 (d) 7

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs2

 Answer Key
1.(d) 2.(b) 3.(b) 4.(a) 5.(c) 6.(a) 7.(b) 8.(a) 9.(d) 10.(a)

11.(d) 12.(b) 13.(c) 14.(d) 15.(b) 16.(a) 17.(c) 18.(b) 19.(a) 20.(c)

21.(a) 22.(c) 23.(a) 24.(b) 25.(a)

r
si
an by
n
ja
R s
a th
ty a
di M
A

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs3

 SOLUTIONS
1. (d) 4 = 44
+  = – 1,  = – 1  2   2  64  22  42
5 +  5 =(2 +  2)(3 +  3)–2 2 ( + )
Again squaring :-
2 + 2 = 1 – 2(–1) = 3
4 +  4 = 1764 – 2  121 = 1522
3 + 3 = –1 –3(–1)(–1) = –4
Qudratic eqn for root 4 and  4
5 +  5 = 3 × (–4)–1(–1) = – 11
x2 – (4 +  4) + (4 4) = 0
 Required equation is = x² – (–11)x –1 = 0
x2 – 1522x + 14641= 0
x² + 11x – 1 = 0
5. (c)
2. (b)
a(b – c) x² + b (c – a)x + c(a – b) = 0 roots are
Q R

r
We know a b  , a.b 
P P
If roots are equal

si
Then, Q2
a 2  b2  2ab 
P2

an by
B² = 4AC
2 2
b² (c – a)² = 4a (b – c) × c(a – b) a b 2ab q2
  
ab ab p(ab)

n
b² (c – a)² = 4ac (b – c) (a – b)
On solving this :- 1 1 a b a 2  b 2 a 2  b2
 2    

ja
2
2ac = ab + bc R s a b b a a 2 b2 ab
2 1 1
 
a 2  b2  1 2 2
a th
b a c  a  b 1  ab 
 1   
3. (b) ab  ab  ab  ab 
Let roots be , 
put value
ty a

B
+ =  Q2 2ab  1  ab 
A  2 –  
 P ab ab   ab 
di M

–= 4
C R
  
A  Q2  1  P 
 – 2 
R R 
B2  P2   
 2   2  2  ...(1)  P  P 
A2
(R  P)(Q2 – 2RP)
 2   2  2  16 ...(2) 
PR 2
Subtracting (2) from (1), we get
6. (a)
B2 x2 – x –1 = 0
4  2  16
A  +  =1,  = –1
A

2 +  2 + 2 (–1) = 1
C B2  16A 2
4  2 +  2 = 3, 4 +  4 = 7
A A2
8 +  8 + 2 (–1)8 = 49
4CA = B2 – 16A2
8 +  8 = 47
B2 – 10A2 = 4CA + 6A2
7. (b)
4. (a)
Let,
 +  = 8,  –  = 2 5 roots of equation  , 
Squaring :- ATQ,
2 + 2 + 2= 64 ...(1) – (+)= 5 &  =  + 1
2 +  2 – 2= 20 ...(2) ( + 1) – ( +  + 1) = 5
Subtracting (2) from (1), we get  2 +  – 2 – 1 = 5

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs4

  2 – 3 + 2 – 6 = 0 let, p(x) = 4x² – kx + 7
( – 3) + 2 ( – 3) = 0 given, –2 is Remainder when
 = – 2,3 p(x) is divided by x – 3
then So, x = 3 and r = –2
=4,–1 By remainder theorem:-
 Required value p(x) = r
 4 × 3, – 1 × –2 = 12 , 2 p(3) = –2
4(3)2 – k(3) + 7 = –2
8. (a)
36 – 3k + 7 = –2
a3 – 7a – 6
3k = 45
a² (a+1) – a(a+1) – 6(a + 1)
k = 15
=(a+1)(a² – a – 6)
14. (d)
(a + 1)(a – 3) (a + 2)
let p(x) = 2x3 + kx² + 4x-12

r
a + 1 + a – 3 + a + 2 = 3a fex) = x3+ x² – 2x + k
9. (d)

si
given, p(x) and f(x) leaves same remainder whe
Let , 2 are roots of given equation divided by (x – 3)

an by
Sum of roots So, x = 3
r = p(x) = f(x)
1  3a

n
 + 2 = 3 = ...(1) then, p(3) = f (3)
a 2  5a  3
2(3)3 + k(3)² +4(3)–12 = 33 +3² – 2(3) +k
And product of roots

ja
R s 54 + 9k + 12 –12 = 27 + 9 – 6 + k
2 54 + 9k = 30 + k
(2) = 22 = ...(2)
a 2  5a  3 8k = –24
a th
By (1) and (2), we have k = –3
put K = –3 in p(x),
9 2 (13a)2 a 2 5a3 p(x) = 2x3 + (–3) x² + 4x – 12
 2 
ty a

2 2
2 (a 5a 3) 2 =2x³ – 3x² + 4x – 12
Now, P(3) = 2(3)³–3(3)² + 4(3) –12
9 (a 2 5 a  3 )  (1 3 a ) 2
di M

=54 – 27 + 12 – 12 = 27
2 K= –3 and 27 remainder
 a 15. (b)
3
10. (a) p(x) = 4x4 – 4x3 – mx² + 12x – 3
x = 1/2 is zero of p(x)
1 – p is root of x2 + px + 1 – p = 0
So, (x –1/2) is a factor of equation)
then
By remainder theorem.
p=1
r = p(a)
x2 + x + 0 = 0
P(1/2) = 0
x2 + x = 0
4 3 2
x(x+1) = 0 1 1 1 1
4    4    m    12    3  0
A

x = 0, – 1 2 2 2 2

11. (d)
4 4 m
x4 + 64 = x4 + 82   63 0
16 8 4
(x² + 8)² –16x2
= (x² + 4x +8) (x² – 4x +8) m 1
 3
12. (b) 4 4
fex) = 2x3 + 5x2 – x – k m  11
(x + 2) is factor. 16. (a)
x = –2 satisfies f(x) = 0 p(x) = (3x – 2) (x – k) – 8
f(–2) = –16 + 20 + 2 – k = 0 Also, it is given that the remainder is 4 when
K = 6' p(x) is divided by (x – 2)

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs5

 Using remainder theorem:-  (x – 1) (4x – 37)
p(x) = r, p(2) = 4 37
[3(2) – 2) (2 – k] – 8 = 4  x = 1,
4
(6 – 2) (2 – k) = 4 + 8 22. (c)
4 (2 – k) = 12 ATQ
12  4(2)3 + 3(2)2 – 8(2) – P = 0
2–k= =3
4  32 + 12 – 16 – P = 0
k = 2 – 3= –1  28 – P = 0
17. (c)  P = 28
let p(x) = mx3 – 24x² +192x – 512 23.(a) Let and are the roots of given equation
given, (x – 8) divides p(x) exactly  c  1
 or 
Let x = 8 r = 0  1  c

r
By remainder theorem:- equation given, ax2 + bx + c = 0
p(x) = r –b

si
+ = __ (1)
P(8) = 0 a
m (8)3 – 24 (8)2 + 192 (8) – 512 = 0 c

an by
 =
512 m – 1536 + 1536 – 512 = 0 a
512 m = 512 Squring (1)

n
m=1 b2
2 +  2 + 2=
18. (b) a2

ja
put x –1 = 0 x = 1
R s
f(x) = x4 + x3 – 2x² + x +1 Divide this equation by .
b2
a th
let f(x) = x4 + x3 – 2x² + x + 1   2
 2 a
f(1) = 1 + 1 – 2 + 1 + 1 = 2   c
Subtracting 2 from given a
ty a

Equation makes it divisible by (x –1) 1

2
19. (a) c + 1  2  b
di M

f(x) = 3x2 + 8x + 5 = 0 c ac
= 3x² + 3x + 5x + 5 = 0 a(c2 + 1 + 2c) = b2
3x (x + 1) + 5 (x + 1) = 0 a(c + 1)2 = b2
(3x + 5)(x+1) = 0 24. (b)
(x +1) is one linear factor. f(x) = 2x3 + 5x2 – x – k
20. (c) (x + 2) is factor
From options x = – 2 satisfied f(x) = 0
P(–1) = p(2) f(–2) = – 16 + 20 + 2 – k = 0
option (c) k=6
P(x) = x² – x – 2 25. (a)
P(–1) = 1 + 1 – 2 = 0 (x + y) = (x – y)² + 4xy .....(1)
A

P(2) = 4 – 2 – 2 = 0 Also,
(x2 – x – 2) is a factor (x – y)² = x² + y² – 2xy
21. (a) 1 = 41 – 2xy
ATQ,  xy = 20 ....(2)
 4x2 – 41x + 37 Put (2) in (1)
 4x2 – 37x – 4x + 37 (x + y) = 1+ 80 = 81 = 9

Aditya Ranjan Sir (Excise Inspector) Selected gSSelection fnyk,axs6

 TRIGONOMETRY /f=kdks.kfefr
(CLASSROOM SHEET-01)
Questions Based on Basic 4. If sin  
1
, (sec – tan)2 is equal to:
3
Trigonometric Ratios 1
;fn sin   gS] rks
(sec – tan)2 fdlds cjkcj gS
4 3
1. If tan A = , 0  A  90, then find the value
3 SSC CPO 05/10/2023 (Shift-01
of sin A. 1 1
(a) (b)
4 3 2
;fn tan A = , 0  A  90, gS] rkssin A dk eku

r
3 2 3
D;k gksxk\ (c) (d)

si
3 4
SSC CPO 03/10/2023 (Shift-01)
3

(a)
4
5
an by (b)
3
5
5. If cos θ=
2
, then tan2 cos2 = ?

n
3
3 4 ;fn cos θ=
2
gS] rks
tan2 cos2 = ?
(c) (d)
4
ja 3
R s
SSC CGL 14/07/2023 (Shift-1
 9  1
a th

1
2. If sin =   , 0° << 90° then what is the (a) (b)
 41  3 4
value of cot ? 1
(c) (d) 3
ty a

2
 9 
;fn sin =  41  , 0° << 90° rkscot  dk eku 6. If 5sin- 4 cos= 0, 0°<<90°, then the valu
 
di M

5sinθ + 2cosθ
D;k gksxk\ of is:
5sinθ + 3cosθ
SSC CGL MAINS (08/08/2022 (Shift-01)
;fn 5sin- 4 cos = 0, 0°<  <90° g S
40 35 5sinθ + 2cosθ
(a)
9
(b)
8 5sinθ + 3cosθ
dk eku Kkr dhft,A
SSC CGL 13/04/2022 (Shift- 02
39 47
(c) (d)
9 8 4 6
(a) (b)
7 7
2 sec2 A
3. If tan A = find the value of . 2 3
5 cosec 2 A (c) (d)
A

7 7
2 sec2 A
;fn tan A = gS] rks dk eku Kkr dhft,A 3 1  sin 
5 cosec 2 A 7. tan θ = , find the value of expression
4 1  sin 
SSC CPO 03/10/2023 (Shift-3)
3 1  sin 
3 4 ;fn tan θ = vfHkO;fÙkQ dk eku Kkr dhft,
(a) (b) 4 1  sin 
5 25
SSC CHSL 09/08/2023 Shift-0
2 9 (a) 4 (b) 3
(c) (d)
5 25 (c) 8 (d) 5

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 5 cosec B + sin B
6sinθ - 3cosθ
;fn tanB = 3 gS rks cos B – sec B dk eku K
is :
7sinθ + 3cosθ dhft,A
SSC CGL 20/04/2022 (Shift- 02
6sinθ - 3cosθ
;fn 5 cot = 3 gS rks7sinθ + 3cosθ dk eku Kkr 177 177
(a) – (b)
125 125
dhft,A
SSC CGL 9/03/2020 (Shift- 02) 59 59
(c) – (d)
15 15
21 44
(a) (b) 12
44 21 12. If cos  = , then the value o
13
11 20 sin  (1 – tan )
(c) (d)
40 41 tan  (1  cosec) is :

r
2 12 sin (1 – tan )
9. If sin A =
5
, Where A is an acute angle, what is ;fn cos = gS rkstan (1  cosec) dk ek

si
13
5 sin A  2 cosec A Kkr dhft,A
the value of
an by 21 sec A
?
SSC CGL MAINS 03/02/202

n
2 25 35
;fn sin A = gS] tgk¡A ,d U;wudks.k gS] rks (a) (b)
5 78 234

ja
R s
5 sin A  2 cosec A 35 25
dk eku D;k gS\ (c)
108
(d)
156
21 sec A
a th

SSC CHSL 10/08/2023 (Shift-01)  (3 sin  – cos ) 

13. If  (cos   sin )  = 1, then the value of cot is
 
5 7
(a) (b)
ty a

4 5  (3 sin  – cos ) 
;fn   = 1 gS] rkscot dk eku K
 (cos   sin ) 
4 5
dhft,A
di M

(c) (d)
5 7
SSC CGL 14/07/2023 (Shift-1
17 (a) 3 (b) 0
10. If sec A = , given that A < 90°, what is the
8 (c) 1 (d) 2
34sinA + 15cotA sin   cos  4
value of the following? 14. If  , then the value o
68cosA - 16tanA
sin   cos  5
17 cos ec2 
;fn sec A = 8
gS rks
A < 90° dk eku D;k gS\ fn;k is:
2  cos ec2 
34sinA + 15cotA
gS68cosA - 16tanA ;fn
sin   cos  4
 rks
cos ec2 
dk eku K
A

sin   cos  5 2  cos ec2 

SSC CGL 11/04/2022 (Shift- 03) djsa
(a) 23 (b) 19
SSC CHSL 09/08/2023 Shift-0
(c) 30 (d) 38
16 40
5 (a) (b)
11. If tanB = , what is the value of 25 41
3
cosec B + sin B 41 31
? (c) (d)
cos B – sec B 40 30

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 12 sinA + cosA 17 1 – cosA
15. If sin =
13
, 0 <  < 90°, then ;fn = gS rks dk ek
cosA 12 sinA
sin2– cos 2 1 Kkr dhft,A
  __________.
2 sin. cos tan2 SSC CGL 07/03/2020 (Shift- 02
(a) – 5 (b) 1
12
; fn sin = , 0 <  < 90° gS r ks 5 1
13 (c) (d)
12 5
sin2 – cos2 1
× = _________. cosec cot
2 sin. cos tan2 19. If =7, then the value o
cosec – cot
SSC CHSL 08/06/2022 (Shift- 2)

295 290 4sin2– 1

(a) (b) is:
3456 3542 4sin2+ 5
cosec cot 4sin2– 1
(c)
695
(d)
595 ;fn = 7 gS rks dk ek
cosec – cot 4sin2+ 5

r
3542 3456
Kkr dhft,A

si
2 SSC CGL 16/08/2021 (Shift- 01
16. If tan = , 0 < < 90º, then the value of
11 1 1
(a) (b) –
an by
2cosec 2 θ – 3sec 2 θ
3cosec 2 θ + 4sec 2 θ
is equal to :
3
1 1
3

n
(c) – (d)
9 9
2 sin ²
; fn g S r ks 20. If
tan =
ja 11
, 0 <  < 90º = 5,is an acute angle, the
R s
tan ² – sin ²
2cosec 2 θ – 3sec 2 θ 24 sin ²– 15 sec2
a th

dk eku Kkr dhft,A the value of is:

3cosec 2 θ + 4sec 2 θ 6cosec²– 7 cot2
CHSL 26/10/2020 (Shift- 03)
sin ²
;fn = 5, gS rks
ty a

11 11 tan ² – sin ²
(a) (b)
45 49
24 sin ²– 15 sec2
di M

13 10 dk eku Kkrdhft,A
(c) (d) 6cosec²– 7 cot2
49 49
SSC CGL 23/08/2021 (Shift- 02
a 1 - tan θ 2 (a) 2 (b) –14
17. If sec = , b  0, then =? (c) 14 (d) –2
b 2 - sin2 θ
sec  – tan  1
a 1 - tan 2 θ 21. If = ,  lies in first quadrant
;fn sec = ,b0 gS rks =? sec  + tan  7
b 2 - sin2 θ
CGL-2019 Tier-II (15/10/2020)
cosec  + cot ²
then the value of is:
cosec  – cot ²
a 2 2b 2 + a 2  a 2 2b 2 + a 2 
(a) (b) sec  – tan  1
b 2 a 2 - b 2  b 2 a 2 + b 2  ;fn = , izFke prqFkkZa'k eas f
A

sec  + tan  7

a 2 2b 2 - a 2  a 2 2b 2 - a 2  cosec  + cot ²

dk eku gSA
(c) (d) cosec  – cot ²
b 2 a 2 + b 2  a 2 a 2 + b 2 
SSC CGL MAINS 03/02/202
sinA + cosA 17 19 22
18. If = , then the value of (a) (b)
cosA 12 5 3
1 – cosA 37 37
is : (c) (d)
sinA 12 19

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 1 a
22. In a ABC, right angled at B, if tanA  , then 26. If sin23° = , then the value of sec23° – sin67
3 b
sinA. cosC + cosA. sinC = ______. is ______.
a
1 ;fn sin23° = gS] rks
sec23° – sin67° dk eku D;k g
,d ABC esa]B ij ledks.k gS] ;fn tanA = b
3
CGL PRE, 14/07/2023 (Shift-3
gS] rks
sinA. cosC + cosA. sinC dk eku Kkr dhft,A
2
a b2 – a 2
SSC CHSL 26/05/2022 (Shift- 1) (a) (b)
2 2
b –a ab
(a) 0 (b) 2
a2 a2
(c) – 1 (d) 1 (c) (d)
b b2  a 2 b b2 – a 2
1
23. In ABC, right angled at B, if tanA = , then
2
1
sinA(cosC + cosA) 27. If sec31° = x, then sin 2 59° + 2
sec 31

r
the value of is :
cosC(sinC - sinA)
1

si
is equal to:
1 sin 2 59cosec2 59
,d ABC esa]B ij ledks.k gS] ;fn tan A = gS]
2
an by
sinA(cosC + cosA)
rks cosC(sinC - sinA) dk eku Kkr dhft,A
;fn sec31° = x gS rkssin259° +
1
sec2 31

n
1
dk eku Kkr dhft,A
ja SSC CGL TIER-II (16/10/2020) sin 2 59cosec2 59
R s
(a) 2 5 (b) 3 SSC CGL 24/08/2021 (Shift- 03
a th

(c) 2 (d) 1 x² – 2 2 – x²
(a) (b)
24. In ABC, B = 90° and AB : BC = 1 : 2. The x x2
value of cos A + tan C is: x² – 2 2 – x²
(c) (d)
ty a

ABC esa] B = 90° vkSjAB : BC = 1 : 2 gSA

cos A x 2
x
+ tan C dk eku______ gSA 28. If sec + tan = 2 + 5 , then the value of sin
di M

SSC CPO 04/10/2023 (Shift-01) + cos is :

1 5 2 5 ;fn sec + tan = 2 + 5 gS rkssin + cos d

(a) (b) eku Kkr dhft,A
2 5 2 5
3
5 5 2 5 (a) (b) 5
(c) (d) 5
2 5 2 5
7 1
(c) (d)
5 5 5
25. If sinA = and 7cot B = 24, then the value
13 7
of (secA cosB)(cosecBtan A) is: 29. If sin θ – cos θ = , 0 < θ < 90º, then th
13
A

5 value of sin θ + cos θ is :

;fn sinA = rFkk7cot B = 24 gS rks(secA
13 7
;fn sin – cos = , 0 <  < 90º gS rkssin
cosB)(cosecBtan A) dk eku Kkr dhft,A 13
SSC CGL MAINS/29/01/2022 cos dk eku Kkr dhft,A

13 65 17 13
(a) (b) (a) (b)
7 42 13 17
15 13 1 1
(c) (d) (c) (d)
13 14 13 17

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs4

 24 ;fn 3 tan = 3 sin gS rkssin² – cos² dk ek
30. If 0º < A, B < 45º, cos (A+B) = and
25 Kkr dhft,A
15 SSC CGL MAINS (08/08/2022 (Shift- 01
sin (A–B) = , then tan2A = ?
17
1 1
24 (a) (b)
2 5
;fn 0º < A, B < 45º, cos (A+B) = rFkksin (A–
25
1 1
15 (c) (d)
B) = gS rkstan2A dk eku Kkr dhft,A 3 4
17
35. If 2ksin 30°cos30°cot60°
SSC CGL 06/03/2020 (Shift- 02)
cot²30°sec60°tan45°
(a) 0 (b) 1 = , then find the value of k.
cosec²45°cosec30°

(c)
416
(d)
213 ;fn 2ksin 30°cos30°cot60°
87 4
cot²30°sec60°tan45°
= gS rksk dk eku Kkr dhft,A

r
Questions Based on Values cosec²45°cosec30°

si
SSC CGL 12/04/2022 (Shift- 03
of Trigonometric Ratios
3
31. an by
Evaluate the follwing:/fuEufyf•r
sin60° + tan 30° + cos45°
dk ewY;kadu djsa% (a)
2
(b) 3

n
(c) 1 (d) 6
NTPC CBT-02 15/06/2022 (Shift-01) 36. If A = 10°, what is the value o

ja
R s
5 23 3 3 25 3 12sin 3A + 5cos (5A – 5)°
(a) (b) 9A
4 4 9sin – 4cos (5A +10)°
a th

2
5 23 3 3 25 3
(c) (d) 12sin 3A + 5cos (5A – 5)°
6 6 ;fn A = 10° gS rks 9A
dk ek
ty a

32. What is the value of sin30° + cos30° + tan30° 9sin – 4cos (5A +10)°
2
sin30° + cos30° + tan30° dk eku D;k gS\
Kkr dhft,A
di M

SSC CGL 14/07/2023 (Shift-3)

SSC CGL 18/04/2022 (Shift- 03
5 3 5– 3 6 2 +5 6 2–5
(a) (b) (a) (b)
2 3 2 3 (9 + 2 2) (9 – 2 2)

(c)
5 3
(d)
5– 3
(c)
6 2 +5
(d)
9 – 2 2 
3 3 (9 – 2 2) 6 2 +5
33. Find the value of the given expression. 37. If 3tan= 2 3 sin, 0° << 90°, then th
fn, x, O;atd dk eku Kkr djsaA cosec ²2  cot ²2
value of is:
4 3 sin ²  tan ²2
tan2 45  3 cos2 30  2 sec2 30  cot2 60
A

3 4 ;fn 3tan= 2 3 sin, 0° << 90° gS

SSC CGL 18/07/2023 (Shift-04)
cosec ²2  cot ²2
2 3 dk eku Kkr dhft,A
(a) (b) sin ²  tan ²2
3 2 SSC CGL MAINS 29/01/202
2 3
(c) (d) 4 20
3 2 (a) (b)
13 39
34. If 3 tan = 3 sin, then what is the value of 4 20
sin² – cos²? (c) (d)
3 27

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs5

 (a) 3 +2 (b) 2(2 - 3)
sin 4 (x– 5)  cos 4 (x– 5)
find the value of 1 – 2 sin ²(3 x– 15) cos ²(3 x– 15)
.
(c) 2( 3 –1) (d) 3 +1
;fn 4 sin²(2x – 10)° = 3,0  (2x–10)  90 gS rks
sin 4 (x– 5)  cos 4 (x– 5) cos2
dk eku Kkr 43. If = 3, 0° <  < 90°, the
1 – 2 sin ²(3 x– 15) cos ²(3 x– 15) cot2 sin2– 1
dhft,A
the value of (tan + cosec) is:
SSC CGL 17/08/2021 (Shift- 03)

5 cos2
(a) 1 (b) ;fn = 3, 0° <  < 90° gS
8 cot  sin2– 1
2

5 (tan + cosec) dk eku Kkr dhft,A

(c) – (d) –1 SSC CGL 13/08/2021 (Shift- 01
8
39. If 2sin(3x – 15°) = 1, 0° < (3x – 15) < 90°, then find
5 3

r
the value of cos² (2x + 15)° + cot² (x + 15)°. (a) 2 3 (b)
3
;fn 2sin(3x – 15°) = 1, 0° < (3x – 15) < 90° gS rks
cos²

si
(2x + 15)° + cot² (x + 15)° dk eku Kkr dhft,A 4 3
(c) 3 3 (d)

(a) 1
an by
SSC CGL 16/08/2021 (Shift- 02)
(b) 5/2
3

n
3
(c) – 7/2 (d) 7/2 44. If cos (A-B) = and secA = 2, 0° < A < 90°, 0
2
40. If (2 cos A + 1) (2 cos A – 1) = 0, 0° < A ? 90°,
ja < B < 90°, then what is the measure of B?
R s
then find the value of A.
;fn (2 cos A + 1) (2 cos A – 1) = 0, 0° < A ? 90° 3
a th

gS rks
A dk eku Kkr dhft,A ;fn cos (A-B) = rFkksecA = 2, 0° < A < 90°, 0
2
SSC CGL 21/04/2022 (Shift- 01) < B < 90° gS rks dks.k
B dk eku Kkr dhft,A
(a) 90° (b) 45°
ty a

SSC CGL 11/04/2022 (Shift- 02

(c) 30° (d) 60° (a) 60° (b) 0°
di M

3 (c) 30° (d) 90°

41. If sin (x + 30º) = , then the value of
12  2A  B   2A – B  3 2A  B
x (0 < x < 90º) is : 45. If sin    cos 
  
 , 0, <9
 2   2  2 2
3
;fn sin (x + 30º) = gS rksx (0 < x < 90º) dk and 0° <
2A  B
< 90° then find the value o
12 2
eku Kkr dhft,A sin[3(A – B)].
SSC CHSL 26/10/2020 (Shift- 02)
 2A + B   2A – B  3
(a) 60º (b) 15º ; fn sin   = cos 
   =
 , 0º
 2   2  2
(c) 45º (d) 30º
A

42. If 0º <  < 90º and cos  = 3 (cot2– cos2), then

2 2A + B 2A  B
< 90 rFkk0° < < 90° gS rkssin[3(
–1
2 2
1 
the value of  secθ + sinθ  is : – B)] dk eku Kkr dhft,A
 2 
SSC CGL 16/08/2021 (Shift- 03
;fn 0º <  < 90º rFkkcos2 = 3 (cot2– cos2) gS rks 1
(a) 1 (b)
–1 2
1 
 secθ + sinθ  dk eku Kkr dhft,A
2  1 3
(c) (d) 2
2 2
SSC CGL 04/06/2019 (Shift- 02)

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs6

 2
right angled at B, then find the value of 51. If tan (90 – ) = , then the value of 2 3 ta
3
sin  A  C   sin  A  C   2 sin B
.  + 1 is:
cot A  cot B  cot C
2
;fn f=kHkqt
ABC ,d ledks.k lef}ckgq f=kHkqt gS]Btks ;fn tan (90 – ) = , rks 2 3 tan  + 1 dk eku K
3
sin  A  C   sin  A  C   2 sin B
ij ledks.k gS] rks dhft,A
cot A  cot B  cot C
SSC CHSL, 11/08/2023 (Shift-2
dk eku Kkr dhft,A
(a) 4 (b) 5
NTPC CBT-1 08/04/2021 (Shift-03)
(c) 3 (d) 6
1 1 52. What is the simplified value of cos2 (90°–)
(a) (b) 
2 2 {cos(90 – ) cos} 
  ?/dk ljyhÑr eku D;k gS\

 cot 
3
(c) (d) 0

r
2 SSC CHSL 31/05/2022 (Shift- 3
(a) 4 (b) 2

si
47. If tan² A – 6 tan A + 9 = 0, 0 < A < 90°, What
is the value of 6cot A + 8 10 cosA ? (c) 0 (d) 1

an by
;fn tan² A – 6 tan A + 9 = 0, 0 < A < 90° gS rks
53. Find the value of

sin 90    cos  90    cot  90   

n
6cot A + 8 10 cosA dk eku Kkr dhft,A
cos 2   1
SSC CGL 20/04/2022 (Shift- 02)
dk eku Kkr dhft,A
ja
R s
(a) 10 10 (b) 8 10 Group D 19/09/2022 (Shift-01
a th

(c) 10 (d) 14 (a) tan (b) 0

(c) – 1 (d) 2sin cos
48. If 3 sec + 4cos – 4 3 = 0 where  is an
acute angle than the value of  is: 54. The following expression is equal to/uhps fn
ty a

x;k O;atd cjkcj gSA

;fn 3 sec + 4cos – 4 3 = 0 tgka ,d U;wudks.k cot85º + cos75º
di M

gS rks
 dk eku Kkr dhft,A
SSC CGL 20/07/2023 (Shift- 02
SSC CGL 13/08/2021 (Shift- 03) (a) tan85º + sin75º (b) tan85º – sin75º
(a) 20° (b) 30° (c) tan5º + sin15º (d) tan5º – sin15º
(c) 60° (d) 45° 3
49. If sin + sin = cos + cos= 1, then sin+ 55. Find  , if cos  = –
2
cos = ?
3
;fn sin + sin = cos + cos= 1 gS rkssin+ ;fn cos  = – gS] rks dk eku Kkr djasA
2
cos = ?
SSC CHSL 02/06/2022 (Shift- 2
SSC CGL 23/08/2021 (Shift- 02)
A

3 5
(a) 2 (b) 0 (a) (b)
2 2
(c) 1 (d) –1 2 4
(c) (d)
3 3
50. Sin (– A) = ? 56. tan ( + ) = ?
SSC CHSL 24/05/2022 (Shift- 02) SSC CHSL 08/06/2022 (Shift- 1

(a) Cos A (b) – Cos A (a) sec (b) cosec

(c) cot (d) tan
(c) – Sin A (d) Sin A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs7

 3
sin73° + cos137° dk eku D;k gksxk\ (a) 3 (b)
2
SSC CHSL 10/06/2022 (Shift- 3)
2 1
(a) sin 13° (b) cos 13° (c) (d)
3 3
(c) cos 18° (d) sin 18°
63. Find the exact value of cos120°.
2 5 3
58. The value of sin² + cos² – tan² is: cos120° dk lVhd eku Kkr dhft,A
3 6 4
SSC CGL 18/07/2023 (Shift-04
2 5 3
sin² + cos ² – tan ² dk eku D;k gS\ (a) –0.5 (b) 0
3 6 4
(c) 0.5 (d) 1
SSC Phase X 03/08/2022 (Shift- 02)
64. Find the value of the following?
1 1 
(a) (b)
2 4 cos2(270 – ) – sin2(180 – ) + sin2   sin2(270 – 
2
(c) 4 (d) 2
fuEu dk eku Kkr dhft,\

r
si
RRB NTPC 08/01/2021 (Shift-01
1 cos (π + x)cos(–x)
59. Simplify. 
cosx
an by π
sin(π – x)cos 
 + x
2



(a) cos2 (b) sin2  
2

n
(c) sin2() – 1 (d) sin2()
1 cos (π + x)cos(–x)
ljy djsa 65. What is the value of
cosx sin(π – x)cos  π + x 
ja [cos(90º + A) ÷ sec(270º – A)] + [sin(270º + A)
R s
 
2 
cosec(630º – A)]
dk eku Kkr dhft,A
a th

SSC Phase X 04/08/2022 (Shift- 03)

(a) tan x (b) cot x (a) 3 sec A (b) tanA secA
(c) sec x (d) cosec x (c) 0 (d) 1
ty a

3 3 66. What is the value of sin(–405°)?

60. If cosx = – and  < x < , then the value
2 2 sin(–405°) dk eku Kkr djsaA
di M

of 2cot2x + 3sec2x is : SSC SSC 20/07/2023 (Shift-03

SSC CHSL 08/07/2019 (Shift- 02)
5 1
(a) 10 (b) 4 (a) (b)
2 2
(c) 8 (d) 16
1 –1 –1
61. Find x if cosx = – . (c) (d)
2 2 2
SSC CHSL 15/10/2020 (Shift- 03) 67. Find the value of tan(–1125°).
3 2 tan(–1125°) dk eku Kkr dhft,A
(a) (b)
2 3 SSC CGL 25/07/2023 (Shift-2
A

5 4
(c) (d) 1
6 3 (a) 1 (b)
2
1 (c) –1 (d) 0
62. If cos x   , x lies in third quadrant, then
2
68. Find the value of tan4384° + cot6814° = ?
tanx = ?
tan4384° + cot6814° dk eku Kkr dhft,A
1
;fn cos x   , x rhljs prqFkk±'k esa fLFkr gS]=rks
tanx ? SSC CGL 26/07/2023 (Shift-3
2
(a) –1 (b) 2
SSC CGL 17/07/2023 (Shift-03)
(c) 0 (d) 1

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs8

 7  3
cos570º sin510º + sin(–330º) cos(–390º) 70. Find is the value of sin sin sin
4 4 4
dk eku Kkr dhft,A
5
(a) 1 sin .
4
(b) 0
7  3 5
(c) 2 sin sin sin sin dk eku Kkr djsaA
4 4 4 4
(d) NOT
RRB JE 22/05/2019 (Shift-02

1 3
(a) (b)
4 16

1 1
(c) (d)
8 16

Answer Key

r
si
1.(a) 2.(a) 3.(b) 4.(b) 5.(b) 6.(b) 7.(a) 8.(a) 9.(b) 10.(b)

11.(a) 12.(b)
an by
13.(c) 14.(c) 15.(d) 16.(b) 17.(c) 18.(d) 19.(d) 20.(b)

n
21.(a) 22.(d) 23.(b) 24.(d) 25.(b) 26.(d) 27.(b) 28.(a) 29.(a) 30.(c)
ja
R s
31.(d) 32.(a) 33.(a) 34.(c) 35.(b) 36.(c) 37.(b) 38.(b) 39.(d) 40.(d)
a th

41.(d) 42.(b) 43.(b) 44.(c) 45.(b) 46.(b) 47.(c) 48.(b) 49.(c) 50.(c)
ty a

51.(a) 52.(c) 53.(c) 54.(c) 55.(c) 56.(d) 57.(a) 58.(a) 59.(d) 60.(a)
di M

61.(c) 62.(a) 63.(a) 64.(c) 65.(d) 66.(d) 67.(c) 68.(c) 69.(b) 70.(d)
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs9

 dks.kfefr
TRIGONOMETRY /f=k
(CLASSROOM SHEET-02)
If  +  = 90º   and  are complimentary. cos 50 3 cosec80°
6. What is the value of  –
tan × tan = 1, cot × cot = 1, sin 40 sec10
cos50°. cosec 40°?
sin × sec = 1, cos × cosec = 1
cos 50 3 cosec80°
1. 2(sin 1° × sec 89°) + 3 (cos 11° × cosec 79°) +  – 2 cos50°. cosec 40° d
5 (tan 21° × tan 69°) = ? sin 40 sec10
SSC CPO 04/10/2023 (Shift-02) eku Kkr dhft,A
(a) 20 (b) 12 SSC CGL MAINS (08/08/2022

r
(c) 11 (d) 10 (a) 3 (b) 4
2. The value of tan5° tan25° tan45° tan65° tan85°

si
(c) 2 (d) 5
is equal to ______.
7. The value of
tan5° tan25° tan45° tan65° tan85° d k eku______
ds cjkcj gSA an by tan13 tan 36 tan 45 tan 54 tan 77
is:

n
SSC CGL 24/07/2023 (Shift-2) 2 sec ²60(sin ²60 – 3 cos 60  2)
(a) 4 (b) 3
ja tan13 tan 36 tan 45 tan 54 tan 77
R s
(c) 1 (d) 2 dk ek
2 sec ²60(sin ²60 – 3 cos 60  2)
3. The value of cot 13° cot 27° cot 60° cot 63° cot
a th

77° is: Kkr djsaA

cot 13° cot 27° cot 60° cot 63° cot 77° dk eku gS% SSC CGL 20/08/2021 (Shift- 03
SSC CHSL 02/08/2023 Shift-02
1 1
ty a

(a) – (b) –
1 4 10
(a) (b) 0
3
di M

1 1
(c) (d) 1 (c) (d)
3 10 4
sin 58 sin 55 sec 35 3 sin 58 3 sin 42
4. The value of 
cos 32 tan 5 tan 45 tan 85 8. What is the value of + ?
cos 32 cos 48
is equal to:
3 sin 58 3 sin 42
sin 58

sin 55 sec 35
dk eku cjkcj gS% cos 32 + cos 48 dk eku Kkr dhft,A
cos 32 tan 5 tan 45 tan 85
SSC CHSL 07/08/2023 Shift-04 SSC CGL MAINS (08/08/2022
(a) 2 (b) 1 (a) 9 (b) 6
A

(c) 0 (d) 3 (c) 7 (d) 8

9. If cos2A = sin 75°, then the smallest positiv
cos 37
5. Find the value of – cos 47° cosec 43°. value of A is:
sin 53
; fn cos2A = sin 75° gS] rksA dk lcls NksVk
cos 37
– cos 47° cosec 43° d k eku Kkr dhft,A ukRed eku D;k gksxk\
sin 53
SSC CHSL 27/05/2022 (Shift- 02
SSC CPO 04/10/2023 (Shift-01)
(a) –1 (b) 1 (a) 15° (b) 7.5°
(c) 2 (d) 0 (c) 30° (d) 37.5°

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 Kkr dhft, cos (40° + x) = sin 30° rksx dk eku ___ gSA then find the value of 2sin²(3x + 15)° – cosec
(2x + 10)°.
SSC CGL 25/07/2023 (Shift-04)
(a) 19° (b) 20° ;fn sin(20 + x)° = cos60°, 0 (20 + x)  90
(c) 23° (d) 22° rks2sin²(3x + 15)° – cosec2 (2x + 10)° dk ek
11. If tan 2 = cot ( – 36°), where 2 is an acute Kkr dhft,A
angle, then the value of  is: SSC CGL 20/08/2021 (Shift- 01
;fn tan 2 = cot ( – 36°), tgk¡ 2 ,d U;wu dks.k gS] (a) 3 (b) –3
rks dk eku Kkr dhft,A
1
SSC CHSL, 11/08/2023 (Shift-2) (c) –2 (d) –
3
(a) 18° (b) 36°
17. If sin3A = cos (A+10º), where 3A is an acut
(c) 30° (d) 42°
3A
12. If cos(2+ 54°) = sin, 0° < (2+ 54°) < 90°, angle then what is the value of 2 cosec
2
1 3
then what is the value of ?
6tan2 3A – tan23A?

r
5
cot 5 sec 2
2
;fn sin3A = cos (A+10º) tgka 3A U;wudks.k gS

si
;fn cos(2+ 54°) = sin, 0° < (2 + 54°) <
3A 3
1 + 6tan2 3A – tan23A dk eku Kkr dhft,
90° gS rks
an by
cot 5 sec
5
2
dk eku Kkr dhft,A cosec
2 2
SSC CGL 2019 Tier-II (15/10/2020

n
SSC CGL 20/08/2021 (Shift- 02) 7 35
(a) (b)
3
ja 1 4 2
R s
(a) (b)
2 3 17
(c) (d) 5
a th

3 2
(c) (d) 2 3 18. If cos = sin(2)  0, what is the value of cos4
3
13. If sin (5x – 25°) – cos (5y + 25°), where 5x – 25° sin4 + cos3 + sin3 + sin2 + cos2 + sin + cos
and 5y + 25° are acute angles, then the value ;fn cos = sin(2)  0, rks cos4 + sin4 + cos3
ty a

of (x + y) is:
+ sin3 + sin2 + cos2 + sin + cos dk eku D;k
;fn sin (5x – 25°) – cos (5y + 25°) gS] tgk¡5x – 25°
di M

SSC Phase X 05/08/2022 (Shift- 03

vkSj5y + 25° U;wudks.k gS] (xrks
+ y) dk eku D;k gksxkA
SSC CGL 18/07/2023 (Shift-03) 18 + 8 3 8+7 3
(a) (b)
(a) 50° (b) 40° 7 18
(c) 18° (d) 16°
14. If tan (5 – 10°) = cot (5 + 20°), then the value 7 +18 3 18 + 7 3
(c) (d)
of  +  is : 8 8
;fn tan (5 – 10°) = cot (5 + 20°) gS] rks
 +  dk eku
If  +  = 90º, then
D;k gS\
SSC CHSL 10/08/2023 (Shift-2) sin2 + sin2 = 1 &
(a) 16° (b) 18° cos2 + cos2 = 1
A

(c) 15° (d) 20°

15. If sec(5–15°) = cosec(15°–2), then the value 19. Simplify the given expression.
of cos + sin2+ tan(1.5) is: fn, x, O;atd dks ljy dhft,A
;fn sec(5–15°) = cosec(15°–2) gS rkscos +
sin 2 32  sin 2 58
sin2+ tan(1.5) dk eku Kkr dhft,A + sin253° + cos53°sin37°
cos 2 32  cos 2 58
SSC CGL 20/08/2021 (Shift- 01)
SSC CHSL 07/08/2023 Shift-0
(a) 2 1 (b) 2 –1 (a) 2 (b) –1
(c) 3 –1 (d) 3 1 (c) –2 (d) 1

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 25. The value of
5cos 2 62° + 5cos 2 28° – 21
20. The value of is:  cos 9  sin 81   sec 9  cosec 81 
7sin 2 35° + 7sin 2 55° + 1
is:
cosec2 71  cos 2 15  tan 2 19  cos 2 75
5cos 2 62° + 5cos 2 28° – 21
dk eku Kkr dhft,A  cos 9  sin 81   sec 9  cosec 81 
7sin 2 35° + 7sin 2 55° + 1
cosec2 71  cos 2 15  tan2 19  cos 2 75 dk ek
SSC CGL 20/04/2022 (Shift- 03)
(a) 3 (b) – 2 Kkr dhft,A
(c) 2 (d) – 3 SSC CGL MAINS 03/02/202
(a) 1 (b) 4
2 tan(60º –) tan(30 ) (c) – 3 (d) 2
21. The value of :
sin ²(45 )  sin ²(45 – ) 26. If x = sec57º, then cot233º + sin257º + sin233º
cosec257ºcos233º + sec233º.sin257º is equal to
2 tan(60º –) tan(30 ) ;fn x = sec57º gS rks
cot233º + sin257º + sin233
sin ²(45 )  sin ²(45 – )
dk eku Kkr dhft,A
+ cosec257ºcos233º + sec233º.sin257º dk eku K
SSC CGL 17/08/2021 (Shift- 02) dhft,A

r
1 SSC CGL TIER-II (16/10/2020
(a) x2 + 2 (b) 2x2 + 1

si
(a) (b) 1
2
1
(c) 2 (d) (c) x2 + 1 (d)
22. an by 2
The value of (sin37° cos53° + cos37° sin53°) 27.
x²  1
Find the value of the following.

n
4 cos ²37 – 7  4 cos ²53 fuEufyf[kr dk eku Kkr dhft,A
– is:
tan ²47  4 – cos ec²43 sin 67 cos 37 – sin 37 cos 67
( si n 37 ° ja
c o s5 3 ° + c o s3 7 ° s in 5 3° )
R s
cos13 cos17 – sin13 sin17
SSC CHSL 27/05/2022 (Shift-03
4 cos ²37 – 7  4 cos ²53
a th

– dk eku Kkr dhft,A 1 4

tan ²47  4 – cos ec²43 (a) (b)
3 3
SSC CGL 18/08/2021 (Shift- 01)
2
(c) (d) 7
ty a

(a) 1 (b) –2 3
(c) 0 (d) 2 sin 2 52º  2  sin 2 38º
di M

28. The value of is:

23. The value of 4 cos 2 43º –5  4 cos 2 47º
4 tan ²30  sin ²30 cos ²45  sec ²48 – cot ²42 sin2 52º  2  sin 2 38º
cos 37 sin 53  sin 37 cos 53  tan18 tan 72 4 cos 2 43º – 5  4 cos 2 47º
dk eku Kkr djasA
4 tan ²30  sin ²30 cos ²45  sec ²48 – cot ²42
dk SSC CPO 24/11/2020 (Shift-2
cos 37 sin 53  sin 37 cos 53  tan18 tan 72
eku Kkr dhft,A (a) 3 (b)
1
SSC CGL MAINS 29/01/2022 3
1
35 59 (c) – (d) –3
(a) (b) 3
48 48 29. The value of
2sin 2 38º sec 2 52º+cos64º sin26º +sin 2 64º
A

49 35
(c) (d) tan 2 23º +cot2 23º – sec2 67º – cosec2 67º is :
24 24
2sin 2 38º sec 2 52º+cos64º sin26º +sin 2 64º
24. The value of
5cos 2 62° + 5cos 2 28° – 21
is: tan 2 23º +cot2 23º – sec2 67º – cosec2 67º dk eku K
7sin 2 35° + 7sin 2 55° + 1 dhft,A
5cos 2 62° + 5cos 2 28° – 21
dk eku Kkr dhft,A SSC CGL 2019 TIER-II (15/10/2020
7sin 2 35° + 7sin 2 55° + 1 3
SSC CGL 20/04/2022 (Shift- 03) (a) – 2 (b)
2
(a) 3 (b) – 2 3
(c) 2 (d) – 3 (c) 2 (d) –
2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 30. The value of 3(cot ²47 – sec ²43 ) – 2(tan ²23 – cos ec²67)
3(cos ec²26 – tan ²64)  (cot ²42 – sec ²48) cos ec²(68  ) – tan(  61) – tan ²(22 – )  cot(29 – )
cot(22 –  ) – cos ec²(62   ) – tan(  68)  tan ²(28 –  )
dk eku Kkr dhft,A
3(cos ec²26 – tan ²64 )  (cot ²42 – sec ²48)
SSC CGL MAINS 29/01/202
cot(22 –  ) – cos ec²(62   ) – tan(  68)  tan ²(28 –  )
(a) 1 (b) 0
dk eku Kkr dhft,A
(c) 5 (d) –1
SSC CGL MAINS 03/02/2022
(a) 3 (b) 4 32. The value of/dk eku crkb,A
(c) –1 (d) –2 sin23°cos67° + sec52°sin38° + cos23°sin67° + cosec52°cos38°
31. The value of cosec²20° – tan²70°
SSC CGL 11/04/2022 (Shift- 01
3(cot ²47 – sec ²43) – 2(tan ²23 – cos ec²67)
cos ec²(68  ) – tan(  61) – tan ²(22 – )  cot(29 – ) (a) 3 (b) 4
(c) 2 (d) 0

r
si
an by ANSWER KEY

n
1.(d) 2.(c) 3.(a) 4.(a) 5.(d) 6.(c) 7.(c) 8.(b) 9.(b) 10.(b)

ja
R s
11.(d) 12.(c) 13.(a) 14.(a) 15.(d) 16.(b) 17.(b) 18.(d) 19.(a) 20.(b)
a th

21.(d) 22.(d) 23.(b) 24.(b) 25.(d) 26.(a) 27.(a) 28.(d) 29.(d) 30.(d)
ty a

31.(d) 32.(a)
di M
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs4

 TRIGONOMETRY /f=kdks.kfefr
(CLASSROOM SHEET-03)
TRIGONOMETRIC IDENTITIES 6. For 0° <  < 90°,
1

1
cos tan – sec
is equal to:
1. Select the INCORRECT formula from the 1 1
following options. 0° <  < 90° ds fy,]  cjkcj gSA
cos tan – sec
fuEufyf[kr fodYiksa esa ls xyr lw=k dk p;u dhft,A SSC CGL 17/08/2021 (Shift- 02
SSC CGL 17/07/2023 (Shift-01) (a) –sec (b) tan
(a) sec2– tan2= 1 (c) sec (d) –tan
(b) sin2+ cos2 = 1

r
 1 1  1 1 
(c) cosec²– cot2 = 1 7. The value of      is:
 sin  tan    sin  tan  

si
(d) sec2+ cos2 = 1
2. Which of the following is equal to  1 1  1 1 
 dk eku D;k gksx
1
tan
 tan?
an by    
 sin  tan    sin  tan  
SSC CGL 18/07/2023 (Shift-02

n
1 (a) 0 (b) 2
fuEu eas ls dkSu  tan ds cjkcj gS\ (c) 3 (d) 1
ja
tan
R s
SSC CGL MAINS (08/08/2022 8. (cosecA – cotA) (1+cosA) =?
a th

SSC CGL 23/08/2021 (Shift- 02

cos ec
(a) (b) tan 2 (a) cosecA (b) cosA
sec
(c) sinA (d) cotA
(c) sec. cosec (d) 1
ty a

2
9. (cosec – cot) =, 0° <  < 90°
3. Which of the following is equal to SSC CHSL 09/06/2022 (Shift- 03
di M

 cos sin 1 – sin 1 – cos

 sin  cos ? (a)
1  cos
(b)
1  cos
 
1  cos 1  cos
 cos sin (c) (d)
fuEu eas ls dkSu
 sin  cos ds cjkcj gS\
1 – cos 1 – sin
  10. The given expression is equal to:
SSC CGL MAINS (08/08/2022) (1  tan 2 A)
(a) cot.sec (b) cosec.sec cos ec2 A. tan A
(c) sec.tan (d) cosec. tan (1  tan 2 A)
nh xbZ vfHkO;fÙkQ blds cjkcj
4. Which of the following is equal to secA – cosA? cos ec2 A. tan A
fuEu eas ls dkSu
secA – cosA ds cjkcj gS\ SSC CHSL 03/08/2023 (Shift-03
2
(a) sec A (b) sec A
A

SSC CGL MAINS (08/08/2022)

(c) tan A (d) tan A
(a) sinA. cotA (b) cotA. cosA
11. If  is an acute angle and sin  + cosec =2
(c) tanA. sinA (d) cosA sinA then the value of sin99 + cosec99 is:
5. [(sin  tan  + cos )² – 1] is equal to: ;fn  U;wu dks.k gS vkSj
sin  + cosec  = 2 gS
[(sin  tan  + cos )² – 1] buesa ls fdlds cjkcj gS\ dk eku D;k gksxk\
sin99 + cosec99
SSC Phase X 02/08/2022 (Shift- 02) SSC Phase X 02/08/2022 (Shift- 02
(a) sec²  (b) sec  (a) 1 (b) –1
(c) cosec  (d) tan²  (c) 0 (d) 2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 1  5 –1
12. If cosec2 + cot2 , where 0 <  < , then 16. The value of sin18° is given as . Fin
3 2 4
the value of cosec4 – cot4is: the value of cosec18°.

1  5 –1
;f osec2 + cot2 = gS] tgk¡0 <  < gS] rks sin18° dk eku ds :i esa fn;k x;k g
3 2 4
cosec4 – cot4 dk eku gSA cosec18° dk eku Kkr dhft,A
SSC CHSL 09/06/2022 (Shift- 01) SSC CHSL 06/06/2022 (Shift- 02

5 –1 5 1
2 1 (a) (b)
(a) (b) – 4 2
3 3
(c) 5 –1 (d) 5 1
1 2 1  sin t 1  sin t
(c) (d) – 17. Simplify: 
3 3 4 – 4 sin t 4  4 sin t

r
13. If cosec + cot = 5, then cosec is equal to 1  sin t 1  sin t
_________.  dk eku Kkr djsaA

si
4 – 4 sin t 4  4 sin t
;fn cosec + cot = 5 gS] rkscosec dk eku Kkr SSC CGL 21/07/2023 (Shift-04
dhft,A
an by
SSC CHSL 03/06/2022 (Shift- 01)
(a) 4tant.sint
(c) tant – sint
(b) tant.sect
(d) tant + sint

n
18. Which of the following is equal t
 1 1 1  tan sec– 1
5  
(a)  (b) 3  

 tan– sec 1 ?
 5
ja 2 3
 
R s
1 1  1  tan sec– 1
fuEu eas ls dkSu
 tan– sec 1 ds cjkcj gS\
a th

(c) 5  
 3  
(d)   
2 5  3
SSC CGL MAINS (08/08/2022
14. If tan + cot = 4, then find the value of tan2
1  cos 1  cot
ty a

+ cot2. (a) (b)

sin tan
;fn tan + cot = 4 gS] rkstan2 + cot2 dk eku 1  sin 1  tan
Kkr djsaA
di M

(c) (d)
cos cot
SSC CHSL 30/05/2022 (Shift- 01) 2
 1 - cotθ 
(a) 10 (b) 12 19. The value of   –1 when 1° << 90° , i
 1 - tanθ 
(c) 16 (d) 14 equal to:
2
 1 - cotθ 
15. The value of sin18° is given as
5 –1
. Using   –1 dk eku cjkcj gS] tgka1° << 90°
4  1 - tanθ 
this value, find the value of cos18°. SSC CGL 12/04/2022 (Shift- 02

(a) cos² – 1 (b) sec²+1

5 –1
sin18° dk eku ds :i esa fn;k x;k gSA bl (c) cot²–1 (d) sin²–1
4
A

eku dk mi;ksx djrs gq,cos18° dk eku Kkr dhft,A cotθ + cosθ

20. The value of 1+ , if 0° <<90°, i
SSC CHSL 30/05/2022 (Shift- 2) cotθ - cosθ
equal to:
10  2 5 5 1 cotθ + cosθ
(a) (b) 1+ dk eku cjkcj gS] ;fn 0° <<90
4 4 cotθ - cosθ
SSC CGL 12/04/2022 (Shift- 02
6 2 10 – 2 5 (a) 1– sec+ tan (b) 1 – sec– tan
(c) (d)
4 4 (c) 1+sec- tan (d) 1 + sec+ tan

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

  sinA 1- cosA   cot 2 A  sec 2 θ(2 + tan 2 θ + cot 2 θ) ÷ (sin 2 θ - tan 2 θ) is:
 +  ÷  1+ cosecA + 1  is : (cosec 2 θ + sec 2 θ)(1 + cot 2 θ)2
 1- cosA sinA   

 sinA 1- cosA   cot 2 A  SSC CGL 2019 Tier-II (15/10/2020

 +  ÷  1+ cosecA + 1  dk eku Kkr
 1- cosA sinA    (a) –1
djsa (b) 1
SSC CGL Tier-II (12/09/2019)
(c) – 2
3 1 (d) 2
(a) (b)
2 2
(c) 1 (d) 2 sinθ + cosθ - 1 1 + sinθ
28. The value of ×
sinθ - cosθ + 1 1 - sinθ
22. What is value of the expression
vfHkO;fÙkQ dk ewY; D;k gS\ sinθ + cosθ - 1 1 + sinθ
dk eku Kkr djsaA

r
×
 sin A   cos A  sinθ - cosθ + 1 1 - sinθ
sin A  1    cos A  1  

si
 cos A   sin A  SSC CGL 2019 Tier-II (16/10/2020
SSC CHSL 04/08/2023 Shift-01 (a) 1 (b) – 1
an by
(a) secA + cosecA (b) sinA + cosA (c) – 2 (d) 2

n
(c) sinA – cosA (d) secA – cosecA
29. I f x sin  + y cos  = sin.cos and x sin =
3 3

23. Find the value cos then the value of x2 + y2 is :

ja ;fn x sin3 + y cos3 = sin.cos rFkkx sin
R s
(sinθ - cosθ) - (1 + tanθ + cotθ)
=?
1 + sinθcosθ y cos gS rksof x2 + y2 dk eku Kkr dhft,A
a th

(sinθ - cosθ) - (1 + tanθ + cotθ) SSC CHSL 20/10/2020 (Shift- 01

1 + sinθcosθ
dk eku Kkr djsa\
(a) 0 (b) 4
ty a

SSC CGL Tier-II (13/09/2019) (c) 1 (d) 2

(a) sec– cosec (b) cosec– sec 30. Simplify the following expression:
di M

(c) sin+ cos (d) tan+ cot fuEUkfyf[kr O;atd dks ljy dhft,A
24. The value of (cosecA + cotA+1)(cosecA–cotA + 1)
– 2cosecA is cos A sin A
 – sin A
1 – tan A 1 – cot A
(cosecA + cotA+1)(cosecA–cotA + 1) – 2cosecA dk
eku cjkcj gSA SSC CGL 19/04/2022 (Shift- 03

SSC CGL 09/03/2020 (Shift- 02) (a) 1 + sinA (b) (1 + sinA) cosA
(a) 4cosecA (b) 2 (c) cosA (d) 1 + cosA
(c) 2cosecA (d) 0
cosecθ cosecθ
25. cosA(secA – cosA)(cotA + tanA) = ? 31.  – tan², 0° < < 90
cosecθ – 1 cosecθ  1
A

SSC CGL 2019 Tier-II (15/10/2020) is equal to:

(a) secA (b) cotA
cosecθ cosecθ
(c) sinA (d) tanA  – tan², 0° < < 90° d
cosecθ – 1 cosecθ  1
 1 1 
26.  -  +
1
-
1
=? eku Kkr djsaA
 cosθ sinθ  cosecθ - cotθ secθ + tanθ
SSC CGL 13/08/2021 (Shift- 01
SSC CGL 2019 Tier-II (15/10/2020)
(a) sec.cosec (b) sin.tan (a) 2 sec2 (b) sec² + 1
(c) cosec.cot (d) sin.cos (c) sec² (d) 1 – tan²

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 cos  cos  (sin sec)2 (cos cosec)2
32.  is equal to:
sec   1 sec   1 37. Simplify: ,
(1  seccosec)2
cos  cos  << 90° is equal to:
 ds cjkcj gS%
sec   1 sec   1
(sin sec)2 (cos cosec)2
SSC CHSL 02/08/2023 Shift-01
(1  seccosec)2
, 0 << 90° ljy
2 2
(a) 2sec  (b) 2cot 
2
djsaA
(c) 2cos  (d) 2sin2
SSC CGL 24/08/2021 (Shift- 03
33. 1 + 2 tan2+ 2 sin sec2, 0°<  <90°, is
(a) 0 (b) 2
equal to :
(c) –1 (d) 1
1 + 2 tan2+ 2 sin sec2, 0°<  <90° dk eku
Kkr djsaA  tan ³

cot ³ 
 2 sin  cos   ÷ (1 + cosec²
38. 
SSC CGL 16/08/2021 (Shift- 01)  sec ²  cosec²  
+ tan²), 0° < < 90° is equal to:

r
1 – sin 1  cos
(a) (b)

si
1  sin 1 – cos  tan ³ cot ³ 
   2 sin  cos   ÷ (1 + cosec²
 sec ²  cosec²  

(c)
1 – cos
1  cos
an by (d)
1  sin
1 – sin
+ tan²), 0° < < 90° dk eku Kkr djsaA
SSC CGL MAINS 29/01/202

n
34. If 1 + 2tan² + 2sin sec²= a/b, 0² << 90°, (a) sincos (b) cosec
a b
ja (c) sec (d) cosecsec
R s
then =?
a –b 39. Let 0° <  < 90°, (1 + cot²)(1 + tan²) × (sin
– cosec)(cos – sec) is equal to:
a th

;fn 1 + 2tan² + 2sin sec²= a/b, 0² <<

;fn 0° <  < 90°, (1 + cot²)(1 + tan²) × (sin
a b
90° gS rksa – b dk eku Kkr dhft,A – cosec)(cos – sec) dk eku Kkr djsaA
ty a

SSC CGL MAINS 29/01/202

SSC CGL MAINS 29/01/2022 (a) seccosec (b) sec + cosec
di M

(a) sin (b) cos (c) sin + cos (d) sincos

(c) cosec (d) sec
(1  sec  cos ec)2 (sec  – tan )2 (1  sin )
40. (sin   sec )2  (cos   cos ec)2
, 0°
cot ³ tan ³
35.  + 2sin cos = ?
cosec2 sec2 <  < 90°, is equal to:
SSC CGL 16/08/2021 (Shift-02)
(1  sec  cos ec)2 (sec  – tan )2 (1  sin )
(a) cosecsec (b) cosec²sec² (sin   sec )2  (cos   cos ec)2
d
(c) sincos (d) sin
eku Kkr djsaA
 sin(1  cos)  SSC CGL MAINS 29/01/202
sec ² cosec2 
36.    2
1  cos– sin 

,
(a) 1 – cos (b) 1 – sin
A

0° < < 90° is equal to: (c) cos (d) sin

 sin(1  cos)  1  cos  – sin ²  sec ²   cos ec 2 

sec ² cosec2  41.  , 0° < 
   2

 , 0° <  <
1  cos – sin  sin (1  cos ) tan   cot 

90° dk eku Kkr djsaA 90°.

SSC CGL MAINS 29/01/202
SSC CGL 17/08/2021 (Shift- 01)
(a) tan (b) sec
(a) cot (b) cosec²
(c) cosec (d) cot
(c) sec² (d) tan

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs4

 42. The expression (tan + cot) (sec + tan)(1
– sin), 0° <A < 90°, is equal to: what is the value of 1 + sec  + tan  ?
O;atd (tan + cot) (sec + tan)(1 – sin), ;fn sin2 – cos2 – 3sin +2 = 0, 0° <  < 90° fn;
0° <A < 90° dk eku Kkr djasA gS rks
1 + sec  + tan  dk eku Kkr dhft,A
SSC CGL MAINS 03/02/2022 SSC CGL 13/04/2022 (Shift- 03
(a) sec (b) cosec (a) –1 – 3 (b) –1 + 3
(c) cot (d) sin
(c) 1 + 3 (d) 1 – 3
43. The expression
48. If 7sin² + 4cos² = 5 and  lies in the firs
tan 6  – sec 6   3 sec 2  tan ² 
0º <  < 90º, is quadrant, then what is the value o
tan ²   cot ²  + 2
equal to: 3 sec   tan 
?
2 cot  – 3 cos 
tan 6  – sec 6   3 sec 2  tan ² 
O;atd 0º <  < 90º
tan ²   cot ²   2 ;fn 7sin² + 4cos² = 5 and igys prqFkkZa'k
dk eku Kkr djsaA

r
3 sec   tan 
SSC CGL MAINS 03/02/2022 gS] rks dk eku D;k gksxk\

si
2 cot  – 3 cos 
(a) sec²cosec² (b) –sec²cosec²

44.
(c) cos²sin²
an by (d) –cos²sin²
If 4sin²= 3 (1+cos), 0° < < 90°, then what (a) 2(1  2)
SSC CGL MAINS 29/01/202

(b) 3 2

n
is the value of (2tan+4sin– sec)?
;fn 4sin²= 3 (1+cos), 0° < < 90° gS rks (c) 2( 2 – 1) (d) 4 2

ja
2tan+4sin–sec) dk eku Kkr dhft,A
R s
SSC CGL 11/04/2022 (Shift- 01) sin 2 
49. If = 1,  lies in the firs
a th

(a) 3 15 - 4 (b) 15 3 - 4 cos ²  – 3 cos   2

(c) 15 3 + 3 (d) 4 15 - 3  
tan2  sin ²
quadrant, then the value of 2 2 is:
45. If tan²A + 2tanA – 63 = 0 Given that 0 < A <
ty a

tan   sin 
π
what is the value of (2sinA+5cosA)?
4 sin 2 
di M

;fn = 1,  prqFkkZa'k esa

π cos ²  – 3 cos   2
;fn tan²A + 2tanA – 63 = 0 fn;k gS0 < A < ,
4  
rks(2sinA+5cosA) dk eku Kkr dhft,A rks tan2  sin ² dk eku Kkr dhft,A
2 2
SSC CGL 11/04/2022 (Shift- 03) tan   sin 
(a) 19 50 (b) 15 50 SSC CGL MAINS 03/02/202

19 15 2 3 5 3
(c) (d) (a) (b)
50 50 27 27
46. If 3sec²+tan–7 = 0, 0°<90°, then what
2 3 7 3
A

 2sin + 3cos  (c) (d)

is the value of  ? 9 54
 cosec + sec 
50. If sec + tan = p, then if sec is equal to:
;fn 3sec²+tan–7 = 0, 0°<90° fn;k gS rks
;fn sec + tan = p gS] rkssec dk eku Kkr djs
 2sin + 3cos 
  dk eku Kkr dhft,A  1  1
 cosec + sec  (a)  P –  , P  0 (b) 2  P –  , P  0
P  P
SSC CGL 12/04/2022 (Shift- 01)
 1 1 1
(a) 10 (b) 5/2 (c)  P +  , P  0 (d)  P +  , P  0
 P 2 P
(c) 5/4 (d) 4 2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs5

 25
51. If sec2 + tan2 =
18
, then the value of sec4 – Questions Based on Condition
tan  is:
4
As in  + Bcos = X;
25
;fn sec2 + tan2 = gS] rks
sec4 – tan4dk eku
18 Acos – Bsin = Y then
Kkr djsaA
SSC CGL 14/07/2023 (Shift-3) A² + B² = X² + Y²
18 25
(a)
25
(b)
12 a sin + b cos = a2 + b2
25 25
(c) (d) 55. If 3sin θ + 4cos θ = 5, then tan θ = ?
9 18

52. If cosec  – cot  =

7
, the value of cosec  is: ; fn 3sin+ 4cos= 5 gS rks
tan dk eku Kkr dhft
2
3 5

r
7 (a) (b)
; fn cosec  – cot  = gS] rks
 d k eku Kkr djasA 4 4

si
2

47 51 4 3
(a)
28 an by (b)
28
(c)
5
(d)
5

n
53 49 56. If 2 xy sin θ + (x2 – y2) cos θ = x2 + y2, then tan θ
(c) (d)
28 28 ?
53.
ja
If sec + tan = 2 + 5, then the value of sin
; fn 2xy sin+ (x2 – y2) cos= x2 + y2 gS rkstan
R s
is (0º  90º)
d k eku Kkr dhft,A
a th

; fn sec + tan = 2 + 5 gS] rks

sin dk eku Kkr
djsaA(0º  90º) 2xy 2xy
(a) (b)
3 x 2 - y2 x + y2
2
ty a

(a)
2
x 2 - y2
di M

2 (c) (d) 1
(b)
5 x 2 + y2

1 57. If 12sin+ 35cos= 37, then cosec= ?

(c)
5 ; fn 12sin+ 35cos= 37 gS rks
cosec dk eku
4 Kkr dhft,A
(d)
5
12 37
54. If sec + tan = 2+5, then the value of sin (a) (b)
37 12
+ cos is:
35 12
; fn sec + tan = 2+5 gS] rkssin + cos dk (c) (d)
37 35
eku Kkr djsaA
A

58. (a2 – 1)sin θ + 2a cos θ = a2 + 1, then tan θ = ?

3
(a)
5
(a2 – 1)sin+ 2a cos= a2 + 1 gS rkstan dk ek
Kkr dhft,A
(b) 5
7 a2 - 1 2a
(a) (b)
(c) 2a a2 - 1
5
1 a 2 +1
(d) (c) (d) 1
5 a2 - 1

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs6

 3cosx – 4sinx is equal to : 11(cos4 + cos²)?
;fn 3 sinx + 4 cosx = 2 gS rks3cosx – 4sinx dk ;fn tan4 + tan2 = 1, rks11(cos4 + cos²) dk ek
eku Kkr dhft,A D;k gksxk\
SSC CGL 2019 Tier-II (18/10/2020) SSC CGL 27/07/2023 (Shift-3
(a) 21 (b) 23 (a) –11 (b) 8
(c) 21 (d) 29 (c) 0 (d) 11
65. If cos x + cos² x = 1, the numerical value o
1 (sin¹²x + 3 sin10x + 3 sin8x + sin6x – 1) is:
60. If 2 cos θ – sin θ = , (0º < θ < 90º) the value
2
;fn cos x + cos² x = 1 gS] rks(sin¹²x + 3 sin10x
of 2sin θ + cos θ is :
3 sin8x + sin6x – 1) dk vkafdd eku gSA
1 (a) –1 (b) 2
;fn 2 cos – sin  = , (0º <  < 90º) gS rks
2
(c) 0 (d) 1

r
2sin + cos dk eku Kkr dhft,A
66. If sin + sin²  = 1, then the value of cos²

si
1 cos4 is:
(a) (b) 2
2 ;fn sin + sin²  = 1 gS] rkscos² + cos4 d
3
an by 3
eku Kkr djsaA

n
(c) (d) (a) 2 (b) 4
2 3
(c) 0 (d) 1

ja 11 67. If sin A+ sin A = 1, then the value of cos4 A

2
R s
61. If si n  + cos = , then the value of (cos –
3 cos6 A is:
a th

sin) is: ;fn sin A+ sin2 A = 1 gS] rks

cos4 A + cos6 A d
eku D;k gksxk\
11
;fn sin + cos = , rks(cos – sin) dk eku gS% SSC CPO 04/10/2023 (Shift-01
3
ty a

(a) 1 (b) cos A

SSC CPO 03/10/2023 (Shift-3)
(c) sin A (d) 0
di M

5 7 68. If sin  + sin²  = 1 then the value of cos 1

(a) (b)
3 3 + 3cos 10  + 3 cos 8 + cos 6 – 1 is:
;fn sin  + sin²  = 1 gS] rkscos12 + 3cos10
5 7
(c) (d) 3 cos8 + cos6 – 1 dk eku Kkr djsaA
3 3
(a) 1 (b) 2
(c) 3 (d) 0
62. If 3 sin  + 5 cos  = 5, then 5sin  – 3cos  is
equal to: 69. If tan A + cot A = 2, then the value of tan10
+ cot10 A is:
;fn 3 sin  + 5 cos  = 5 gS rks5sin  – 3cos 
;fn tan A + cot A = 2 gS] rkstan10 A + cot10 A d
dk eku Kkr djsaA
eku Kkr djsaA
A

(a) ± 3 (b) ± 5
(a) 4 (b) 2
(c) 1 (d) ± 2
10
63. If cosA + cos A = 1, then the value of sin4A
2 (c) 2 (d) 1
+sin6A is : 70. If sin  + cosec  = 2, then the value of sin1
;fn cosA + cos2A = 1, rkssin4A +sin6A dk eku D;k  + cosec100  is equal to:
gksxk\ ;fn sin  + cosec  = 2 gS] rkssin100  + cosec1
 dk eku Kkr djsaA
SSC CGL 26/07/2023 (Shift-1)
(a) 2 (b) cosA (a) 1 (b) 2
(c) 1 (d) sinA (c) 3 (d) 100

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs7

 + cosec9  is: cotn  (0º   < 90º, n is an integer) is:
; fn sin  + cosec  = 2 gS] rks
sin9  + cosec9  ; fn tan  + cot  = 2 gS] rks
tann  + cotn  dk ek
d k eku Kkr djsaA Kkr dhft,A (0º   < 90º, n ,d iw.kkZad gSaA)
(a) 3 (b) 2 (a) 2 (b) 2n
(c) 4 (d) 1 (c) 2n (d) 2n+1

ANSWER KEY
1.(d) 2.(c) 3.(b) 4.(c) 5.(d) 6.(d) 7.(a) 8.(c) 9.(b) 10.(c)

11.(d) 12.(c) 13.(c) 14.(d) 15.(a) 16.(d) 17.(b) 18.(c) 19.(c) 20.(d)

r
21.(d) 22.(a) 23.(a) 24.(b) 25.(b) 26.(a) 27.(a) 28.(a) 29.(c) 30.(c)

si
31.(b) 32.(b) 33.(d) 34.(c) 35.(a) 36.(c) 37.(d) 38.(a) 39.(a) 40.(b)

41.(d) 42.(b)
an by
43.(d) 44.(a) 45.(c) 46.(c) 47.(c) 48.(a) 49.(d) 50.(d)

n
51.(d) 52.(c) 53.(b) 54.(a) 55.(a) 56.(a) 57.(b) 58.(a) 59.(a) 60.(c)
ja
R s
61.(d) 62.(a) 63.(b) 64.(d) 65.(c) 66.(d) 67.(c) 68.(d) 69.(b) 70.(b)
a th

71.(b) 72.(a)
ty a
di M
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs8

 TRIGONOMETRY /f=kdks.kfefr
(CLASSROOM SHEET-04)
Multiple and Sub-multiple 1. If 2(cos2 – sin2) = 1 ( is a positive acut
angle), then cotis equal to :
Trigonometric function of 2A
1
2A ds xq.kd rFkk mixq.kd f=kdks.kferh; iQyu (a) – 3 (b)
3
2 tan A
 sin 2A  2 sin A. cos A  (c) 1 (d) 3
1  tan2 A
A sin2θ - 2sin4 θ
2 tan 2. The value of sec2 – is :
A A 2 2cos4 θ - cos2 θ
sin A  2 sin cos 

r
 2 2 2 A
1  tan (a) 1 (b) 2
2

si
(c) – 1 (d) 0
 cos 2A  cos2 A  sin2 A  2 cos2 A  1
3. Simplify the following expression
an by
 1  2 sin 2 A 
1  tan 2 A
1  tan 2 A
sin  2 sin 3
2 cos 3  cos

n
A A A
 cos A  cos2 – sin2  2 cos2 – 1 SSC CGL 21/07/2023 (Shift-03
2 2 2
ja (a) tan  (b) sin 
R s
A
1 – tan2 (c) sec  (d) cos 
2 A 2
= 1 – 2 sin 
a th

2 2 A
4. Evaluate the following:
1  tan
2
2  2  2  cos 8
A 2
 1 - cos A  2 sin
ty a

SSC CGL 27/07/2023 (Shift-3

2
(a) 2cos (b) 2cos2
A
 1  cos A  2 cos2
di M

(c) sin2 (d) cos2

2
2 tan A 2 tan A
 tan 2A  5. ?
1  tan2 A 1  tan2 A
A SSC CGL 20/07/2023 (Shift-02
2 tan
2 (a) cos2A (b) sinA
tan A 
 2 A (c) cosA (d) sin2A
1 - tan
2
sin 4
cot2 A  1 6. The value of (1 – cos 4) is:
 cot 2A 
2 cot A
SSC CHSL 01/06/2022 (Shift- 03
Trigonometric function of 3A
A

(a) cot  (b) cot 2

3A ds f=kdks.kferh; iQyu
(c) tan  (d) tan 2
 sin 3A  3 sin A  4 sin A 3
7. What is the value of cos²15º?
 cos 3A  4 cos3 A  3 cos A SSC CGL 06/12/2022 (Shift- 04

 tan 3A  
 3 tan A  tan3 A  2 + 3 
1  3 tan2 A  (a) (2 + 3) (b)
4
 cot3 A  3 cot A 
 cot 3A   3 cot2 A  1  (c)
2 + 3  (d)
1 + 3 
 
2 2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 4
16. What will be the value of sin10º – sin³ 10º
SSC CGL 13/12/2022 (Shift- 02) 3
(a) 0.5 (b) 4 SSC CGL 06/12/2022 (Shift-01
(c) 2 (d) 1 1 1
9. What is the value of the expression 100(sin (a) (b)
3 3 6
15° cos 15°)? 1 3
SSC CHSL 03/06/2022 (Shift- 03) (c) (d)
2 3 6
(a) 50 (b) 75
Identities based on
(c) 100 (d) 25
Sum & Difference of angle of T-Ratio
º
1
1 - tan 2 22
2 is :
f=kdks.kfefr vuqikr ds dks.kksa
10. Find the value of
1 + tan 2 22
1
º
dk ;ksx rFkk varj ij vk/kfjr loZlfedk,¡
2
 sin(A  B)  sin A cos B  cos A sin B
3 1
(a) (b)  sin(A  B)  sin A cos B  cos A sin B

r
2 2
 cos(A  B)  cos A cos B  sin A sin B

si
1
(c) (d) 3  cos(A  B)  cos A cos B  sin A sin B
2
11. an by
3 sinA – 4sin3A = ?
SSC CHSL 02/06/2022 (Shift- 01)
 tan(A  B) 
tan A  tan B
1  tan A tan B

n
(a) cot3A (b) sin3A tan A  tan B
 tan(A  B) 
1  tan A tan B
(c) cos3A
ja (d) tan3A
R s
cot A cot B  1
1  sin 3  cot(A  B) 
12. Find the value of cot B  cot A
a th

1  sin 3
cot A cot B  1
SSC CGL TIER I 21/07/2023 (Shift-02)  cot(A  B) 
cot B  cot A
(a) sec3 – tan3 17. Which of the following options gives a
ty a

(b) (sec3 – tan3)2 expression equivalent to sin(A + B)?

(c) (sec3 – tan3)3 SSC CGL 24/07/2023 (Shift-2
di M

(d) sec3 + tan3

(a) cosAcosB – sinAsinB
13. What is the value of (4sin3x – 3sinx + sin3x)?
(b) sinAcosB + cosAsinB
SSC CHSL 31/05/2022 (Shift- 02)
(c) cosAcoSB + sinAsinB
(a) 1 (b) 0
(d) sinAcosB-cosAsinB
(c) 4 (d) 3
4 15
18. If sinA = and sin B = , what is the valu
3 5 17
14. If sin , and  is an acute angle, find of sin(A – B)?
2
SSC CGL 02/12/2022 (Shift- 04
the value of cos3 .
SSC CHSL 10/06/2022 (Shift- 01) 32 13
(a) – (b) –
A

45 85
1
(a) (b) 1 13 32
2
(c) (d)
(c) – 1 (d) 0 85 45
19. What is the value of cos45º sin15º?
1
15. If sin = , then the value of (3cos – 4cos³) is: SSC CGL 07/12/2022 (Shift- 02
2
SSC CGL 06/12/2022 (Shift- 03)
(a)
 3 –1  (b)
 3 –1 
(a) 0 (b) 1 2 4
(c) 2 (d) –1 (c) (3 + 1) (d) 23 – 1

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 20. What is the value of sin75º + sin15º?
tanA?
SSC CGL 09/12/2022 (Shift- 04)
SSC CGL 02/12/2022 (Shift- 01
1 3 1 – tanB 1 + tanB
(a) (b) (a) (b)
2 2 1 + tanB 1 – tanB
1 + secB 1 – cosecB
3 3 (c) (d)
(c) (d) 1 – secB 1 + cosecB
2 2
21. Evaluate the following Transformation of sum or
sin25° sin65° – cos25° cos65°. difference into product
SSC CGL 14/07/2023 (Shift-2)
;ksx vFkok vUrj dk xq.ku esa ifjorZ
(a) 40 (b) 4
(c) 0 (d) 1 C  D C  D
 sin C  sin D  2 sin   cos  
22. Using cos (A + B) = cosA cosB – sin A sin B, find 2 2 
the value of cos75° C  D C  D
 sin C  sin D  2 cos   sin 

r

SSC CGL 17/07/2023 (Shift-03) 2  2 

si
5 1 5 1 C  D C  D
(a) (b)  cos C  cos D  2 cos   cos  
4 4 2  2 

(c)
6 2 an by (d)
6 2 C  D
 cos C  cos D  2 sin 
2
D  C
 sin 
2 


n
4 4
28. If sinC + sinD = x, then the value of x is :
tan A  tan B
C +D C – D
23.
ja
Using tan (A – B) =
1  tan A tan B
, find the
(a) 2sin   sin  
R s
2  2 
value of tan 15°.
a th

SSC CGL 18/07/2023 (Shift-04) C - D C +D

(b) 2sin   cos  
2 2 
(a) 3 1 (b) 3 –1
C +D C - D
(c) 2 – 3 (d) 2  3 (c) 2cos   cos  
2  2 
ty a

tan45º –tan15º C +D C - D

24. What will be the value of ? (d) 2sin   cos  
di M

1 + tan45º tan15º 2 2 
SSC CGL 07/12/2022 (Shift-03) 29. What is sin – sin ?
1 SSC CGL 01/12/2022 (Shift- 02
(a) 3 (b)
2  + –
(a) 2 cos sin
1 2 2
(c) (d) 2  + –
3 (b) 2 sin sin
25. If tan( + ) = a, tan( – ) = b, then the value 2 2
of tan2 is: –  +
(c) 2 cos sin
SSC CGL 13/12/2022 (Shift- 04) 2 2
1+ b 1+ b  + –
(a) (b) (d) 2 cos cos
1 – ab 1 + ab 2 2
A

30. If cosC – cosD = y, then the value of y is :

1– b 1– b
(c) (d) C +D C - D
1 + ab 1 – ab (a) 2sin   sin  
2 2 
cot A cot B  1
26. Using cot (A – B) = , find the C +D C - D
cot B  cot A (b) 2cos   cos  
2 2 
value of cot15°.
SSC CGL 19/07/2023 (Shift-02) C + D C - D
(c) –2sin   sin  
2 2 
(a) 2  3 (b) 2 – 3
C +D C - D
(c) 3 1 (d) 3 1 (d) –2cos   cos  
2  2 

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 cos7x + cos5x 37. What will be value of 2cos45º. sin15º
31. Find =?
sin7x - sin5x
3 1 3–2
(a) cot2x (b) cot3x (a) (b)
2 2
(c) cotx (d) cot6x
sin5x - 2sin3x + sinx 3 –1 2
32. Find =? (c) (d)
cos5x - cosx 2 3 1
(a) tan4x (b) tan3x 38. Evaluate the following:
(c) tan2x (d) tanx cos(36° + A).cos(36° – A) + cos(54° + A).cos(54° – A
33. What is the value of SSC CGL 14/07/2023 (Shift-2

(sinx + siny)(sinx - siny) (a) sin 2A (b) cos A

is : (c) sin A (d) cos 2A
(cosx + cosy)(cosx - cosy)
39. If A + B = C, then tanA tanB tanC = ?
(a) 0 (b) 1
SSC CGL 08/12/2022 (Shift- 01
(c) – 1 (d) 2

r
(a) tanC + tanA – tanB
34. What is the value of sin75º + sin15º ?

si
(b) tanC + tanA + tanB
2 (c) tanA – tanB – tanC
(a) (b)
3
an by 3
40.
(d) tanC – tanA – tanB
What is the value of the expression cos2

n
3 3 cos2B + sin²(A – B) – sin²(A + B)?
(c) (d)
2 2 SSC CGL 12/12/2022 (Shift- 01
35.
ja
Simplify the following:
R s
(a) sin (2A – 2B) (b) sin (2A + 2B)
sin 2x + 2sin4x + sin 6x (c) cos (2A + 2B) (d) cos (2A – 2B)
a th

SSC CGL 18/07/2023 (Shift-02)

41. If cot 75° = 2 – 3 . Find the value of cot15°.
2
(a) 4 cos x sin 4x (b) 4 cos2x sinx
SSC CHSL 24/05/2022 (Shift- 03
(c) 2 cos2x sin 4x (d) 4 sin2x sin 4x
ty a

(a) 2 – 3 (b) 2  3
Transformation of a product
into sum or difference (c) 3 1 (d) 3 –1
di M

(xq.ku dk ;ksx ;k vUrj esa ifjorZu)  If A + B = 45º = 225º,

(i) (1 + tanA) (1 + tanB) = 2
(i) sin(A + B) sin(A – B) = sin² A – sin² B
(ii) (1 – cotA) (1 – cotB) = 2
= cos2 B – cos²A
42. The value of (1 + tan 10°) (1 + tan 35°) is:
(ii) cos(A+B) cos(A – B) = cos² A – sin² B
= cos² B – sin² B SSC CHSL 25/05/2022 (Shift- 03
(iii)2 sin A cos B = sin(A + B) + sin(A – B) 1 3
(a) (b)
(iv) 2 cos A sin B = sin(A + B) – sin(A – B) 2 4
(v) 2 cos A cos B = cos(A+B) + cos(A – B) (c) 1 (d) 2
(vi) 2 sin A sin B = cos(A – B) – cos(A + B)
A

  1o    1o 
36. Using 2cosA cosB = cos(A + B) + cos(A – B), find 43. 1  tan  22  x – y   1  tan  22  y – x 
  2     2  
the value of cos75°cos15°.
SSC CHSL 01/06/2022 (Shift- 02) (a) 0 (b) 1
(c) 2 (d) 4
1 1
(a) (b) 44. Find the value of (1 + tan11º) (1 + tan 21º) (1
2 4
tan 24º) (1 + tan 34º) (1 + tan 45º)
3 (a) 0 (b) 1
3
(c) (d) (c) 2 (d) 4
2 4

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs4

 Some Important Results
(oqQN egRoiw.kZ ifj.kke) (a)
5
8
(b)
7
8
1
(i) sin A. sin 2A. sin 4A  sin 3A 3 1
4 (c) (d)
8 8
1
(ii) cos A. cos 2A. cos 4A  cos 3A 50. The value of (cos15º.cos45º.cos75º) is :
4
(iii) tan A. tan A. tan 4A  tan 3A 1 1
(a) (b)
3 2 4 2
1
(iv) sin(60  A). sin A. sin(60  A)  sin 3A
4 1 3
(c) (d)
1 8 8
(v) cos(60  A). cos A. cos(60  A)  cos 3A
4 51. cos10º.cos30º.cos50º.cos70º =?
(vi) tan(60  A). tan A. tan(60  A)  tan 3A 1 3
(a) (b)
16 16

r
2 2 2 3
(vii) sin (60  A)  sin A  sin (60  A) 
2 5 1

si
(c) (d)
3 16 8
2 2 2
(viii) cos (60  A)  cos A  cos (60  A)  52. cos42º. cos78º. cos36º =?
an by 2

(a)
1
(b)
1

n
45. Find sin20º.sin40º.sin60º.sin80º. 2 4
1 5 1 1
(a)
ja (b) (c) (d)
R s
16 16 8 16
53. The value of tan6º.tan42º.tan66º.tan78
3 1
a th

(c) (d) is :
16 8
1
46. The value of (tan20º.tan40º.tan80º) is : (a) 1 (b)
2
(a) 1 (b) 0
ty a

1 1
(c) 3 (d) 3 (c) (d)
4 8
di M

47. Find 1 – sin10º.sin50º.sin70º. 54. tan6º. tan42º. tan66º. tan78º =?

5 7 1
(a) (b) (a) 1 (b)
8 8 4
3 1 1 1
(c) (d) (c) (d)
8 8 16 9
48. sin6º sin42º sin66º sin78º = ?
ab
55. If sin6º sin42º sin54º sin66º sin78º =
1 1 c
(a) (b) where a, b, c is relatively co-prime number
4 2
then find a + b + c?
1 1
A

(c) (d) (a) 60 (b) 58

8 16 (c) 72 (d) 70

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs5

 ANSWER KEY
1.(d) 2.(a) 3.(a) 4.(a) 5.(d) 6.(b) 7.(b) 8.(b) 9.(d) 10.(b)

11.(b) 12.(a) 13.(b) 14.(c) 15.(a) 16.(b) 17.(b) 18.(b) 19.(b) 20.(b)

21.(c) 22.(c) 23.(c) 24.(c) 25.(a) 26.(a) 27.(a) 28.(d) 29.(a) 30.(c)

31.(c) 32.(d) 33.(c) 34.(c) 35.(a) 36.(b) 37.(c) 38.(d) 39.(d) 40.(c)

41.(b) 42.(d) 43.(c) 44.(d) 45.(c) 46.(c) 47.(b) 48.(d) 49.(d) 50.(b)

51.(b) 52.(c) 53.(a) 54.(a) 55.(d)

r
si
an by
n
ja
R s
a th
ty a
di M
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs6

 Trigonometry/f=kdks.kfefr
( Practice Sheet With Solution)
Level-01 ;fn 7sin2 + 3cos2 = 4 vkSj 0°   
 

  g
2 
(tan69º + tan66º) rkstandk eku gS%
1. Find the value of (1 – tan69º tan66º) ?
3  2 
(a) 
  (b) 
 
(tan 69º + tan 66º ) 7  7 
(1 – tan 69º tan 66º ) dk eku Kkr dhft,\ 1 1
(c) (d)

r
(a) 1 (b) – 1 3 7
(c) 2 (d) – 2 7. Find the value of/dk eku Kkr dhft,%

si
 cos 60º   cos 65º cosec 25º
2. Show that the value of 3sin15° – 4 sin315° is?  
 sin 30º  + 

 tan10º tan 30º tan 45º tan 60º tan 80º

an by
n'kkZb, fd3sin15° – 4 sin315° dk eku D;k gksxk\ 
(a) 1

(b) –1
(c) 0 (d) 2

n
1 1
(a)
3
(b)
4 8. Find the value of:/dk eku Kkr dhft,%
 1 – sin   

ja
1  1  sin  
R s
cos2   
  
  

(c) (d)
2 3  1 – sin  
 1  sin   
a th
3. What is the value of sin79° cos19° – cos79° sin19°. (a) cos (b) cos/2

sin79° cos 19° – cos79° sin19° dk eku D;k gksxk\ (c) 2 cos (d) 2cosθ
9. Evaluate: sin10º sin30º sin50º sin70º
1 ewY;kadu djsa%
ty a

3 sin10º sin30º sin50º sin70º

(a) (b)
2 2 1 1
(a) (b)
di M

(c) (d) 8 16
2 3
4. What is the value of cos70° cos40° + sin70° sin40°? 1 1
(c) (d)
4 32
cos70° cos40° + sin70° sin40° dk eku D;k gS\ 10. Evaluate: sin20º sin40º sin60º sin80º
1 ewY;kadu djsa%
sin20º sin40º sin60º sin80º
(a) (b) 2
2 1 3
(a) (b)
16 16
3 5 7
(c) (d) 3 (c) (d)
2 16 16
5. If 0 <  < 90°, solve the following equations: 11. Find tan20º tan40º tan80º
tan20º tan40º tan80º dk eku Kkr djsaA
A

2 cos2 + sin – 2 = 0.
;fn 0 <  < 90°, fuEufyf•r lehdj.kksa dks gy djsa% (a) 1 (b) 3
2
2 cos  + sin – 2 = 0 1
(c) 2 (d)
16

(a)  = 0 (b)  = 12. Find the value of/dk eku Kkr djsaA
6
x = cos10º cos20º cos40º
(c)  = 1 (d) None of these
1 1
(a) tan10º (b) tan10º
  4 8
6. If 7sin2 + 3cos2 = 4 and 0°    
  , then
2  1 1
the value of tanis: (c) cot10º (d) cot10º
4 8

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 ;fn 7sin²+ 3cos² = 4(0  90) gS] rks d
1 – sin10° sin50° sin70° dk eku Kkr djsa\ eku Kkr dhft,A
3 4  
(a) (b) (a) (b)
5 5 2 3
 
7 5 (c) (d)
(c) (d) 6 4
8 8
14. The value of (cos15° cos45° cos75° 5 2
21. The numberical value of +
sec2  1 + cot2 
(cos15° cos45° cos75° dk eku Kkr djsa\ 3sin² is?
1 1 5 2
(a) (b) + + 3 sin² dk la[;kRed ek
3 2 4 2 sec2  1 + cot2 
D;k gksxk\
3 1

r
(c) (d) (a) 5 (b) 2
8 8
(c) 3 (d) 4

si
15. The value of cos10º cos30º cos50º cos70º cos90º 22. If A and B are complementary to each othe
cos10º cos30º cos50º cos70º cos90º dk eku Kkr djas\ then the value of sec²A + sec²B – sec²A sec²B

an by
(a) 3 (b) 0 ;fn A vkSj B ,d nwljs ds iwjd gS]
sec²A + sec²B – sec²A sec²B dk eku Kkr dhft,\

n
(c) 5 (d) 1
16. If  +  = 90°, then the value of (1 – sin2) (a) 1 (b) –1

ja
(1 – cos2) × (1 + cot2) (1 + tan2) is? (c) 2 (d) 0
R s
;fn  +  = 90°, rks (1 – sin2) (1 – cos2) × 23. If 0º <  < 90º, the value of (sin + cos) is
a th
(1 + cot2) (1 + tan2) dk eku Kkr djsa\ ;fn 0º <  < 90º gS] rks(sin + cos) dk ek
(a) 1 (b) –1 Kkr djsa\
(c) 0 (d) 2 (a) equal to 1 (b) greater than 1
ty a

17. Find the value of tan4º tan43º tan47º tan86º (c) less than 1 (d) equal to 2
tan4º tan43º tan47º tan86º dk eku Kkr dhft,\ sec + tan 5
di M

24. If = then sin is equal to?

2 sec – tan 3
(a) (b) 1
3 sec + tan 5
;fn = gS] rks
sin dk eku Kkr dhft,\
sec – tan 3
1
(c) (d) 2
2 1 1
(a) (b)
18. Simplify cos(36º – A) cos(36º + A) + cos(54º – A) 4 3
cos(54º + A)
2 3
fuEu dk eku Kkr dhft,\ (c)
3
(d)
4
cos(36º – A) cos(36º + A) + cos(54º – A) cos(54º + A)
(a) cosA (b) sin2A
 sec – 1
A

25. For any real value of  = is.

(c) cos2A (d) sinA
 sec +1
19. If sec = cosec 0º  (, )  90º
 sec – 1
 sec +1 dk okLrfod eku Kkr djsaA
Then find cos² ( + ) =
;fn sec = cosec 0º  (, )  90º
(a) cot – cosec (b) sec – tan
rc cos² ( + ) dk eku Kkr djsa\
(c) cosec – cot (d) tan – sec
(a) 0 (b) 1
(c) 2 (d) 3 cos 2 
26. If = 3 and 0º <  < 90º then th
20. If 7sin²+ 3cos² = 4 (0  90) then value cot 2  – cos 2 
of  is. value of  is.

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 cos 2 
;fn = 3 vkSj 0º <  < 90º gS] rks (c) 1 (d) 2
cot  – cos 2 
2

 dk eku Kkr dhft,A 34. What is the expression

(a) 30º (b) 45º (sin4x – cos4x + 1) cosec2x equal to?
(c) 60º (d) None (sin4x – cos4x + 1) cosec2x vfHkO;fDr fdlds cjkcj g
27. Find the maximum and minimum value of (a) 1 (b) 2
8cosA + 15sinA + 15 is: (c) 0 (d) –1
8cosA + 15sinA + 15 dk vf/dre vkSj U;wure 35. If x + y = 90°, then what is
eku Kkr dhft,A
cos x cosecy  cos x sin y equal to?
(a) 11 2 + 15 (b) 30, 8
;fn x + y = 90°, rks cos x cosecy  cos x sin y
(c) 32, –2 (d) 23, 8 fdlds cjkcj gS\
28. If cosx + cosy = 2 then value of sinx + siny is. (a) cosx (b) sinx

r
;fn cosx + cosy = 2 gS] rkssinx + siny dk eku
(c) cos x (d) sin x
Kkr dhft,A

si
2 2 2
36. If tan y cosec x – 1 = tan y; then which on
(a) 0 (b) 1

an by
of the following is correct?
(c) 2 (d) –1
;fn tan2y cosec2x – 1 = tan2y; rks fuEu esa ls

n
29. Find the simplest numerical value of
3(sinx – cosx)4 + 4(sin6x + cos6x) + 6(sinx + cosx)² lk lgh gS\
(a) x – y = 0 (b) x = 2y

ja
3(sinx – cosx)4 + 4(sin6x + cos6x) + 6(sinx + cosx)²
R s
dk lcls ljy la[;kRed dk eku Kkr dhft,\ (c) y = 2x (d) x – y = 1°
a th
(a) 12 (b) 10 37. For what value of  is (sin + cosec) = 2.5
where 0 <  < 90°?
(c) 21 (d) 13
 ds fdl eku ds fy, (sin + cosec) = 2.5, g
30. If tan15º = 2 – 3 then find the value tan15º
tgk¡ 0 <  < 90° gSA
ty a

cot75º + tan75º cot15º.

(a) 30° (b) 45°
;fn tan15º = 2 – 3 gS] rkstan15º cot75° +
di M

(c) 60° (d) 90°

tan75º cot 15º dk eku Kkr dhft,A
Level -02
(a) 14 (b) 12
(c) 10 (d) 8 38. Evaluate : cos2 45° – sin2 15°.

31. If A = tan11º tan29º and B = 2cot61º cot79º ewY;kadu djsa%

cos2 45° – sin2 15°
then: 1 2
;fn A = tan11º tan29º vkSjB = 2cot61º cot79º rc% (a) (b)
2 3
(a) A = 2B (b) A = – 2B 3
(c) 2A = B (d) 2A = –B (c) (d) 4
4
32. If cos + sec = k, then what is the value of
A

tan57º +cot37º
sin2 – tan2? 39. The expression is equal to?
tan33º +cot53º
;fn cos + sec = k gS] rkssin2 – tan2 dk eku tan57º +cot37º
D;k gS\ O;atd ds cjkcj gksxk\
tan33º +cot53º
(a) 4 – k (b) 4 – k2 (a) tan33º cot57º (b) tan57º cot37º
(c) k2 – 4 (d) k2 + 2 (c) tan33º cot53º (d) tan53º cot37º
33. If   ,   90° such that cos( – ) = 1, then 40. Evaluate : (Cot  – Cosec4 + Cot2 + Cosec2)
4

what is sin – sin + cos – cos equal to? ewY;kadu djsa%

(Cot4 – Cosec4 + Cot2 + Cosec2)
;fn   ,   90° ,slk gS fd cos( – ) = 1, rks (a) 1 (b) 0
sin – sin + cos – cos fdldscjkcj gS\ (c) – 1 (d) 2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 2
 x 3  y 3
2
1
41. 3 3
If x = acos  and y = bsin , then      = ? (a) (b) 3
2
a  b

2 2
1
 x 3  y  3 (c) (d) 1
3
;fn x = acos3vkSjy = bsin3, rks      = ?
a  b 47. The value of expression (1 + sec22º + cot68
(a) 1 (b) 0 (1 – cosec22º + tan68º) is:
(c) 2 (d) 4 O;atd (1 + sec22º + cot68º) (1 – cosec22º + tan68
42. Find the value of sin210° + sin220° + sin230° dk eku gS
+ ....... + sin280°? (a) 0 (b) –1
sin210° + sin220° + sin230° + ....... + sin280° (c) 1 (d) 2
dk eku Kkr dhft,\ 48. Find the value of expression:-
(a) 2 (b) 3 O;atd dk eku Kkr dhft,A

r
(c) 1 (d) 4

si
tanA cotA 2
x + –
1 – cot A 1 – tanA sin2A

an by
43. If sin 21° = y , then sec 21° – sin 69° is equal to
(a) –1 (b) 0

n
x (c) 1 (d) 2
;fn sin 21° = y , rkssec 21° – sin 69° fdlds cjkcj gS\ 49. If sec( + ) + sec( – ) = 2sec [  0], The

ja
the value of sin² = ?
R s
x2 y2 ;fn sec(+ ) + sec( – ) = 2sec [  0] g
(a) (b)
a th
y (y² – x ²) y (y² – x ²) rkssin² = ?
(a) –sec (b) –sec
x2 y2
(c) (d) (c) –cos (d) –sin cos
ty a

y (x ² – y²) x (x ² – y²)
50. Find the value of 8cos10º cos20º cos40º
44. In ABC A, B, C are angles then evaluate:
dk eku Kkr dhft,A
di M

8cos10º cos20º cos40º

sin2A + sin2B + sin2C
(a) sin10º (b) cos10º
ABC esaA, B, C dks.k gSa rks ewY;kadu djsa%
sin2A + sin2B + sin2C (c) 1 (d) 4 cos10º  3

(a) 4sin A sin B sin C 51. For all i {i = 1, 2, 3 ...... 20}, O  90º. Give
that sin1 + sin2 + ...... + sin20 = 20 The
(b) 4sin A cos B sin C
value (in degree ) of (1 + 2 + 3 +....+ 20)
(c) 4sin A sin B cos C
lHkh ds fy, i {i = 1, 2, 3 ...... 20}, O  90
(d) 4cos A cos B cos C
fn;k gS rkssin1 + sin2 + ...... + sin20 = 2
x (1 + 2 + 3 +....+ 20) dk eku fMxzh eas Kkr d
45. If sin = x² – 2x + 2, then the value of x is
A

2 (a) 1800º (b) 900º

x (c) 0º (d) 20º
;fn = x² – 2x + 2 rks x dk eku Kkr djasA
2
3cos2 – 1
(a) 0 (b) 1 52. If ,  are acute angle and cos2 =
3 – cos2
(c) –1 (d) None of these
then the value of tan cot is:
46. If cosA, sinA, cotA are in geometric progression,
then the value of tan6A – tan²A is: 3cos2 – 1
;fn ,  U;wudks.k gS cos2
vkSj = r
3 – cos2
;fn cosA, sinA, cotA T;kferh; çxfr esa gSa]tan
rks6A
– tan²A dk eku Kkr dhft,A tan cot dk eku Kkr dhft,A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs4

 (a) 3 (b) 2 4 b
(a) (b)
a 2 + b2 a
2
(c) 1 (d)
3 a 4
(c) (d)
53. If sin = acos and cos = bsin then the value b a 2 – b2
of (a² – 1) cot² + (1 – b²) cot² is equal to. 58. If 0° < x < 45º and 45º < y < 90º, Then whic
;fn sin = acos vkSj cos = bsin gS] rks one of the following is correct?
(a² – 1) cot² + (1 – b²) cot² dk eku cjkcj gSA ;fn 0° < x < 45º vkSj45º < y < 90º rks fuEufyf
a2 + b2 a2 – b2
eas ls dkSu&lk lgh gS\
(a) (b) (a) sinx = siny (b) sinx < siny
a2 b2
(c) sinx  siny (d) sinx  siny
a2 – b2 a2 – b2 59. In a right angle triangle XYZ, right angle a
(c) (d)
b2 a2 Y, if XY = 26 and XZ – YZ = 2 then secx

r
54. Find the value/dk eku Kkr dhft,A tanx is.
,d ledks.k f=kHkqtXYZ eas]Y ij ledks.k ;fn X

si
1 + sec cosec2 .  sec – tan2 1 + sin = 26 vkSj XZ – YZ = 2 rks secx + tanx gSA

an by
 sin + sec 2 + cos + cosec 2
1
(a) sec + tan (b) 1 – sin (a) (b) 6

n
6
(c) 1 + sin (d) None

ja
6
2sin (c) 26 (d)
R s
55. If x = then the value of 2
1 + cos + sin 
a th
60. ABCD is a rectangle of which AC is a diagona
1 – cos + sin The value of (tan² CAD + 1) sin² BAC
is:
1 + sin  ABCD ,d vk;r gS ftldk AC ,d fod.kZ gSA(tan
dk eku gSA
ty a

2sin 1 – cos + sin CAD + 1) sin² BAC

;fn x = gS] rks
1 + cos + sin  1 + sin  1
di M

dk eku Kkr dhft,A (a) 2 (b)

4
x
(a) (b) x (c) 1 (d) 0
1 + x 
4 5  –  
1 1 + x  61. If sin =
5
and cos =
13
then cos
2
(c) (d)
x x
4 5  –  
sinA ;fn sin = vkSj cos = rks cos
56. In ABC, If cosB = , then the triangle is 5 13 2
2sinC
4 63
sinA (a) (b)
ABC eas ;fn cosB = , rks f=kHkqt gSA 65 65
2sinC
(a) isosceles triangle 4 8
A

(c) (d)
(b) Equilateral " 65 65
(c) Right angle " 62. If xsin³ + ycos³ = sin cos and xsin
– ycos = 0, then the value of (x² + y²).
(d) Scalene triangle
57. If sinx + siny = a and cosx + cosy = b then ;fn xsin³ + ycos³ = sin cos vkSj xsin
x + y
– ycos = 0gS] rks(x² + y²) dk eku Kkr dhft,
tan  is:
 2  1
(a) 1 (b)
;fn sinx + siny = a vkSj cosx + cosy = b gS] rks 2
x + y 3
tan  dk eku Kkr dhft,A (c) (d) 2
 2  2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs5

 ;fn x = cosec – cos vkSj y = sec – cos r
;fn (a² – b²) sin + 2ab cos = a² + b² rc tan = ? x vkSjy ds chp laca/ gSA
(a) x² + y² + 3 = 1
2ab a2 – b2
(a) (b) (b) x² y² (x² + y² + 3) = 1
 a – b2 
2
2ab
(c) x² (x² + y² – 5) = 1
ab 2
a –b 2 (d) y² (x² + y² – 5) = 1
(c) (d) 69. tanx = a and tany = b
a – b2
2
ab
64. If x, y are acute angles 0 < (x + y) < 90º and What is the value of (tanx + tany
sin(2x – 20) = cos(2y + 20) then the value (1 – cotxcoty) + (cotx + coty) (1 – tanx tany
of tan(x + y) equal to?

;fn x, y U;wudks.k0 gSa

< (x + y) < 90º vkSjsin(2x – 20) = tanx = a vkSjtany = b
cos(2y + 20) rkstan(x + y) dk eku Kkr dhft,A (tanx + tany) (1 – cotx coty) + (cotx + coty)
(1 – tanx tany) fdlds cjkcj gS\

r
1
(a) 3 (b) a–b ab – 1

si
3
(a) (b)
ab ab

an by
3 (c) 0 (d) 1
(c) (d) 1
2 70. A rectangle is 48 cm long and 14 cm wide.

n
2sinx 1 – cosx + sinx the diagonal makes an angle  with the longe
65. If 1 + cosx + sinx  = t then 1 + sinx  an side, then what is (sec + cosec) equal to?

ja
,d vk;r 48 lseh yack vkSj 14 lseh pkSM+k gSA ;f
R s
be written as.
2sinx 1 – cosx + sinx cM+h Hkqtk ls dks.k cukrk gS] rks
(sec + cosec
;fn gS] rks
a th
=t
1 + cosx + sinx  1 + sinx  fdlds cjkcj gS\
,d ds :i esa fy[kk tk,xkA 775
(a) (b) 2
1 168
ty a

(a) (b) t
t 771 770
(c) (d)
di M

107 107
t
(c) t secx (d) 71. If cosec – sin = m and sec – cos = n, the
sinx
4 2 2 4
66. sin75º + sin15º can be expressed as. what is m 3 n 3  m 3 n 3 equal to?

sin75º + sin15º ds :i esa O;Dr fd;k tk ldrk gSA ;fn cosec – sin = m vkSj sec – cos = n] r
fdlds cjkcj gS\
4 2 2 4
3 2
(a) (b) m3 n3  m3 n3
2 3
(a) 0 (b) 1
2 3 1 1
(c) (d) (c) (d)
3 2 2 4
A

67. If 3sinx + 4cosx + r is always greater than

72. If sin + cos = 2 then what is sin6 + cos6
or equal to 10. Then what is the smallest
value 'r' can to take? 6sin2 cos2 equal to?

;fn 3sinx + 4cosx + r ges'kk 10 ls vf/d ;k mlds ;fn sin + cos = 2 rkssin6 + cos6 + 6sin2 cos2
cjkcj gS] rks'r' dk lcls NksVk eku D;k fy;k tk fdlds cjkcj gS\
ldrk gS\
(a) 5 (b) –5 1 3
(a) (b)
4 4
(c) 4 (d) 3
68. If x = cosec – cos and y = sec – cos then 7
the relation between x and y is. (c) 1 (d)
4

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs6

 ;fn x + y = 90° vkSj sinx : siny = 3 : 1 r
: BC = 3 : 4. What is sinA + sinB + sinC
equal to? x:y fdlds cjkcj gS\
(a) 1 : 1 (b) 1 : 2
ABC ,d f=kHkqt gS ftldk dks.k
B ledks.k gS vkSj
(c) 2 : 1 (d) 3 : 2
AB : BC = 3 : 4 gSAsinA + sinB + sinC fdlds
cjkcj gS\ 79. P = tan 2x + cot 2x, then which one of th
following is correct?
11 P = tan2x + cot2x] rks fuEufyf•r esa ls dkSu lk lgh
(a) 2 (b)
5
(a) P  2 (b) P  2
12
(c) (d) 3 (c) P < 2 (d) P > 2
5
80. If  is the angle of first quadrant such tha
74. If x = acos and y = bcot, then (ax–1 – by–1) cosec4 = 17 + cot4, then what is the valu
(ax–1 + by–1) is equal to of sin?
;fn x = acos vkSjy = bcot, rks (ax–1 – by–1)

r
;fn çFke prqFkkZa'k dk dks.k
 bl çdkj gS fd cosec4
(ax–1 + by–1) ds cjkcj gS = 17 + cot4, rks sindk eku D;k gS\

si
(a) 0 (b) 1

an by
1 1
(c) tan 2
(d) sin2 (a) (b)
3 4

n
75. If  is an acute angle and sincos = 2cos3
– 1.5 cos, then what is sin equal to? 1 1
(c) (d)
9 16

ja
;fn  ,d U;wudks.k gS vkSj sincos = 2cos 3
R s
– 1.5 cos, rks sinfdlds cjkcj gS\ p psecθ – qcosecθ
a th
81. If tan = q , then what is psecθ + qcosecθ
5 1 1 5 equal to?
(a) (b)
4 4
p psecθ – qcosecθ
  ;fn tan = q rks psecθ + qcosecθ fdlds cjkcj gS
ty a

5 1 5 +1
(c) (d) –
4 4
pq q 2  p2
di M

76. What is the expression (a) p  q (b)

q 2  p2
tan x tan x p2  q 2

1  sec x 1  sec x equal to: (c) 2 (d) 1
p  q2
tan x tan x 82. The value of cosec2 – 2 + sin2 is always
 ds cjkcj gS%
1  sec x 1  sec x
cosec2 – 2 + sin2dk eku lnSo gksrk gS
(a) cosecx (b) 2cosecx (a) less than zero (b) non-negative
(c) 2sinx (d) 2cosx (c) zero (d) 1
77. In the given figure, BC = 15 cm and sinB 83. If  and  are complimentary angles, then wha
A

1
4 –
= , what is the value of AB?  sin cos 2
5 is cosec.cosec  + equal to?
 sin cos
4 ;fn  vkSj  iwjd dks.k gSa] rks
nh xbZ vkÑfr esa]
BC = 15 cm lseh vkSjsinB = ] cosec.cosec
5 1
–
AB dk eku D;k gS\  sin cos 2
 sin + cos fdlds cjkcj gS\
(a) 25 cm (b) 20 cm
(c) 5 cm (d) 4 cm (a) 0 (b) 1
(c) 2 (d) None of these
78. If x + y = 90° and sinx : siny = 3 : 1 then 84. (1–tanA)2 + (1+tanA)2 + (1–cotA)2 + (1+cotA
what is x : y equal to? is equal to?/fdlds cjkcj gS\

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs7

 (c) 2sec2A cosec2A (d) None of these measure is 1 and their sum in circula
85. If secx.cosecx = 2 then what is tannx + cotnx measure is also 1. What are the angles i
equal to? circular measure?
;fn secx.cosecx = 2 rkstannx + cotnx fdlds cjkcj gS\ nks dks.kksa dk va'kh; eki esa varj 1 gksrk gS
(a) 2 (b) 2n+1 eki esa mudk ;ksx Hkh 1 gksrk gSA o`Ùkh; e
(c) 2n (d) 2n–1
dks.k gksrs gSa\
1 1 1
86. What is the value of sin x  ?   1  
1  cos x 1 – cos x  –
(a)  , 
  + 

 2 360°   2 360° 
1 1
sin x  dk eku D;k gS\ 1
–
90   1 90 
1  cos x 1 – cos x (b) 
 , 
  + 

2   2  
(a) 2 (b) 2 2
1   1  
(c) 2 tan x (d) 0  –
(c)  , 
  + 

 2 180°   2 180° 

r
sin 45 – sin 30 sec 45 – tan 45 (d) None of these
87. If A = and B =

si
cos 45  cos 60 cosec 45 cot 45
92. If ABC is a right angled triangle at C an
then which one of the following is correct? having u units, v units and w units as th

an by
sin 45 – sin 30 sec 45 – tan 45 length of its sides, opposite to be vertices A
;fn A= vkSj B = cosec 45 cot 45 B and C respectively, then what is tanA + tan
cos 45  cos 60

n
fiQj fuEu esa ls dkSu lk lgh gS\ equal to?
(a) A = B (b) A > B > 0 ;fn ABC, C ij ,d ledks.k f=kHkqt gS vkSj b

ja
Hkqtkvksa dh yackbZ dsu:ibdkb;k¡]
esa v bdkb;k¡ v
R s
(c) A < B (d) B < A < 0
Level-03 w bdkb;k¡ gSa] tks Øe'k% A,'kh"kZ
B vkSj C ds foijh
a th
88. If cosec – sin = P3 and sec – cos = q3 then gSa] rks
tanA + tanB fdlds cjkcj gS\
what is the value of tan?
u2
;fn cosec – sin = P3 vkSj sec – cos = q3 gS (a) (b) 1
ty a

uw
rkstan dk eku D;k gS\
w2
di M

p q
(a) q (b) p (c) u + v (d)
uv
(c) pq (d) p2q2
1  cos B 2 tan A
89. If sinx + cosx = C then sin6x + cos6x is equal to 93. If tanA = then what is
sin B 1  tan 2 A
vxj sinx + cosx = C rkssin6x + cos6x ds cjkcj gS
equal to?
1  6c2  3c4 2
1  6c  3c 4
(a) (b) 1  cos B 2 tan A
16 4 ;fn tanA = sin B
rks 1  tan2 A fdlds cjkcj gS
1  6c2  3c4 1  6c2  3c 4
(c) (d) tan B
16 4 (a) (b) 2 tanB
2
A

cosx cosx
90.  = 2, then which one of (c) tanB (d) 4 tanB
1  cosecx cosecx  1
the following is one of the values of x? 1  2 sincos
94. If a² = , then what is the valu
1 – 2 sincos
cosx cosx
 = 2, rks fuEufyf•r esa ls dkSu a 1
1  cosecx cosecx  1 of ?
lk x ds ekuksa esa ls ,d gS\ a –1

  1  2 sincos a 1
(a) (b) ;fn a² = gS] rks dk eku D;k g
2 3 1 – 2 sincos a –1
  (a) sec (b) 1
(c) (d)
4 6 (c) 0 (d) tan

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs8

 x ytanθ xtanθ y
which one of the following is correct? 98. If – = 1 and + = 1 then th
a b a b
;fn sinA + cosA = P vkSjsin³A + cos³A = q, rks
fuEufyf•r esa ls dkSu lk lgh gS\ value of
x² y²
 is.
a² b²
(a) p³ – 3p + q = 0 (b) q³ – 3q + 2p = 0
(c) p³ – 3p + 2q = 0 (d) p³ + 3p + 2q = 0 x ytanθ xtanθ y
;fn – = 1 vkSj + = 1 gks]
sec² – tan a b a b
96. If x = , then which one of the
sec ² tan
x² y²
following is correct?  dk eku gSA
a² b²
sec² – tan
;fn x= gS] rks fuEu esa ls dkSu lk lgh gS\ (a) 2 sec²  (b) sec²
sec ² tan
(c) cos² (d) 2 cos²
1
(a) x3 99. If p = cot + tan and q = sec – cos the
3

r
2 2

1  (p²q) 3 – (q²p) 3 is equal to?

si
(b) x   , 3
3 
 ;fn p = cot + tan vkSj q = sec – cos r

an by
2 2
1 (p²q) 3 – (q²p) 3 fdlds cjkcj gS\
(c) – 3 < x < –

n
3
(a) 0 (b) 1
1 (c) 2 (d) 3

ja
(d)  x  3
R s
3 100. If A, B and C are interior angles of a triangl
B  C 
97. If cot (1 + sin) = 4m and cot (1 – sin) =
a th
ABC then sin   will be equal to.

4n, then which of the following is correct?  2 
;fn cot (1 + sin) = 4 m vkSj cot (1 – sin) = ;fn A] B vkSjC ,d f=kHkqtABC ds vkarfjd dks.k
4n, rks fuEu esa ls dkSu lk lgh gS\
rkssin  B  C  cjkcj gksxkA
 
ty a

(a) (m² + n²)² = mn  2 

di M

(b) (m² – n²)² = mn (a) cosA (b) cos2A

(c) (m² – n²)² = m²n² 3A A
(d) (m² + n²)² = m²n² (c) cos (d) cos
2 2
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs9

 Answer Key
1.(b) 2.(c) 3.(c) 4.(c) 5.(b) 6.(c) 7.(d) 8.(c) 9.(b) 10.(b)

11.(b) 12.(d) 13.(c) 14.(b) 15.(b) 16.(a) 17.(b) 18.(c) 19.(a) 20.(c)

21.(a) 22.(d) 23.(b) 24.(a) 25.(c) 26.(c) 27.(c) 28.(a) 29.(d) 30.(a)

31.(c) 32.(b) 33.(b) 34.(b) 35.(b) 36.(a) 37.(a) 38.(c) 39.(b) 40.(b)

41.(a) 42.(d) 43.(a) 44.(a) 45.(b) 46.(d) 47.(d) 48.(c) 49.(c) 50.(d)

r
51.(a) 52.(d) 53.(d) 54.(b) 55.(b) 56.(a) 57.(c) 58.(b) 59.(b) 60.(c)

si
61.(d) 62.(a) 63.(b) 64.(d) 65.(b) 66.(d) 67.(a) 68.(b) 69.(c) 70.(a)

an by
n
71.(b) 72.(d) 73.(c) 74.(b) 75.(a) 76.(b) 77.(a) 78.(c) 79.(b) 80.(a)

ja
81.(c) 82.(b) 83.(b) 84.(c) 85.(a) 86.(a) 87.(a) 88.(b) 89.(b) 90.(c)
R s
a th
91.(a) 92.(d) 93.(c) 94.(d) 95.(c) 96.(d) 97.(b) 98.(d) 99.(b) 100.(d)
ty a
di M
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 SOLUTIONS
1. (b) 7. (d)
tan69º + tan66º cos 60º cos 65º cosec 25º
1 – tan 69º tan 66º + tan10º tan 30º tan 45º tan 60º tan 80º
sin 30º
tan(69° + 66°) = tan135º
cos 60º cos(90º –25) cosec 25º
= tan(90° + 45º) = – cot45º = –1 +
sin(90º – 60º ) tan10º tan(90 – 10º ) tan 45º
2. (c)
tan(90 – 30º ) tan 30º
3 sin15º – 4 sin³15º
Formula:- sin3 = 3sin – 4sin3 1
=1+ =1+1=2
1 ×1 ×1
1
sin(3 × 15º) = sin45º = 8. (c)

r
2

 1 + sin  1 – sin  
3. (a)

si
cos²  1 – sin   1  sin  
sin79º cos19º – cos79º sin19º 
 


an by
Formula:- sin(A – B) = sinA.cosB – cosA.sinB 
1 + sin   1 – sin   2
cos2  2
 = cos2×
3 
 1 – sin   cos 

n
sin(79° – 19°) = sin60º =
2 = 2cos

ja
4. (c) 9. (b)
R s
cos70º cos40º + sin70º sin40º sin10° sin30° sin50° sin70° = ?
a th
Formula:- cos(A – B) = cosA.cosB + sinA.sinB
We know,
3
cos(70º – 40º) = cos30º = 1
2  sin sin(60 – ) sin(60 + ) = sin3
ty a

4
5. (b)
Then,
2cos² + sin – 2 = 0
di M

2(1 – sin²) + sin – 2 = 0  sin30° [sin10° sin50° sin70°]

2 – 2sin² + sin – 2 = 0 1 1 
   sin 3  10
2sin² – sin = 0 2 4 
sin(2sin – 1) = 0
2sin – 1 = 0 1 1
  sin 30 =
8 16
1
sin = 10. (b)
2
sin20º sin40º sin60º sin80º

So,  = We know,
6
A

6. (c) 1
 sinsin2sin4 = sin3
7sin² + 3cos² = 4 4
7sin² + 3 – 3sin2 = 4 Then,
4sin² = 1
sin60º sin20º sin40º sin80º
1
sin = 3 1
2   sin(3  20)
 = 30º 2 4

1 3 1 3 3
So, tan30º =   
3 2 4 2 16

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 tan20º tan40º tan80º ATQ,
We know,  +  = 90º
 tan tan2 tan4 = tan3 (1 – sin²)(1 – cos²) (1 + cot²) (1 + tan²)
Then, cos² sin² cosec² sec²
cos² sin² cosec² (90° – )sec²(90° – )
tan(3 × 20) = tan60° = 3
cos² sin² sec²cosec² = 1
12. (d)
17. (b)
1   tan4° tan43° tan47° tan86°
cos10º cos 20º cos 40º cos 80º
  
cos 80º 
  tan4° tan86° tan43° tan47°
tan4° tan(90° – 4)° tan43° tan(90º – 43°)
We know,
tan4° × cot4º × tan43º × cot43°
1 1× 1=1
cos3θ = θ.2θ.4θ

r
3
SMART APPROACH:-

si
cos10º 1
 cos (3 × 20º)
cos 80º 4 If A + B = 90

an by
Then,
1
cot10º × cos60º  TanA.TanB = 1

n
4
 CotA.CotB = 1
Where: cos80° = sin10°

ja
18. (c)
R s
1 cos(36º – A) cos(36º + A) + cos(54º – A) cos(54º + A
cot10º
8
a th
 Using,
13. (c) cos(A + B) cos(A – B) = cos²A – sin²B
1 – sin10° sin50º sin70º  cos²36º – sin²A + cos²54º – sin²A
ty a

1  cos²36º – sin²A + sin²36º – sin²A

Formula:- sin3 = sin.sin(60 – ).sin(60 + )
4  1 – 2sin²A
di M

1 – sin (60º – 10°) sin10º sin(60º + 10º)  cos2A

Where: cos2 = 1 – 2sin²A
1
1– sin(3 × 10º) 19. (a)
4
ATQ,
1 7 sec = cosec
1– =
8 8 If we take  =  = 45°
14. (b) Then,
cos15° cos45º cos75º
sec45° = cosec45°  2  2 verify
1 cos² ( + ) = cos290° = 0
Formula :- cos3 = cos.cos(60 – ).cos(60 + )
A

4 20. (c)
cos(60º – 15º) cos15º cos(60º + 15º) ATQ,
1 1 1 7sin² + 3cos² = 4
1  
cos(3 × 15°) = Use option (c)
4 4 2 4 2
15. (b) 
Where:  
cos10º cos30º cos50º cos70º cos90º 6
Where, 1 3
7× + 3 × = 4 verify
cos90° = 0 4 4
cos10º cos30º cos50º cos70º × 0 = 0 So, option (c)

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 8 cosA + 15 sinA + 15
5 2
+ + 3sin2 
sec  1 + cot2 
2
Max. = + 82 +152 + 15 = 32
5cos² + 2sin² + 3sin² Min. = – 82 +152 + 15 = –2
5cos² + 5sin² 28. (a)
5×1=5 cosx + cosy = 2
22. (d) Put x = y = 0
ATQ, 1 + 1 = 2 verify
A + B = 90º So, sinx + siny
Put A = 45º & B = 45º 0+0=0
Then, 29. (d)
sec²A + sec²B – sec²A sec²B 3(sinx – cosx)4 + 4(sin6x + cos6x) + 6(sinx + cosx

r
2+2– 2×2=4– 4=0 Put x = 0
23. (b)

si
3(–1)4 + 4(1) + 6(1) 2
0º <  < 90º (sin + cos)
3 + 4 + 6 = 13

an by
Put  = 45º
30. (a)
1 1

n
=  tan15º = 2–3
2 2
then cot 15º = 2+3

ja
2 tan15º cot75º + tan75º cot15º
R s
= 2 = 1.414
2 tan²15º + cot²15º
a th
greater than 1 (2–3)² + (2 + 3)²
24. (a) 2(4 + 3) = 14
sec + tan 5 31. (c)
ty a

=
sec – tan 3 ATQ,
 3sec + 3tan = 5sec – 5tan A = tan11º tan29º & B = 2cot61º cot79º
di M

2sec = 8tan B = 2cot (90° – 29º) cot(90° – 11º)

1 B = 2tan29º tan11º
sin =
4 Then, we get
25. (c)
B = 2A
Rationalization
32. (b)
 sec – 1  sec – 1 cos + sec = k
×
 sec +1  sec – 1 Square both sides,
cos² + sec² + 2cos sec = k²
 sec – 12
(sec – 1) 1 – cos 
=
= 1 – sin² + 1 + tan² + 2 = k²
A

2
sec  – 1 tan sin
= cosec – cot tan² – sin² = k² – 4
26. (c) sin² – tan² = 4 – k²
cos² 33. (b)
=3
cot  – cos2
2 cos( – ) = 1 = cos0°
 –  = 0°
cos2 
 =3 =
cos2   cosec2  – 1
sin – sin + cos – cos
 tan = 3
 = 60º = sin – sin + cos – cos = 0

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 Put x = 90° sin57º cos37º cos 57º – 37º
+
4 4 tan57 + cot 37º cos57º sin37º
(sin x – cos x + 1) cosec²x = = cos57º sin37º
tan33º + cot53º sin33º cos53º cos 53º – 33º
= (1 – 0 + 1) × 1 +
cos33º sin53º cos33º sin53º
=2×1=2
Where: cos(A + B) = cosA.cosB – sinA.sinB
35. (b)
cos20 cos33º sin53º
= × = tan57º cot37º
 cosx cosecy  cosx siny cos57º sin37º cos20º
40. (b)
x + y = 90°  y = 90° – x
cot4 – cosec4 + cot2 + cosec2.
cos x cosec (90 – x ) – cos x sin (90 – x ) (cot² + cosec²)(cot²– cosec²) + (cot² + cosec²
(cot² + cosec²)(–1) + (cot² + cosec²)
= cos x sec x – cos x cos x
Where: cosec2 – cot2 = 1
= 1 – cos ² x –(cot² + cosec²) + (cot² + cosec²) = 0

r
41. (a)

si
= sin ² x
x = acos³, y = bsin³
= sinx
x y

an by
36. (a) = cos³, = sin³.
a b
tan²y cosec²x – 1 = tan²y

n
2 2
x  3 y 3
tan²y cosec²x – tan²y = 1 
  
 

a  b

ja
tan²y [cosec²x – 1] = 1
R s
2 2
tan²y cot²x = 1 3 3
 sin3  3
cos   
a th
tan²y = tan²x
cos² + sin² = 1
x=y
42. (d)
x–y=0
sin²10º + sin²20º + sin230º + --------- + sin²80º
ty a

37. (a)
sin²10º  sin ²(90º –10º )  sin²20º  sin ²(90º –20º )  ....
sin + cosec = 2.5   
di M

use option (a) Total terms = 8

 = 30 1+1+1+1=4
43. (a)
1 1
sin =  2 verify Given,
sin 2
x
38. (c) sin21º = y
cos²45º – sin²15º

3 –1 3 1
 sin15  , sin 75  y
2 2 2 2 x
A

2
 1 2  3 – 1  21º
   
  –  
 2  2 2  y² – x ²
1 31– 2 3 sec21º – sin69º
 – sec21º – cos21º
2 8
y y² – x ²
 –

1 2 2– 3  y² – x ²
–
y
2 8
y 2 – (y² – x ²) x²
2 – 2 3 3 =
  y y² – x ² y (y² – x ²)
4 4

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 sin2A + sin2B + sin2C
tanA cotA 2
+ –
XY X–Y 1 – cot A  1 – tanA  sin2A
Formula:- sinX + sinY = 2sin . cos
2 2
sin A cos A 2
(2A  2B) (2A – 2B)  –
 2 sin cos  sin 2C cos A sin A sin 2A
2 2 cos A cos A
2sin(A + B) cos(A – B) + sin2C 1– 1–
sin A sin A
2sin(180 – C) cos(A – B) + 2sinC cosC
2sinC [cos(A – B) – cos(A + B)] sin 2 A cos 2 A 2
+ –
 2sinC [(cosA cosB + sinA sinB) – (cosA cosB cos  sinA – cosA  sinA  cosA – sinA  sin2A
+ sinA sinB)]
 2sinC [2sinA sinB] 1  sin 3 A – cos3 A  2
–
 4sinA sinB sinC  sinA – cosA   cosA.sinA  sin2A

r
45. (b)
Given 1   sinA – cosA   sin2 A + cos2 A + sinAcosA   2
 –

si
 sinA – cosA   cos A sin A   sin2A
x
sin = x² – 2x + 2

an by
2 1 1
+1 – =1
Put x = 1 sinAcosA sinAcosA

n
 'OR'
Then sin =1–2+2
2 tanA cotA 2

ja
+ –
R s
 1 – cot A  1 – tanA  sin2A
sin = 1 (verify)
2 Let, A = 135
a th
So, x = 1
–1 –1 2
46. (d)   –
–  –1 –  –1 sin 270
Given, cosA , sinA & cotA are in G.P.
ty a

So, sin²A = cosA.cotA 1 1 2

2
– – –
cos A 2 2 sin 180  90
di M

sin²A =
sinA
2
sin³A = cos²A  tan²A = cosecA  –1 – 1
ATQ, –1
tan6A – tan²A = tan²A (tan4A – 1) 49. (c)
= tan²A × (cosec²A – 1) sec( + ) + sec( – ) = 2sec
= tan²A × cot²A = 1
47. (d) 1 1
  2 sec 
 (1 + sec22º + cot68º) (1 – cosec22º + tan68º) cos      cos   –  
(1 + sec22º + tan22º) (1 – cosec22º + cot22º)
cos   –   + cos    
cos22º +sin22º +1  sin22º +cos22º –1 cos   +   cos  –   = 2sec
A

×
cos22º sin22º
Formula:- (A + B)(A – B) = A2 – B2 2cos.cos
= = 2sec
cos 2  – sin2 
cos22º +sin22º2 –1 cos².cos = cos² – sin²

cos22º sin22º cos² (1 – cos) = sin²


cos 2
22º +sin2 22º  2 cos 22  sin 22  – 1 sin2 
cos² = 1 – cos = (1 + cos)
cos22º sin22º  
2  cos 22  sin 22 1 – sin² = 1 + cos
 = 2
cos 22  sin 22 sin² = – cos

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 8cos10º cos20º cos40º ATQ,
4[2cos40º cos20º] cos10º sin = acos & cos = bsin
4[cos60º + cos20º] cos10º
Put  =  = 30º
1 
4  + cos20º cos10º 1
2  a= & b = 3
3
1 
4  cos10º +cos20º cos10º (a² – 1) cot² + (1 – b²) cot²
2 
= 2[cos10º + cos30º + cos10º] 1 
Then  – 1 × 3 + (1 – 3) × 3
 3 3
= 2  2cos10º + 
 2 
–2
× 3 – 6 = –8
= 4 cos10º + 3 3

r
51. (a) Now, put the value on option (d)

si
sin1 + sin2 +......sin20 = 20
Here 1 = 2 = 3 .........= 90º 1 

an by
So, 1 + 2 + 3 +........+ 20 = 20 × 90  – 3 
3
= (–8) verify
= 1800 1

n
52. (d) 3
3cos2 – 1

ja
54. (b)
cos2 =
R s
3 – cos2
1 + sec cosec2 .  sec – tan2 1 + sin
a th
Put  = 30º
3   sin + sec 2 + cos + cosec 2
 – 1 1 1
2
cos2 = = 2= 2
1 5 1 – sin 
ty a

 5
 3 –  2 1  sec .cosec 2 .   . 1  sin  
2 cos 
 sin + sec 2 + cos + cosec 2
di M

1
2cos² – 1 =
5
1 – sin 2 . 1  sin  
6 1  sec .cosec 2 .
2cos² =
5
1 – sin   1  sin  
 sin + sec  + cos + cosec 2
2

3
 cos² = Put  = 45°
5
3  1 
 cos = 1  22 . 1 – 
5  2
2 2
 1   1 
 2    2
A

   
2 2

5  1   1 
2 9  1 –  9  1 – 
 2  2 1
 1–
9 9 9 2

 2 2
3 Use option (b)

2 2 2 1
tan cot = ×  (1 – sin45º) = 1 – (verify)
3 3 3 2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 2sin 1 – cos + sin 0 < x < 45º & 45º < y < 90º
x= find We know sin increase when angle increas
1 + cos + sin 1 + sin 
Put  = 90º Here y > x
So, siny > sinx
2 ×1 59. (b)
x= = 1
1 + 0 +1 Let, YZ = X, XZ = X + 2
XZ – YZ = 2
1 – cos90º + sin90º 2
Now = =1 = x
1 + sin90º 2 Z

56. (a)

2)
+
x

(x
A

r
si
c b X 26 Y

an by
(x + 2)² = x² + (2 6 )²
B a C x² + 4 + 4x = x² + 24

n
Given, 4x = 20
x= 5

ja
sinA a c
R s
cosB = Where, = 7 5 12
2sinC sinA sinC secX + tanX = + = = 6
2 6 2 6 2 6
a th
a 2 + c 2 – b2 a sinA a
= = 60. (c)
2ac 2c sinC c
a² + c² – b² = a²
ty a

c² = b²  c = b
So, isoscales triangle
di M

57. (b)
ATQ,
sinx + siny = a & cosx + cosy = b
put x = y = 45º (tan² CAD + 1) sin² BAC
(tan² + 1) sin² (90º – )
a = 2
(tan² + 1) cos²
b = 2 sec² × cos² = 1
61. (d)
x + y
tan   = tan45º = 1 4 5
2 
If sin = & cos =
5 13
A

By putting the value of (a) and (b) in option

(a), (b) and (c) we get the same result. 3 12
then. cos = &sin =
5 13
For other angle value
Put x = 0º , y = 60º  –  
cos( – ) = 2 cos2 
 –1

 2 
3 3
a ,b  –  
2 2
cos cos + sin sin = 2 cos² 
 
 –1
x + y 1  2 
tan  =
 2  3 3 5 4 12  –  
    2 cos ² 
 –1

option (c) 5 13 5 13  2 

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

   –   63
2 cos ² 
 =
 1
 2  65 sin75º + sin15º
sin(45° + 30º) + sin(45° – 30°)
  –   128 64
cos ² 
 =
 
 2  130 65  1 3 1 1  1 3 1 1
 × + ×  +  × – × 
 2 2 2 2  2 2 2 2
 –      
8
cos 
 =

 2  65
( 3  1) ( 3 – 1)

62. (a) 2 2 2 2
x sin³ + ycos³ = sin cos .........(1)
x sin – ycos = 0 2 3 3

2 2 2
sin  y
 ........(2)
cos  x 67. (a)

r
from (i) & (ii)

si
3 sin x + 4 cos x
   r  10
xy³ + yx³ = xy (P ) Let

an by
xy (x² + y²) = xy
Theory max value in sin and cosis
So, x² + y² = 1

n
63. (b) asin + bcos = a 2  b2

ja
If (a² – b²) sin + 2ab cos = a² + b² For minimum value of r we have to mak
R s
maximum value of P
(a² – b²) 2ab
a th
sin  + a²  b² cos  = 1
(a²  b²) Maximum (3sinx + 4cosx) = 3²  42 = 5
According to theory, So, Min. of r = 5
68. (b)
ty a

a² – b² 2ab
sin  = & cos  = If x = cosec – cos& y = sec – cos
a²  b² a²  b²
di M

Put  = 45º
sin  a² – b²
tan = 
cos  2ab 1 1 1 1
x= 2– = & y= 2– 
64. (d) 2 2 2 2

sin(2x – 20°) = cos(2y + 20°) Now, put the value  = 45° in option (b)
(2x – 20°) + (2y + 20°) = 90º
1 1 1 1  1
    3
 = 4 =1
2(x + y) = 90º 2 2 2 2  4
x + y = 45º
69. (c)
tan (x + y) = tan 45º = 1
Let, tan x = a and tan y = b
A

65. (b)
 1  1 1 
2sin x (a  b) 1 –      (1 – ab)
Given = t 
 ab   a b 
1  cos x + sin x
put x = 0 ab – 1  a  b 
= (a  b)    (1 – ab)
 ab  
  ab 
2 0
then t = = 0
11 0 ab – 1  1 – ab 
= (a  b)  

 ab 
1 – cosx + sinx 1 – 1+ 0 0
Now = = = 0 = t
1 + sinx 1+ 0 1 = (a + b) × 0 = 0

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 A B
A

Length  14 cm
3x 5x

D Base  48 cm C B C
4x

Diagonal (BD) = 482  14 2  50 sinA + sinB + sinC

sec + cosec 4x 3x 4 3
 1 =  1
5x 5x 5 5

r
50 50 25 25  24  7 
=  =  = 25 
 

48 14 24 7  24  7 

si
7 12
= 1 =
5 5
25  31 775

an by
= =
168 168 74. (b)
71. (b) (ax–1 – by–1) (ax–1 + by–1)

n
Put  = 45°  a b  a b 

ja
=
x – y x  y 
 
R s
1 1   
m= 2– =
2 2
a th
a 
 = 1 and b – 1 
1 1  x cos y cot
 
n= 2– =
2 2 (sec – tan) (sec + tan)
ty a

m=n = sec² – tan² = 1

75. (a)
di M

4 2 2 4
3 3 3 3
m n  m n
sin cos = 2 cos³ – 1.5 cos
Put the value in this equ. m = n Divide by cos
4 2 2 4
sin = 2cos² – 1.5
= m 3 m3  m3 m 3
sin = 2 (1 – sin²) – 1.5
= m² + m² = 2m²
sin = 2 – 2 sin² – 1.5
2
 1  1 2sin² + sin – 0.5 = 0
= 2
 
 = 2 =1
 2 2 Discriminant of quadratic equ.
72. (d)
–1  1  4 –1  5
Put  = 45°, sin = =
A

22 4
LHS = sin + cos = sin45° + cos45°
76. (b)
1 1
=  = 2 = RHS tanx tanx
2 2 
1  secx 1  secx
sin6 + cos6 + 6 sin² cos²
 1 6  1 6 1 1  1 – sec x – 1 – sec x 
=
 
 
   6 2  2
 tan x  
 2  2 (1  sec x ) (1 – sec x ) 


1 1 6 1  1  12 14 7  – 2 sec x 
=   = = = = tan x  
8 8 4 8 8 4 1 – sec ²x 


Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

  – 2secx 
= tanx   Let a = tan²x and b = cot²x
 –tan² x 

We know,
2
 Am > Gm
cosθ
sinθ ab
cosθ  ab
2
= 2 cosecx
tan ² x  cot² x
77. (a)  tan ² x cot ² x
2

A
tan²x + cot²x > 2
P>2
80. (a)
cosec4 = 17 + cot4

r
cosec4 – cot4 = 17

si
(cosec²)² – (cot²)² = 17

an by
(cosec² + cot²) (cosec² – cot²) = 17
B 15 cm. C cosec² + cosec² – 1 = 17

n
2 cosec² = 18
4

ja
sin B = cosec² = 9 in first quadrant all functions ar
R s
5 positive
a th
4 AC cosec = + 3
=
5 AB 1
sin =
Let AC = 4x and AB = 5x 3
ty a

Then BC = 3x (Using Pythagorean triplet) 81. (c)

3x = 15
di M

p sec– q cosec
x=5
p sec q cos ec
AB = 5x = 5 × 5 = 25
Divide by cosec in this equ.
78. (c)
p p² p² – q²
sin x 3 p
= p tan – q –q
q  q
–q
sin y 1 =   = q
p² =
p tan q p
q p²  q²
p  q
q  q q
sin x 3  
=
sin(90 – x ) 1
p² – q²
=
A

sin x p²  q²
= 3
cos x
82. (b)
tan x = 3 N = cosec² – 2 + sin²
tanx = tan 60° N = sin² + cosec² – 2
x = 60° N = sin² + cosec² – 2 sin cosec
x + y = 90° N = (sin – cosec)²
y = 90° – 60° = 30° Square of a number is always greater than o
x : y = 60° : 30° equal to zero
=2:1  N > O ie Non-Negative

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

  +  = 90° sin 45 – sin 30
A=
Let, cos 45  cos 60
 = 45° and  = 45° 1 1 2– 2
–
Put the value in this equ.
A= 2 2
 2 2 = 2– 2 = 2 –1
1 1 2 2 2 2 2 1
1 
 sin 45 cos 45 – 2 2 2 2 2
cosec 45 cosec 45 
  

 sin 45 cos 45  sec 45 – tan 45
B=
1
cosec 45  cot 45
–
–
1 1
= cosec45° × (1  1) 2
= 2  (2) 2 = 2 =1 2 –1
2 B=
2 1
84. (c) A=B
2 [1 + tan²A] + 2 [1 + cot²A] 88. (b)

r
= 2 sec²A + 2 cosec²A cosec – sin = p³
[  (a + b)² + (a – b)² = 2 (a² + b²)]

si
1
= 2 [sec²A + cosec²A] – sin = p³
sin

an by
 1 1  = 2 sin ²A  cos ²A  1 – sin ²
= 2     = p³
cos ²A sin ²A   sin ²A cos ²A 
 sin

n

cos ²
1 = p³ ..........(1)
= 2

ja
= 2 sec² A cosec²A sin
cos ² A sin ² A
R s
sec – cos = q3
85. (a) 1
a th
sec x × cosec x = 2 – cos  q 3
cos
1 1 1 – cos2
 =2  q3
cos x sin x cos
ty a

2 sin x cos x = 1
sin²
= q³
di M

sin 2 x = 1 .........(ii)
cos
sin 2 x = sin 90°
(ii) ÷ (i)
2x = 90°
q³ sin² sin³
x = 45° –  = tan³
p³ cos cos³
tannx + cotnx = tann45° + cotn45°
cos ²
= 1n + 1n =2 sin
86. (a)
q
= tan
1 1 p
sin x 
1  cos x 1 – cos x 89. (b)
Put x = 0°, sinx + cosx = C
A

1 – cos x  1  cos x
sin x C = sin0° + cos0°
(1  cos x ) (1 – cos x )
C=0+1
2 C=1
= sin x sin6x + cos6x
1 – cos² x
= sin60° + cos60° = 06 + 16 = 1
2 Check options by putting the value of C = 1,
= sin x
sin ² x option (b) is the right answer.
2 1  6C² – 3C 4 1  6 – 3 7 – 3 4
= sin x × (b) = = = =1
sin x = 2 4 4 4 4

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 cosx cosx 2 tan A (1 – cos B)
 =2 = 2
cosecx  1 cosecx – 1 1 – tan ²A sin B
 cosecx – 1  cosecx  1  1 – cos B 2
1– 
 

cosx  =2  sin B 
(cosecx + 1) (cosecx – 1) 


2 cosecx  2 (1 – cos B) sin ²B – (1 – cos B)²

= ÷
cosx  =2 sin B sin ²B
 cot² x 

2 (1 – cos B) sin B
2 sin²x =
cosx   =2 (1 – cos ²B) – (1 – cos B)²
sinx cos²x
2 tanx = 2 2 (1 – cos B) sin B
=
tanx = 1 (1 – cos B)[1  cos B – 1  cos B]

r
 2 sin B
x = 45 = =
4 2 cos B = tan B

si
91. (a) 94. (d)
Let angles (circular measurement) be x and y

an by
1  2 sincos
x+y=1 ......(i) a² =
1 – 2 sincos

n
180 x 180y
– =1 sin ² cos ² 2 sincos
  a² =
sin ² cos ² – 2 sincos

ja
R s

x–y= .......(ii) (sin cos)²
180 a² =
a th
(sin– cos)²
From equ. (i) to (ii)
Put the value, y = 1 – x a sin cos
=
 1 sin – cos
ty a

2x = 1  By Comp. & Div.

180
a  1 sin cos sin – cos
di M

1   =
x =
  
 a – 1 sin cos – sin cos
 2 360 
Again put the value, x = 1 – y a  1 2 sin
=  tan
a – 1 2 cos
1  
y =1– 
  
 95. (c)
 2 360  sinA + cosA = P
1   Cube both side.
y=
 – 
 sin³A + cos³A + 3sinA cosA (sinA + cosA) = p³
 2 360 
q + 3 sinA cosA (p) = p³
92. (d) .......(1)
A Also, sinA + cosA = P
A

Square both side.

sin²A + cos²A + 2 sinA cosA = p²
v w
1 + 2 sinA cos A = p²
p² – 1
sinA cosA =
2
Put this value in equ. (1) we get
C B
u
w² = u² + v² (p² – 1)
q3 p = p³
2
u v u²  v² w² 2q + 3p³ – 3p = 2p³
tanA + tanB =  = =
v u uv uv p³ – 3p + 2q = 0

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 8 2 16
1 sec ² tan mn =  =
= 15 15 225
x sec ² – tan  64 2
 60 2  4 2
4  16
1  x 2 sec ² (m² – n²)² = 
 –  =
   =
    =
=  225 225   225  15
  225
1– x 2 tan (m² – n²)² = mn
1
1 x cos ² 98. (d)
=
1– x sin x y tan
cos – =1
a b
1 x 1
= x² y² tan ² x y tan
1 – x sincos  –2×  =1 ......(1)
a² b² a b
1– x
= sincos x tan y
1 x  =1
a b
2 – 2x
= 2 sincos x ² tan ² y² x tan y

r
1 x   2  =1 ......(2)
a² b² a b
2 – 2x

si
sin 2 = (1) + (2), we get
1 x
– 1 < sin 2 < + 1 x² y²
(1  tan²)  (tan² 1) = 2

an by
a² b²
 x ² y² 

n

   (1  tan²) = 2

 a² b² 
 x ² y² 

ja
   sec² = 2

2 – 2x
R s
2 – 2x 
–1  1  a² b² 
1 x 1 x
x ² y² 2
a th
 =
– 1 – x < 2 – 2x 2 – 2x < 1 + x a² b² sec ²
x<3 1 < 3x x ² y²
 = 2 cos ²
a² b²
ty a

1
x 99. (b)
3 p = cot + tan
1
di M

 x  3 Put  = 45°
3 p=1+1=2
97. (b) q = sec – cos
Let,
1 1
q= 2– =
2 2
2 2
2 2
3 5
 1 3  1 3
3 3
(p²q) – (q²p) = 
4   –
   2

 2  2 
2 2
= (2 2) 3 – (1) 3
4 2
ATQ,
4m = cot (1 + sin) = ( 2³) 3 – 1 = 2 – 1 = 1
A

4  3  32 100. (d)
4m =   1    A + B + C = 180°
3  5  15
8
B + C = 180° – A
m= B  C A
15 = 90 –
Again, 2 2
4n = cot (1 – sin) B  C   A
sin 
  = sin 
 90 – 

4  3 8  2   2
4n = 1 – 
 
3  5  15
B  C  A
2 sin 
  = cos

n=  2  2
15

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 TRIGONOMETRY/f=kdks.kfefr
MAXIMA/MINIMA
(Class Room Sheet)
1. Find the min & max value of 3 – 2 sin3 6. Find minimum and maximum value of
3 – 2 sin  3
dk U;wure vkSj vf/dre eku Kkr dk U;wure vkSj vf/dre eku Kkr dhft,
dhft, (i) 5sin² + 4 cos²
(a) –1, 3 (b) 1, 5 (a) 0, 4 (b) None of these

r
(c) 1, 3 (d) 0, 5 (c) 0, 5 (d) 4, 5

si
2. Find the minimum and maximum value of (ii) 3sin² – 5 cos²
3 sin + 4 cos (a) 3, 5 (b) –5, 3

an by
3 sin + 4 cos dk U;wure vkSj vf/dre eku (c) –3, 5 (d) None of these
Kkr dhft,

n
1 1
(a) –5, 5 (b) 3, 4 (iii) sin 2  + cos 2 
2 3
(c) 0, 5 (d) –3, 4

ja
3.
expressions.
R s
Find the min & max value of the following
(a) 0,
1
3
(b) –
1 1
,
3 2
a th
fuEufyf•r O;atdksa dk U;wure vkSj vf/dre eku
1 1
Kkr dhft,A (c) ,
3 2
(d) None of these
(i) 7 sin – 24 cos
ty a

(iv) – 5sin² – 3 cos²

(a) 7, 24 (b) –25, 25
(a) –5, –3 (b) 5, 3
di M

(c) –7, 24 (d) –25, 0

(c) –5, 3 (d) 3, 5
(ii) sin + cos
7. The minimum value of 2 sin² + 3 cos² is:
(a) –2, 2 (b) –2, 2
2 sin² + 3 cos² dk U;wure eku gSA
(c) –1, 1 (d) 0, 2
(a) 0 (b) 3
(iii) 5 sin + 12 cos – 3 (c) 2 (d) 1
(a) –16, 10 (b) –13, 13 8. If Y = 4 tan² + 9 cot2 then find Ymin = ?
(c) –16, 16 (d) 10, 13
;fn Y = 4 tan² + 9 cot2 gS] rks
Y dk U;uwre
4. Find minimum and maximum value of the
eku Kkr dhft,\
following
(a) 12 (b) 4
fuEufyf•r dk U;wure vkSj vf/dre eku Kkr dhft,
(c) 6 (d) 9
A

(i) 3 sin + cos

9. If Y = 8 tan² + 2 cot2then find Ymin = ?
(a) –10, 10 (b) –5, 5
(c) 0, 3 (d) –3, 3 ;fn Y = 8 tan² + 2 cot2 gS] rks
Y dk U;uwre
(ii) –7 sin + 24 cos eku Kkr dhft,\
(a) –25, 25 (b) –7, 24 (a) 8 (b) 4
(c) 7, 24 (d) None of these (c) 2 (d) 6
2
5. Find minimum and maximum value of 10. If Y = 3 sin² + 12 cosec  then find Ymin = ?
dk U;wure vkSj vf/dre eku Kkr dhft, ;fn Y = 3 sin² + 12 cosec2gS] rks
Y dk U;uwre
Y = 3 sin² + 4 cos² eku Kkr dhft,\
(a) 3, 4 (b) 0, 3 (a) 12 (b) 15
(c) 0, 4 (d) None of these (c) 3 (d) 9
11. If Y = 12 sin² + 3 cosec2then find Ymin = ? 17. Find minimum value of the following
;fn Y = 12 sin² + 3 cosec gS] rks
2
Y dk U;uwre fuEufyf•r dk U;wure eku Kkr dhft,
eku Kkr dhft,\ (a) 4 sec² + 25 cosec²
(a) 36 (b) 49
(a) 12 (b) 15
(c) 25 (d) 16
(c) 3 (d) 9
(b) 100 sec² + 9 cosec²
2
12. If Y = 8 cos² + 18 sec then find Ymin = ? (a) 144 (b) 121
;fn Y = 8 cos² + 18 sec2gS] rks
Y dk U;uwre (c) 169 (d) 189
eku Kkr dhft,\ 18. Find the minimum and Maximum value of
(a) 24 (b) 12 2 sin cos
(c) 26 (d) 18 2 sin cos dk U;wure vkSj vf/dre eku Kkr
dhft,

r
13. If Y = 18 cos² + 8 sec2then find Ymin = ?
;fn Y = 18 cos² + 8 sec2gS] rks
Y dk U;uwre

si
1 1
(a) 0, 1 (b) – ,
eku Kkr dhft,\ 2 2

an by
(a) 24 (b) 12 (c) –1, 1 (d) None of these
19. Find the minimum and Maximum value of
(c) 26 (d) 18

n
Y = sin3 . cos3
14. Find minimum value of the following.
Y = sin3 . cos3 dk U;wure vkSj vf/dre eku
fuEufyf•r dk U;wure eku Kkr dhft,A

ja
Kkr dhft,
R s
(i) 4 sin² + 9 cosec²
1 1 1 1
a th
(a) 13 (b) 4 (a) – , (b) – ,
8 8 3 3
(c) 9 (d) 5
(ii) 8 cos² + 18 sec² 1
(c) 0, (d) None of these
8
ty a

(a) 8 (b) 26
20. The least value of cos sin is:
(c) 18 (d) 10
cos sin dk U;wure eku gksxkA
di M

(iii) 12 sin² + 3 cosec²

SSC Phase X 05/08/2022 (Shift- 03)
(a) 3 (b) 15
(c) 12 (d) 9 1
(a)   (b) 0
(iv) 4 sec² + 9 cos² 2
(a) 4 (b) 9
 1
(c) 5 (d) 12 (c) (–1) (d)  – 
 2
15. Find the minimum value of:
21. Find the minimum and Maximum value of
dk U;wure eku Kkr dhft,% Y = sin4 . cos4
(i) 32 cos² + 2 tan² dk U;wure vkSj vf/dre eku Kkr dhft,
A

(a) 4 (b) 14
1 1 1
(c) 2 (d) None of these (a) 0, (b) – ,
16 16 16
(ii) 4 sin² + 64 cot²
1
(a) None of these (b) 8 (c) None of these (d) – ,1
16
(c) 4 (d) 6
22. Find the minimum and Maximum value of
(iii) 4 sec² + 9 cosec²
Y = 32 sin5 . cos5
(a) 4 (b) 9
dk U;wure vkSj vf/dre eku Kkr dhft,
(c) 6 (d) 25
16. Y = 16 sec² + 25 cosec2 Ymin = ?? 1 1
(a) – , (b) –1, 1
(a) 81 (b) 64 32 32

(c) 49 (d) 36 (c) 0, 1 (d) None of these

23. Max(sin6 + cos10) 3
(a) 1  A  2 (b) A1
(a) 2 (b) 1 4
(c) 3 (d) 4 13 3 13
(c) A1 (d) A
24. Max(sin20 + cos40) 16 4 16
28. Find the minimum value of (sin + cosec)²
(a) 1 (b) 2 + (cos + sec)².
(c) 0 (d) 3 (sin + cosec)² + (cos + sec)² dk U;wure
25. The greatest value of sin4 + cos4 is: eku dhft,A
sin4 + cos4 dk vf/dre eku gSA (a) 8 (b) 7
(a) 2 (b) 3 (c) 9 (d) 4
1 29. Find the maximum and minimum values
(c) (d) 1 of 7cos + 24sin.
2

r
26. The minimum value of sin² + cos² + sec² 7cos + 24sin dk U;wure vkSj vf/dre eku

si
+ cosec² + tan² + cot² is: dhft,A
sin² + cos² + sec² + cosec² + tan² + cot² (a) –7 and 7 (b) –25 and 25

an by
dk U;wure eku gSA (c) –24 and 24 (d) –31 and 31
(a) 1 (b) 3 30. Find the maximum and minimum values

n
(c) 5 (d) 7 of 5sin² + 10cos² + 12sin cos
27. If A = sin² + cos4, for any value of , then 5sin² + 10cos² + 12sin cos dk U;wure vkS

ja
the value of A is: vf/dre eku Kkr dhft,A
R s
;fn  ds fdlh eku ds fy, A = sin² + cos4 (a) (1,12) (b) (0,14)
a th
gS rks
A dk eku gSA (c) (1,14) (d) (–1,14)
ty a
di M

Answer Key
1.(b) 2.(a) 3.i (b) ii.(a) iii.(a) 4.i (a) ii.(a) 5.(a) 6.i (d) ii.(b)

iii.(b) iv.(a) 7.(c) 8.(a) 9.(a) 10.(b) 11.(a) 12.(c) 13.(a) 14.i (a)

ii.(b) iii.(c) iv.(d) 15. i.(b) ii.(c) iii.(d) 16.(a) 17.i (b) ii.(c) 18.(b)

19.(a) 20.(d) 21.(b) 22.(b) 23.(b) 24.(a) 25.(d) 26.(d) 27.(b) 28.(c)
A

29.(d) 30.(c)
 HEIGHT AND DISTANCE
(Å¡pkbZ vkSj nwjh)
Important Concept voueu dks.k % ;fn izs{kd fdlh ,slh oLrq dk voyksdu
(i) Observer : Who observes the object. djrk gS tks mlls uhps gks vkSj ml oLrq dks ns[kus
izs{kd % tks oLrq dk voyksdu djrk gSA izs{kd dks viuh utj uhps dh vksj djuh iM+rh gks
(ii) Object : Which is observed by the observer. bl izdkj n`f"Vjs[kk }kjk {kSfrt js[kk ls cuk, x, d
oLrq % ftldk voyksdu izs{kd }kjk fd;k tkrk gSA dks voueu dks.k dgrs gSaA
(iii) Line of sight : Line of sight is a straight line,
which joins the observer to the object. Object
n`f"ViFk js[kk % n`f"V js[kk oLrq dks izs{kd ls tksM+us okyh
Angle of depression
Horizontal line
lhèkh js[kk gksrh gSA

r
Line of sight
(iv) Horizontal line : If a line drawn horizontally

si
from an observer, it called the horizontal
line.

an by
{kSfrt js[kk % ;fn izs{kd ls dksbZ {kSfrt js[kk [khaph tkrh
gS rks mls {kSfrt js[kk dgrs gSaA

n
(v) Angle of elevation : If an observer observes Observer

ja
an object, which is above the observer and
R s
to see the object observer has to elevate or (vii) Relation in angle of elevation and depression
raise his line of sight. Thus the angle formed : To solve the problems of the height and
a th

by the line of sight and horizontal line is disatnce, we take angle of elevation instead
called the angle of elevation. of angle of depression and which is same in
mÂ;u dks.k % ;fn izs{kd fdlh ,slh oLrq dk voyksdu measurement. In the given figure, if angle
djrk gS tks mlls Åij gks vkSj ml oLrq dks ns[kus ds of depression (q) is given, we can take q as
ty a

angle of elevation. i.e. Mathematically both

fy, izs{kd dks viuh utj Åij dh vksj mBkuh iM+rh gks]
are equal.
di M

rks bl izdkj n`f"Vjs[kk }kjk {kSfrt js[kk ls cuk, x, dks.k

dks mÂ;u dks.k dgrs gSaA mÂ;u dks.k vkSj voueu dkss.k esa laca/ % Å¡pkbZ
ds iz'uksa dks gy djus ds fy, ge voueu dks.k ds LFk
Object ij mÂ;u dks.k ysrs gSa tks fd eki esa leku gksrk gS
x;s fp=k esa ;fn voueu dks.k (q) fn;k x;k gks rks ge
q dks mÂ;u dks.k ds :i esa fy[k ldrs gSaA xf.krh; :i
Line of sight esa nksuksa cjkcj gSaA

Angle of
Angle of depression
elevation
A

Observer
Horizontal line Angle of elevation

(vi) Angle of depression : If an observer observes

an object, which is below the observer and
to see the object the observer has to depress (viii) Sun and Shadow : If a person standing
or lower his line of sight. Thus the angle infront of sun, then its shadow is formed.
formed by the line of sight and the lw;Z vkSj Nk;k % ;fn dksbZ O;fDr lw;Z ds vfHk
horizontal line is called the angle of
gks rks mldh Nk;k curh gSA
depression.
 (ii) Angle of elevation is 45º
A ,sls f=kHkqt ftlesa mÂ;u dks.k
45º gksrk gSA
If one of the angle of a right-angled triangle
is 30º, the other acute angle must be 60º and the
ratio of the sides will be shown as follows :
Person
Angle of ;fn fdlh ledks.k f=kHkqt dk ,d dks.k30º gS] rks
elevation nwljk U;wu dks.k
60º gksxk vkSj Hkqtkvksa dk vuqikr bl
fn•k;k tk,xk%
B O
Shadow of the person
The shadow of a building or tower is formed 60º 30º
2 2
as follows : 1 3
fdlh bZekjr ;k ehukj dh Nk;k fuEu izdkj curh gS % 30º 60º

3 1
A
If one of the angle of a right-angled triangle
is 45º, then the other acute angle will also be 45º
Building and the ratio of the sides will be as shown below :
Angle of
elevation of sun ;fn fdlh ledks.k f=kHkqt dk ,d dks.k45º gS] rks
nwljk U;wu dks.k45ºHkh gksxk vkSj Hkqtkvksa dk vuqi
B O çdkj fn•k;k tk,xk%
Shadow of building
(ix) The height and distance problems are solved
with the help of trigonometric ratios of angle
of elevation in a right-angled triangle. 45º
2
Å¡pkbZ ,oa nwjh ds iz'u ledks.k f=kHkqt ds mÂ;u dks.k ds 1
f=kdks.kferh; vuqikr dh enn ls gy fd;s tkrs gSaA
45º
A
1

Example-1
Angle of
elevation
In the given right-angle ABC, C = 30º and
BC = 18 cm, what will be the value of AB and AC?
fdlh ledks.k f=kHkqt
ABC esa
C = 30º vkSjBC = 18
B C lseh gS rks
AB vkSjAC ds eku D;k gksxsa\
AB A
Trigonometric ratios are sinq = ,
AC
BC AC AC
cosq = , tanq = , secq =
AC AB BC
BC AB 30º
cosecq = , cotq = B C
AB AC
18 cm
(x) In most of the problems of height and
distance, we come across the two special Example-2
right-angled traingles as follow : In the given right-angle triangle ABC, C =
ÅapkbZ vkSj nwjh dh vf/dka'k iz'uksa esa] gesa60º nksand
fo'ks"k
BC = 6 cm, what will be the value of AB
ledks.k f=kHkqt feyrs gSa tks bl çdkj gSa% and AC?
(i) Angle of elevation is 30º or 60º. fdlh ledks.k f=kHkqt
ABC esa
C = 60º vkSjBC = 6
,sls f=kHkqt ftlesa mÂ;u dks.k lseh
30º ;k 60º gksrk gSA
gS rks
AB vkSjAC ds eku D;k gksxsa\
 A (b)

3
1
60º 30º 60º
B C
3
6 cm
Example-3
(c)
In the given right angled triangle ABC, C =
45º and BC = 8 cm, what will be the value of AB
and AC? 3
fdlh ledks.k f=kHkqt
ABC esa
C = 45º vkSjBC = 8
lseh gS rks
AB vkSjAC ds eku D;k gksxsa\ 60º 30º
A 1 3

r
si
an by
45º

n
B C
8 cm
3
30º-60º Relation
ja
R s
In more than 50% of height and distance (d)
a th

problems in which two right-angled triangles are 1

formed, angles of elevation or angles of 60º 30
º
depression will be 30º and 60º.
3
Å¡pkbZ vkSj nwjh dh 50 izfr'kr ls vf/d iz'uksa esa ftlesa
ty a

nks ledks.k f=kHkqt curs gSa] mUu;u dks.k ;k voueu

30º dks.k
vkSj60º gksaxsA
di M

In these height and distance problems in

which two angles of elevation or angles of 1
depression are 30º or 60º. In these problems two
30 º
right-angled triangles will be formed. If one of 3
the side of these two right angled triangle is
common or of equal length, the other side will
be in the ratio 1 : 3. (e)
Å¡pkbZ vkSj nwjh ds ,sls iz'u ftuesa nks mUu;u dks.k ;k
voueu dks.k 30º ;k 60º gksrs gSaA ,sls iz'uksa esa nks ledks.k 60º
f=kHkqt curs gSaA ;fn bu nks ledks.k f=kHkqt dh ,d Hkqtk
3
A

mHk;fu"B ;k leku yackbZ dh gks] rks nwljh Hkqtk 1% 3 ds

vuqikr esa gksxhA (f)

(a) 1 1
30º
60º
60º
3 30º

3 3

60º 30º
1 3
3 3
 Type-1 10 ehVj yach ,d lh<+h nhokj ds lgkjs [kM+h g
tehu ls 30º ds dks.k ij >qdh gqbZ gSA nho
In this type, figure formed will be a right- lh<+h ds ikn dh nwjh (ehVj esa) gS (fn;k
angled triangle and the angle of elevation will be
given or asked as shown below : 3 = 1.732)
bl çdkj ds iz'uksa esa cuus okyh vkÑfr ,d ledks.k (a) 8.16 (b) 7.32
(c) 8.26 (d) 8.66
f=kHkqt gksxh vkSj mUu;u dks.k ;k rks fn;k x;k gksxk ;k iwNk
tk,xk] tSlk fd uhps fn[kk;k x;k gSA 5. The shadow of tower is 3 times its
height. Then the angle of elevation of the
top of the tower is :
fdlh ehukj dh Nk;k bldh Å¡pkbZ dh3 xquh gS]
Height

rks ehukj ds 'kh"kZ dk mÂ;u dks.k gS %

(a) 45º (b) 30º
Angle of elevation
(c) 60º (d) 90º
Distance 6. A man 6 ft tall casts a shadow 4 ft long,

r
1. 129 meter from the foot of a cliff on level at the same time when a flag pole casts
of ground, the angle of elevation of the a shadow 50 ft long. The height of the flag

si
top of a cliff is 30º, the height of this cliff pole is :
is : tc ,d èot LraHk dh Nk;k 50 iQhV yach gks r

an by
fdlh LraHk ds ikn ls 129 ehVj nwj /jkry ij mlh le; 6 iQhV yacs fdlh O;fDr dh Nk;k 4 iQhV
fLFkr fdlh fcUnq ls LraHk ds 'kh"kZ dk mUu;u dks.k
yach gSA èot LraHk dh Å¡pkbZ gS &

n
30º gS] LrHk dh ÅapkbZ gS % (a) 80 ft (b) 75 ft
(c) 60 ft (d) 70 ft
(a) 50 3 metre

ja(b) 45 3 metre
R s
7. In the length of the shadow of a girl is
(c) 43 3 metre (d) 47 3 metre
same as her height, then the angle of
a th

2. From a point P on a level ground, the elevation of the sun is :

angle of elevation of the top of a tower is ;fn fdlh yM+dh dh izfrNk;k mldh Å¡pkbZ ds cjk
30°. If the tower is 110 3 m high, what gS] rks lw;Z dk mÂ;u dks.k gS %
ty a

is the distance (in m) of point P from the (a) 30º (b) 45º
foot of the tower? (c) 60º (d) 75º
lery Hkwfe ij fdlh fcanqP ls ,d ehukj ds 'kh"kZ dk8.
di M

A kite is attached to a string. Find the

mUUk;u dks.k
30° gSA ;fn ehukj length of the string (in m) when the
110 3 ehVj Åaph gS]
height of the kite is 90 m and the string
rks ehukj ds ikn ls fcanq
P dh nwjh (ehVj esa) D;k gS\ makes an angle of 30° with the ground.
SSC CGL Pre (2021) dksbZ irax ,d Mksjh ls tqM+h gSA Mksjh dh y
(a) 330 (b) 220 esa) Kkr dhft,] tc irax dh ÅapkbZ 90 ehVj g
(c) 115 (d) 110 vkSj Mksjh tehu ds lkFk
30° dk dks.k cukrh gSA
3. What is the angle of elevation of the sun, SSC CGL Pre (2021)
when the shadow of a pole of height x m
x (a) 180 (b) 90 3
is m?
3 (c) 45 (d) 60 3
x
A

9. From a point P ona level ground, the

;fn x ehVj Å¡ps fdlh [kaHks dh Nk;k ehVj angle of elevation of the top of the tower
3
gks] rks lw;Z dk mÂ;u dks.k D;k gS\ is 30°. If the distance of point P from the
(a) 30º (b) 45º foot of the tower is 510 m, then 50% of
(c) 60º (d) 75º the height of the tower (in m) is:
4. A 10 metre long ladder is placed against lery Hkwfe ij ,d fcanqP ls ehukj ds f'k[kj dk
a wall. It is inclined at an angle of 30º to mUUk;u dks.k° gSA30 ;fn ehukj ds ikn ls fcanq
P
the ground. The distance (in m) of foot of dh nwjh 510 ehVj gS] rks ehukj dh Å¡pkbZ d
the ladder from the wall is (given 3 = (ehVj esa) Kkr dhft,A
1.732) SSC CGL Pre (2021)
 14. The length of the shadow of a vertical
85 3 pole on the ground is 24m. If the angle
(a) 85 (b)
3 of elevation of the sun at that time is ,
(c) 85 3 (d) 150 3 5
such that sin = , then what is the
13
10. A kite is flying at a height of 50 metre. height of the pole?
If the length of string is 30 metre, then
the angle of inclination of string of the ,d yacor [kaHks dh Hkwfe ij ijNkbZ dh yack
horizontal ground in degree measures is : ehVj gSA ;fn ml le; lw;Z ds mÂ;u dk dks.k

dksbZ irax 50 ehVj dh Å¡pkbZ ij mM+ jgh gSA ;fnbl izdkj gS fd sin = 5
gS] rks [kaHks dh Å
Mksj dh yackbZ 30 ehVj gS] rks {kSfrt /jkry ls Mksj 13
Kkr djsaA
ds >qdko dh eki fMxzh esa gS &
SSC CPO 13 March 2019 (Evening)
(a) 90º (b) 60º
(a) 8 m (b) 10 m
(c) 30º (d) 45º
(c) 12 m (d) 18 m
11. A boy is standing near a pole which is 2.7
15. A girl 1.2 m tall can just see the sun over
m high and the angle of elevation is 30º. a 3.62 m tall wall which is 2.42 m away
The distance of the boy from the pole is :

r
from her. The angle of elevation of the
,d yM+dk ,d LraHk ds ikl [kM+k gS tks sun is :

si
2-7 ehVj Å¡pk gS vkSj bldk mÂ;u dks.k
30º gSA 1-2 ehVj yach ,d yM+dh 3-62 ehVj yach nho
yM+ds dh LraHk ls nwjh Kkr dhft,A
 3 = 1.73 ds Åij dsoy lw;Z dks ns[k ikrh gS tks mlls 2-

an by
SSC CPO 16 March 2019 (Morning)
ehVj nwj gSA lw;Z dk mÂ;u dks.k gS %

n
SSC CPO 16 March 2019 (Afternoon)
(a) 4.42 m (b) 4.53 m
(a) 60º (b) 30º
(c) 4.68 m (d) 4.63 m

ja (c) 90º (d) 45º

R s
12. At a certain time of a day a tree 5.4 m 16. A ladder attached to the wall makes an
height casts a shadow of a 9 m. If a pole angle of 60º to the horizontal of the land.
a th

casts a shadow of 13.5 m at the same If the lower end of the ladder is 10 meters
time, the height of the pole is : away from the wall, what will be the
fnu ds fdlh le; esa 5-4 ehVj Å¡ps isM+ dh 9 length of the ladder?
ehVj dh ijNkbZ curh gSA ;fn blh le; ,d [kaHks nhokj ds lgkjs yxh gqbZ ,d lh<+h] Hkwfe ds
ty a

60º dk dks.k cukrh gSA ;fn lh<+h dk fupyk fl

dh ijNkbZ 13-5 ehVj dh gS] rks [kaHks dh Å¡pkbZ Kkr djsaA
nhokj ls 10 ehVj nwj gS] rks lh<+h dh yEckbZ D;
di M

SSC CPO 16 March 2019 (Morning)

SSC CPO 16 March 2019 (Morning)
(a) 8.1 m (b) 9.9 m
(a) 20 m (b) 40 m
(c) 7.2 m (d) 6.3 m
(c) 17.3 m (d) 34.6 m
13. A ladder leaning against a wall makes an 17. The length of the shadow of a vertical
angle  with the horizontal ground such pole on the ground is 36m. If the angle
12 of elevation of the sum at that time is ,
that sin = . If the foot of the ladder
13 13
is 7.5 m from a wall, then what is the such that sec = , then what is the
12
height of the point from where the top of height (in cm) of the pole?
the ladder touches the wall?
Hkwry ij ,d v/ksyac [kaHks dh ijNkbZ dh yack
fdlh nhokj ij >qdh gqbZ ,d lh<+h {kSfrt Hkwfe ds ehVj gSA ;fn ml le; lw;Z dh Å¡pkbZ dk mÂ
A

lkFk dks.k cukrh gS tks bl izdkj gS fd

sin = 13
12
dks.k ,sls gSa] tSlssec
fd = gS] rks [kaHks
12
gSA ;fn lh<+h dk ry nhokj ls 7-5 ehVj gS] rks Å¡pkbZ (ehVj esa) D;k gS\
13
ml fcUnq dh Å¡pkbZ Kkr djsa tgk¡ ls lh<+h dk 'kh"kZ SSC CPO 24/11/2020 (Morning)
Hkkx nhokj dks Li'kZ djrk gSA (a) 12 (b) 18 (c) 9 (d) 15
SSC CPO 12 March 2019 (Evening) 18. Asha and Suman's mud forts have heights
9 cm and 16 cm. Their tops are 25 cm
(a) 15 m (b) 8 m
part from each other, then the distance
(c) 18m (d) 12 m between two forts is :
 vk'kk vkSj lqeu feV~Vh ds fdyks dh Å¡pkbZ 9 lseh fdlh ehukj ds vk/kj ls 70 ehVj nwj fLFkr fdlh
vkSj 16 lseh gSA muds 'kh"kZ ,d nwljs ls 25 lseh fcUnq dk voueu dks.k60º gSA ehukj dh Å¡pkbZ
vyx gS] fiQj nks fdyks ds chp dh nwjh gS & (a) 35 3 m (b) 70 m
SSC CPO 24/11/2020 (Morning)
(a) 16 (b) 25 (c) 7 (d) 24 70 3
(c) m (d) 70 3 m
3
Type-2 22. From the top of a light house at a height
20 m above the sea level, the angle of
In this type, we will study the question in depresion of a ship is 30º. The distance
which the figure formed will be right-angle and of the ship from the foot of the light
the angle of depression will be given or asked as house is :
shown below.
leqnz ry ls 20 ehVj Å¡pkbZ ij fLFkr fdlh izdk'k?
bl çdkj ds iz'uksa esa cuus okyh vkÑfr ,d ledks.k ds 'kh"kZ ls tgkt dk voueu dks.k
30º gSA izdk'k?k
f=kHkqt gksxh vkSj voueu dks.k ;k rks fn;k x;k gksxk ;k iwNk
ds vk/kj ls tgkt dh nwjh gS %
tk,xk] tSlk fd uhps fn[kk;k x;k gSA
(a) 20 m (b) 20 3 m

r
(c) 30 3 m (d) 30 m
Angle of desperssion

si
23. Pintu is flying a kite with a 60 m long
thread. If the angle of depression of Pintu

an by from kite is 45º, the height of the kite

above the ground is :
height

fiaVw ,d irax mM+k jgk gS ftlds /kxs dh yack

n
ehVj gSA ;fn irax ls fiaVw dk voueu dks.k
45º
gS] rks Hkwfe ls irax dh Å¡pkbZ gS %
ja
R s
Angle of elevation
(a) 60 m (b) 60 3 m
a th

Distance
(c) 30 m (d) 20 3 m

In this type, we will take

Type-3
bl izdkj ds iz'uksa esa]
ty a

In this type, a telegraph post or a pole is

angle of desperssion = Angle of elevation
bent or broken at a certain height and the top or
voueu dks.k = mÂ;u dks.k
di M

upper part meets or touch the ground at a certain

19. From 125 metre high tower, the angle of distance.
depression of a car is 45º. How far the car bl izdkj ds iz'uksa esa dksbZ VsyhiQksu [kaHkk
is from the tower? fdlh fuf'pr Å¡pkbZ ls eqM+ ;k VwV tkrs gSa vkSj bldk
125 ehVj Å¡ph ehukj ls fdlh dkj dk voueu Åijh Hkkx fdlh fuf'pr nwjh ij tehu dks Li'kZ djrk gSA
dks.k45º gSA dkj] ehukj ls fdruh nwjh ij gS \ Pole or post = AB which gets broken at point
(a) 125 m (b) 75 m C and the top meets the ground at point D.
(c) 95 m (d) 60 m
[kaHkk ;k LraHk
= AB tks fcUnqC ls VwV@eqM+ x;k g
20. The angle of depression of a stone
situated at a distance of 80 m from the 'kh"kZ tehu dks fcUnq
D ij Li'kZ djrk gSA
base of a pole is 30º, the height of the A
A

pole is :
fdlh [kaHks ds vk/kj ls 80 ehVj nwj fLFkr fdlh
iRFkj dk voueu dks.k30º gSA [kaHks dh Å¡pkbZ gS %
(a) 40 m (b) 160 m
80 C
(c) 80 3 m (d) m
3
21. The angle of depression of a point
situated at a distance of 70 m from the
base of a tower is 60º. The height of tower B D
is :
 AB = CD = Same part of the pole or post. m from its foot and make an angle of 30º,
In this type of question, in most of the cases, the height of the post is :
the angle of elevation will be 30º and in this case ,d VsyhiQksu dk [kaHkk rwiQku ds dkj.k Hkw
bl izdkj ds iz'u esa] vf/dka'k fLFkfr;ksa esa mÂ;u dks.k fdlh fcUnq ls eqM+ tkrk gSA bldk 'kh"kZ] b
30º gksxk vkSj bl fLFkfr esa ls 8 3 ehVj nwjh ij fLFkr fdlh fcUnq 30º ij
A dk dks.k cukrs gq, Li'kZ djrk gSA [kaHks dh Å¡
(a) 16 m (b) 23 m
(c) 24 m (d) 10 m
27. The top of a broken tree touches the
ground at 60 degree angle, 45 meters
C away from the root of the tree. What will
be the total height of the tree?
VwVs gq, ,d isM+ dk 'kh"kZ isM+ dh tM+ ls
30º nwj60º ds dks.k ij Hkwfe dks Li'kZ djrk gSS
B D dh dqy Å¡pkbZ D;k gksxh\

r
Height = 3 × Distance Use 3 = 1.73 and 
2 = 1.41

si
AB = 3 × AD SSC CPO 14 March 2019 (Evening)
24.

an by
A telegraph post gets broken at a point
against storm and its top touches the
(a)
(b)
153.45 m
141.3 m

n
ground at a distance 20 m from the base (c) 167.85 m
of the post making an angle 30º with the (d) 137.24 m
ground. What is the height of the post?

ja
R s
VsyhiQksu dk ,d [kaHkk rwiQku ds dkj.k 'kh"kZ ls VwVdj Type-4
blds vk/kj ls 20 ehVj dh nwjh ij fLFkr fdlh
a th

In This type, we will study the questions in

fcUnq ij30º dk dks.k cukrs gq, Li'kZ djrk gSA
which two angles of elevation with same height
[kaHks dh Å¡pkbZ D;k gS\ will be given.
40 bl izdkj esa ge ,sls iz'uksa ds ckjs esa i<+saxs ftle
ty a

(a) m (b) 20 3 m
3 Å¡pkbZ ds lkFk nks mÂ;u dks.k fn, x, gksaxsA
(c) 40 3 m (d) 30 m
di M

In the given figure two angles of elevation

25. A straight trees breaks due to storm and 1 and 2 are given.
the broken part bends so that the top of
the tree touches the ground making an fn;s x;s fp=k esa] nks mÂ;u dks.k
1 vkSj2 fn;s x;s gSaA
angle 30º. The distance from the foot of In most of the problems, of this type
the tree to the point, where the top bl izdkj ds vf/dka'k iz'uksa esa]
touches the ground is 10 m. Find the
total height of the tree (in m)? (i) If AB will be given, then CD is asked
,d lh/k o`{k rwiQku ds dkj.k VwV tkrk gS vkSj VwVk;fn AB fn;k gks] rks CD dh yackbZ iwNh tkrh gS
gqvk Hkkx bl izdkj eqM+rk gS fd o`{k dk Åijh Hkkx (ii) If CD will be given, then AB is asked.
Hkwfe ij30º dk dks.k cukrs gq, Li'kZ djrk gSA o`{k ;fn CD fn;k gks] rks
AB dh yackbZ iwNh tkrh gS
ds vk/kj ls ml fcUnq dh nwjh ftl ij 'kh"kZ Hkwfe
A

dks Li'kZ djrk gS] 10 ehVj gSA o`{k dh dqy Å¡pkbZ A

(ehVj esa) Kkr dhft,A
(a) 10 3 (b) 10  3 +1
h
10 3
(c) 10  3  1 (d)
3
26. A telegraph post is bent at a point above D
B C d
the ground due to storm. Its top just
meets the ground at a distance of 8 3
First Method/çFke fof/ (d) A

This type of problems are solved by

considering the two right-angled triangles ABC
and ABD. 1

bl izdkj ds iz'uksa dks nks ledks.k f=kHkqt

ABC vkSj
ABD ekurs gq, gy djrs gSaA
30º 15º
B C D
Second Method/f}rh; fof/ 3 2
(e) A
A

h
h

2
D B C D

r
B C
x y
d = h(cot1 – cot2)

si
h2 = y2 – x2
Third Method/r`rh; fof/ (f) If both angles of elevation are

an by
For this we will see the relation between AB
and CD in different cases.
complementary
;fn nksuksa mÂ;u dks.k lEiwjd gksa rks

n
blds fy, ge AB vkSjCD ds eè; fofHk fLFkfr;ksa esa h2 = xy or h = xy
laca/ ns[krs gSaA

ja
R s
(a) A
a th

h
1
ty a

(90º– )
45º 30º x
B D
C
di M

1 3 –1
28. If the angle of elevation of the sun
(b) A changes from 30º to 45º, the length of the
shadow of a pillar decreases by 20 metres.
The height of the pillar is :
1 ;fn lw;Z dk mÂ;u dks.k
30º ls 45º gksrk gS] rk
fdlh [kaHks dh Nk;k dh yackbZ 20 ehVj de
60º 30º tkrh gSA [kaHks dh Å¡pkbZ gS %
B D
1 C 2
3 3
(a) 20  3  1 m (b) 20  3 +1 m

(c) 10  3  1 m (d) 10  3 +1 m

A

A
29. The shadow of a tower standing on a level
plane is found to be 50 m longer when the
1 sun's elevation is 30º. What is the height
of the tower when it is 60º?
(c)
60º
tc lw;Z dk mÂ;u dks.k60º ls 30º gks tkrk gS rk
45º
B C D lery Hkwfe esa [kM+s fdlh ehukj dh Nk;k dh
1  1 
3
1 -
3
 esa 50 ehVj dh o`f¼ gks tkrh gSA ehukj dh Å
Kkr djsaA
 (a) 25 m (b) 25 3 m fcanqM vkSjN ,d bekjr ds vk/kj ls ,d lh/h
25 js[kk esa Øe'k% 72 ehVj vkSj 128 ehVj dh nw
(c) m (d) 30 m
3 fLFkr fcanq gSa vkSj bu fcanqvksa ls ml bekj
30. A and B are two points on the same side
of a ground, 50 metres apart. The angles dk mUUk;u dks.k iwjd gSaA ml bekjr dh Å¡pkb
of elevation of these points to the top of eas) fdruh gS\
a tree are 60° and 30°, respectively. What SSC CGL Pre (2021)
is 40% of the height of the tree (in m)? (a) 84 (b) 96
A vkSjB fdlh isM+ ds ,d gh rjiQ fLFkr nks fcanq (c) 80 (d) 90
gSa] ftuds chp dh nwjh 50 ehVj gSA bu fcanqvksa
34. ls
The length of the shadow of a vertical
isM+ dh pksVh ds mUUk;u dks.k °Øe'k%
vkSj 30
60
° tower on level ground increases by 8.4 cm
when the altitude of the sun changes
gSA isM+ dh Å¡pkbZ dk 40» (ehVj esa) fdruk gS\
from 45° to 30°. What is the height of the
SSC CGL Pre (2021) tower (in m)?
(a) 10 3 (b) 15 3 tc lw;Z dk mUu;u dks.k °45
ls 30° gks tkrk gS] rk
(c) 5 3 (d) 25 3
lery Hkwfe ij ,d ÅèokZ/j ehukj dh Nk;k dh

r
yackbZ 8-4 lseh c<+ tkrh gSA ehukj dh Å¡pkb
31. The angle of elevation of the top of a
esa) fdruh gS\

si
tower from two points A and B lying on
the horizontal through the foot of the SSC CGL Pre (2021)

an by
tower are respectively 15º and 30º. If A
and B are on the same side of the tower
and AB = 48 metre, then the height of the
(a) 4.2  3 –1  (b) 8.4  33 

n
tower is : (c) 4.2  3  3 (d) 4.2  3  1
ehukj ds vk/kj ds {kSfrt lery esa fLFkr nks fcUnq

ja 35. When the sun's angle of depression

R s
A vkSjB ls ehukj ds 'kh"kZ dk mÂ;u dks.k Øe'k% changes from 30º to 60º, the length of the
15º vkSj30º gSA ;fnA rFkkB ehukj ds ,d gh shadow of a tower decreases by 70 m.
a th

fn'kk esa gks vkSj

AB = 48 ehVj gks] rks ehukj dh What is the height of the tower?
Å¡pkbZ gS % tc lw;Z dk voueu dks.k 30º lss cnydj 60º gks
(a) 24 3 m (b) 24 m t5krk gS] rks ,d ehukj dh ijNkbZ dh yackbZ
ty a

(c) 24 2 m (d) 96 m
ehVj de gks tkrh gSA ehukj dh Å¡pkbZ D;k gS
SSC CPO 16 March 2019 (Evening)
di M

32. A tower standing on a horizontal plane

subtends a certain angle at a point 160 (a) 36.55 m (b) 65.55 m
m apart from the foot of the tower. On (c) 45.65 m (d) 60.55 m
advancing 100 m towards it, the tower is 36. A and B standing on the same side of a
found to subtend an angle twice as before. wall and observe that the angle of
The height of the tower is : elevation to the top of the wall are 45º
{kSfrt ry esa [kM+k dksbZ ehukj vius vk/kj ls 160 and 60º respectively. if the height of the
ehVj nwj fLFkr fdlh fcUnq ij dksbZ fuf'pr dks.k wall is 50 m, the distance between A and
cukrk gSA Vkoj dh vksj 100 ehVj pyus ij dks.k B is :
nksxquk gks tkrk gSA ehukj dh Å¡pkbZ Kkr djsaA A vkSjB fdlh nhokj ds ,d gh rjiQ [kM+s gSa rF
(a) 80 m ns[krs gSa fd nhokj ds 'kh"kZ dk mÂ;u dks.k
(b) 100 m 45º vkSj 60º gSA ;fn bl nhokj dh Å¡pkbZ 5
A

(c) 160 m ehVj gS] rks

A vkSjB ds chp dh nwjh Kkr djsaA
(d) 200 m
33. The angle of elevation of the top of a tall Use 3 = 1.73 and 2 = 1.41 
building from the points M and N at the SSC CPO 15 March 2019 (Morning)
distances of 72 m and 128 m, (a) 25.07 m
respectively, from the base of the building
(b) 21.10 m
and in the same straight line with it, are
complementary. The height of the (c) 17.38 m
building (in m) is: (d) 14.65 m
37. If the height of a pole and the distance rFkkQ ,d [kaHks ds fdlh Hkh ,d fjiQ Hkwfe
P
between the pole and a man standing fLFkr nks fcUnqP gSA
vkSj Q ls ns[kus ij [kaHks d
nearby are equal, what would be the
'kh"kZ dk mÂ;u dks.k Øe'k%60º vkSj 30º gS rFkk
angle?
muds chp dh nwjh 84 3 gSA bl [kaHks dh Å¡
;fn fdlh [kaHks dh Å¡pkbZ rFkk [kaHks ,oa ikl gh esa
fdruh gS\
[kM+s ,d O;fDr ds chp dh nwjh cjkcj gS] rks dks.k
SSC CGL Tier-II (13 September 2019)
D;k gksxk\
(a) 63 (b) 73.5
SSC CPO 15 March 2019 (Morning)
(c) 52.5 (d) 60
(a) 60º (b) 90º
41. As observed from the top of a lighthouse,
(c) 30º (d) 45º
120 3 m above the sea level, the angle
38. The angle of elevation of a flying drone
of depression of a ship sailing towards it
from a point on the ground is 60º. After
from 30º to 60º. The distance travelled by
flying for 5 seconds the angle of elevation
the ship during the period of observation
drops to 30º. If the drone is flying
is :
horizontally at a constant height of
,d izdk'k LraHk ds 'kh"kZ ls tks fd leqnz ry

r
1000 3 m, the distance travelled by the
120 3 ehVj Åij gS] mldh vksj vk jgs tgkt

si
drone is :
dk voueu dks.k 30º ls 60º gks tkrk gSA voyksd
Hkwfe ij fLFkr fdlh fcUnq ls ,d mM+rs gq, Mªksu dk
dh vof/ ds nkSjku tgkt }kjk r; dh tkus okyh
an by
mÂ;u dks.k60º gSA 5 lsd.M rd mM+us ds ckn
mÂ;u dks.k de gksdj30º gks tkrk gSA ;fn Mªksu
nwjh Kkr dhft,A

n
SSC CGL 2019 Tier-II (15/11/9)
{kSfrt :i ls 1000 3 ehVj dh Å¡pkbZ ij mM+ku
(a) 240 m (b) 240 3 m
Hkj jgk gS] rks Mªksu }kjk r; dh x;h nwjh Kkr djsaA
ja
R s
SSC CPO 16 March 2019 (Afternoon) (c) 180 3 m (d) 180 m
a th

(a) 2000 m (b) 1000 m 42. The length of the shadow of the vertical
(c) 3000 m (d) 4000 m tower on level ground increases by 10m
when the altitude of the sun changes
39. From the top of a hill 96 m high, the from 45º to 30º. The height of the tower
angles of depression of two cars parked
ty a

is :
on the same side of the hill (at same level
as the base of the hill) are 30º and 60º lery Hkwfe ij [kM+h ehukj dh Nk;k dh yackb
di M

respectively. The distance between the ehVj rd c<+ tkrh gS tc lw;Z dh Å¡pkbZ
45º ls
cars is : 30º rd cny tkrh gSA VkWoj dh Å¡pkbZ gS %
96 ehvj Å¡ph igkM+ dh pksVh ls igkM+h dh ,d gh SSC CGL Tier-II (18/11/2020)

fn'kk esa [kM+h dh xbZ nks dkjksa ds voueu dks.k(a) 10  3 +1 m (b) 10 3 m
(igkM+h ds vk/kj :i esa leku Lrj ij) Øe'k%
30º vkSj60º gSA dkjksa ds chp dh nwjh fdruh gS\ (c) 5 3 m (d) 5  3 +1 m

( 3 = 1.73 dk iz;ksx dhft, vkSj fudVre iw.kZ Type-5

la[;k esa iw.kk±fdr dhft,A)
In this type, we will study the question in
SSC CPO 15 March 2019 (Evening) which the figure will be formed as shown below :
A

(a) 220 m (b) 165 m

bl izdkj esa ge ,sls iz'uksa dk vè;;u djsaxs ftlesa fp
(c) 111 m (d) 243 m
fuEufyf[kr izdkj ls curs gSaA
40. P and Q are two points on the ground on
A
either side of a pole. The angles of C
elevation of the top of the pole as
observed from P and Q are 60º and 30º
respectively and the distance between
them is 84 3 . What is the height (in m)
of the pole?
B D
(i)  +  = 90º or both the angle of elevation are 44. Two vertical poles 12 m and 4m high
complementary. stand apart on a horizontal plane. What
 +  = 90º ;k nksuksa mÂ;u dks.k lEiwjd gksrs gSaA is the height of the point of intersection
of the lines joining the top of each pole
x2 = H1H2 or x = H1H 2 to the bottom of the other pole?
A C 12 ehVj vkSj 4 ehVj Å¡ps nks yacor~ LraHk
{kSfrt ry esa fdlh fuf'pr nwjh ij fLFkr gSaA
H1 H2
LraHk ds vk/kj ls nwljs LraHk ds 'kh"kZ d
okyh js[kkvksa ds izfrPNsn fcUnq dh Å¡pkbZ
(90º – ) ls fdruh gS\
B D
x (a) 5 m (b) 3 m
(ii) Height of the point of intersection of the (c) 2 m (d) 1 m
lines joining from foot of one to top of the 45. The angle of elevation of the top of a
other from horizontal line is h. tower 12 m high from the foot of another
tower in the same plane is 45º and the
,d ds ikn dks nwljs ds 'kh"kZ ls feykus okyh js[kkvksa
angle of elevation of the top of the second
ds izfrPNsn fcUnq dh {kSfrt js[kk lshšpkbZ
gks rks

r
tower from the foot of the first tower is
1 1 1 30º. If the height of the second tower is
= +

si
h H1 H 2 4 3 m, what is the distance between the
two towers is ?

an by 12 ehVj Å¡ps fdlh ehukj ds 'kh"kZ dk mÂ;.k d

mlh ry esa fLFkr fdlh nwljs ehukj ds vk/kj

n
H1 H2
45º gSA vkSj nwljs ehukj ds 'kh"kZ dk mÂ;
h igys Vkoj ds vk/kj ls 30º gSA ;fn nwljs ehuk

ja dh Å¡pkbZ4 3 ehVj gS] rks nksuksa ehukjksa

R s
(iii) If both angles of elevation are not
dh nwjh D;k gS\
a th

complementary, then we will solve these

problem by using the right-angled triangles (a) 12 m (b) 4 3 m
ABC and BCD. (c) 4 m (d) 12 3 m
;fn nksuksa mÂ;u dks.k lEiwjd u gks rks ge ,sls46.
iz'uksaTwo poles are such that angles of
ty a

dks nks ledks.k f=kHkqtksa

ABC vkSj BCD dk iz;ksx elevation of a pole of height 18 m from
djrs gq, gy djrs gSaA the bottom of the other pole is  and the
di M

angle of elevation of top of other pole of

A height 90 m is from the bottom of the
C
pole is 3. What is the height of the point
H1 of intersection of the lines as shown by
H2
dotted line in the given figure.
nks LraHk bl izdkj gSa fd 18 ehVj Å¡ps fdlh e
B D ds 'kh"kZ dk mÂ;.k dks.k mlh ry esa fLFkr 90
x
43. The distance between two pillars of length Å¡ps nwljs ehukj ds vk/kjlsgSA vkSj nwljs eh
16 m and 9 m is x meters. If two angles ds 'kh"kZ dk mÂ;u dks.k igys Vkoj ds vk/kj
of elevation of their respectively top from 3 gSA Hkqtkvksa ds çfrPNsn fcUnq dh Å¡pk
the bottom of the other are tSlk fd fp=k esa fn[kk;k x;k gS
A

complementary to each other, the value

of x (in m) is :
16 ehVj vkSj 9 ehVj yacs nks LraHkksa ds chp dh nwjh
x ehVj gSA ;fn ,d ds vk/kj ls nwljs ds 'kh"kZ ds 90 m
mÂ;u dks.k ,d&nwljs ds lEiwjd gksax dkrkseku
(ehVj esa) gS % 18 m
(a) 15 (b) 16 3
(a) 12 m (b) 15 m
(c) 12 (d) 9
(c) 9 m (d) 8 m
 49. Two poles of equal height are standing
Type-6 opposite to each other on either side of
In this type, we will study the questions in a road which is 100 m wide. From a point
which the figure will be formed as shown below : between them on the road angles of
bl izdkj esa ge ,sls iz'uksa dk vè;;u djsaxs ftlesa fp=k elevation of their topes are 30º and 60º.
The height of each pole (in metre) is :
fuEufyf[kr izdkj ls curs gSaA
leku ÅapkbZ ds nks [kaHks ,d nwljs ds foijh
A
ehVj pkSM+h ,d lM+d ds nksuksa rjiQ •M+s
E A chp ,d fcanq ls muds 'kh"kZ dh mÂ;u dks.kk
or eki 30º vkSj60º gSaA çR;sd [kaHks dh Åapkb
esa) gS%
B C D B C D
(a) 25 3 (b) 20 3
These types of questions are solved by taking
right-angled triangles separetely. (c) 28 3 (d) 30 3
bl izdkj ds iz'uksa dks ledks.k f=kHkqt dks vyx ysdj
50. From a light house, the angles of
gy djrs gSaA

r
depression of two ships on opposite sides
47. Two points are x m apart and the height of the light house are observed 30º and

si
of one is double of the other. If from the 45º respectively. If the height of light
mid-point of the line joining their feet house is h m, what is the distance

an by
and the observer finds the angular
elevations of their tops to be
between the ship?
,d izdk'k ?kj ls] çdk'k ?kj ds foijhr fn'kkvksa e

n
complementary, the height (in m) of the
shorter post is : nks tgktksa ds voueu dks.k Øe'k% 30º vkSj 45º
nks fcUnq
x ehVj dh nwjh ij gSa vkSj ,d dh Å¡pkbZ gSaA ;fn çdk'k ?kj dh ÅapkbZ
h ehVj gS] rks tgktk

ja
R s
nwljs dh nksxquh gSA ,d i;Zos{kd dks muds vk/kjksads chp dh nwjh D;k gS\
dks feykus okyh js[kk ds eè; fcUnq ls muds 'kh"kks±
a th

(a)  3 +1 h (b)  3  1 h

dh dks.kh; Å¡pkbZ lEiwjd izkIr gksrh gSA NksVs LraHk
dh Å¡pkbZ (ehVj esa)gS% 
1+
1 
h
(c) 3h (d)  
3
ty a

x x
(a) (b) 51. From an aeroplane just over a river, The
2 2 4
di M

angles of depression of two points on the

x opposite to bank of the river are found to
(c) x 2 (d)
2 60º and 30º respectively. if the breadth
48. A pole 23 m long reaches a window which of the river is 400 m, the height of
aeroplane above the river at that instant
is 3 5 m above the ground on one side
of a street. Keeping its foot at the same is (assume 3 =1.732)
point, the pole is turned to the other side fdlh unh ds Åij ,d gokbZ tgkt lss unh ds foijhr
of the street to reach a window 4 15 m fdukjksa ij fLFkr nks fcUnqvksa ds voueu dks.
high. What is the width (in m) of the 60º vkSj 30º ik, tkrs gSaA ;fn unh dh pkSM
street? 400 ehVj gS] rks ml le; esa unh ds Åij gokb
23 ehVj yack [kaHkk] ,d f[kM+dh tks xyh ds ,d tgkt dh ÅapkbZ gS %
A

rjiQ Hkw&ry 3ls 5 ehVj Åij fLFkr gS rd igqaprk (a) 173.2 m (b) 346.4 m
gSA [kaHks ds ikn dks leku fcanq ij j[krs gq, xyh ds (c) 519.6 m (d) 692.8 m
nwljh vksj
4 15 ehVj Åij fLFkr f[kM+dh rd igqapkus
52. From the top of 75 m high tower, the
angle of depression of two points P and
ds fy, ?kqek;k tkrk gSA lM+d dh pkSM+kbZ (ehVj esa)
Q on opposite side of the base of the
fdruh gS\ tower on legvel ground is  and , such
SSC CGL Pre (2021) 3 5
(a) 17 (b) 35 that tan = and tan = . What is the
4 8
(c) 39 (d) 22 distance between the points P and Q?
 75 ehVj Å¡ph ehukj ds 'kh"kZ ls ehukj ds ry ds This type is same as Type 4. Only orientation
foijhr fn'kk esa Hkwfe ij fLFkr nks fcUnqvksa
P vkSj is different.
Q dk voueu dks.k  rFkk  bl izdkj gS fd ;g izdkj&4 ds leku gh gSA fliZQ vfHkfoU;kl esa v
3 5 To solve this type of question, we can use
tan = vkSjtan=
gSA fcUnq
P vkSjQ ds
following methods.
4 8
chp dh nwjh Kkr djsaA bl izdkj ds iz'uksa dks gy djus ds fy, ge fuEufyf[kr
SSC CPO 13 March 2019 (Morning)
fofèk;ksa dk iz;ksx dj ldrs gSa &
(a) 190 m (b) 200 m
(c) 180 m (d) 220 m First Methods : In this method we use the
53. From the top of 120 m high lighthouse, two right-angled triangles separately.
the angle of depression of two ships on çFke fof/ % blesa ge nks ledks.k f=kHkqt vyx&
opposite side of the base of the lighthouse
is 30º and 60º. What is the distance ekurs gSaA
between the ships? (rounded off) Second Method : Given ADB =1, and ADC
120 ehVj Å¡ph ykbV gkml ds 'kh"kZ ls blds vk/kj
= 2
ds foijhr fn'kkvksa esa nks tgktksa dk voueu dks.k
30º vkSj60º gSA tgktksa ds chp dh nwjh Kkr djsaA C

r
SSC CPO 14 March 2019 (Morning) h

si
(a) 327 m (b) 127 m
(c) 277 m (d) 177 m B
54.

an by
From a point exactly midway between the
foot of two towers P and Q, the angles of a

n
elevation of their tops are 30º and 60º A
respectively. The ratio of the height of P x D
to that of Q is :
nks ehukjksa
ja
P vkSj Q ds Bhd chp fLFkr ,d fcUnq In most of the problems of this type
R s
ls muds 'kh"kZ ds mÂ;u dks.k Øe'k%
30º vkSj60º bl izdkj ds vf/dka'k iz'uksa esa &
a th

gSAP vkSjQ dh Å¡pkbZ dk vuqikr gS % (i) If h will be given, x is asked

SSC CGL Tier-II (12 September 2019)
(a) 1 : 3 (b) 1 : 2
;fn h fn;k x;k gksxk rks
x iwNk tk,xk

(c) 1: 2 3 (d) 2 : 3 3 (ii) If x will be given, h is asked

ty a

55. Exactly midway between the foot of two ;fn x fn;k x;k gksxk rks
h iwNk tk,xk
towers P and Q, the angles of elevation
di M

of their tops are 45° and 60°, respectively. a tanθ1

=
The ratio of the heights of P and Q is: h tanθ2  tanθ1 and
nks ehukjksa
P vkSjQ ds ikn ds Bhd chp ls muds
h = x(tan2 – tan1 )
'kh"kksZa ds mUu;u dks.k Øe'k% ° vkSj 45
60
° gSaa
P
56. The angle of elevation of the top of an
vkSjQ dh Å¡pkbZ dk vuqikr D;k gS\ unfinished tower at a point distant 78 m
SSC CGL Pre (2021) from its base is 30°. How much higher
(a) 1 : 3 (b) 3 : 1 must the tower be raised (in m) so that
(c) 1 : 3 (d) the angle of elevation of the top of the
3 :1
finished tower at the same point will be
Type-7 60°?
In this type, we will study the questions in ,d v/wjh ehukj ds vk/kj ls 78 ehVj dh nwjh ls
A

which the figure will be formed as shown below :

v/wjh ehukj ds f'k[kj dk mUu;u dks.k
30° gSA ehuk
bl izdkj esa ge ,sls iz'uksa dk vè;;u djsaxs ftlesa fp=k
fuEufyf[kr izdkj ls curs gSaA dks fdruk Åapk (ehVj esa) cuk;k tkuk pkfg, rkf
mlh fcanq ls rS;kj ehukj ds f'k[kj dk mUu;u d
D
60° gks tk,\
SSC CGL Pre (2021)
A
(a) 78 3 (b) 80
B C (c) 52 3 (d) 26 3
57. A poster is on top of a building. A person
is standing on the ground at a distance
Type-8
of 50 m from the building. The angles of In this type, we will study the questions in
elevation to the top of the poster and which figure will be formed as shown below :
bottom of the poster are 45° and 30°,
bl izdkj esa ge ,sls iz'uksa dk vè;;u djsaxs ftlesa fp
respectively. What is 200% of the height
(in m) of the poster? fuEufyf[kr izdkj ls curs gSaA
,d bekjr ds 'kh"kZ ij ,d iksLVj yxk gSA ,d O;fDr A
bekjr ls 50 ehVj dh nwjh ij tehu ij [kM+k gSA
iksLVj ds Åijh vkSj iksLVj ds fupys fljs ds mUUk;u
D
dks.k Øe'k% 45° vkSj30° gSaA iksLVj dh ÅapkbZ (ehVj E
esa) dk 200» fdruk gS\
SSC CGL Pre (2021)

25 75 B C
(a)
3
3– 3  (b)
3
3– 3 

r
We can solve these types of questions by
taking the two right-angled triangles ABC and

si
50 100
(c)
3

3– 3  (d)
3
3– 3  AED.
bl izdkj ds iz'uksa dks ge nks ledks.k f=kHkqtksa
ABC vkSj
58.
an by
The angle of elevation of the top of a
building and the top of the chimney on AED dks ysdj gy djrs gSaA

n
the roof of the building from a point on 60. There are two vertical poles, one on each
the ground are x and 45º respectively. side of a road, just opposite to each other.

ja
The height of building is h m. The height One pole is 108 m high. From the top of
R s
of the chimney (in metre) is : this pole, the angles of depression of the
top and the foot of the other poles are 30º
,d bekjr ds 'kh"kZ vkSj bekjr dh Nr ij fLFkr
a th

and 60º respectively. The height of the

fpeuh ds 'kh"kZ dk mÂ;u dks.k Øe'k%
xº vkSj45º other pole (in m) is :
gSaA bekjr dh ÅapkbZ
h ehVj gSA fpeuh dh ÅapkbZ ,d lM+d ds nksuksa rjiQ ,d&nwljs ds foijhr
(ehVj esa) gS %
ty a

yacor [kaHks gSaA ,d [kaHks dh Å¡pkbZ 10

(a) h cotx + h (b) h cotx – h bl [kaHks ds 'kh"kZ ls] nwljs [kaHks ds 'kh"kZ
di M

(c) h tanx – h (d) h tanx + h voueu dks.k Øe'k%30º vkSj 60º gSaA nwljh [k
59. From a point P on the ground, the angle dh ÅapkbZ (ehVj esa) gS%
of elevation of the top of a 10 m tall (a) 36 (b) 72
building is 30º. A flag is hoisted at the top (c) 108 (d) 110
of the building and the angle of elevation
61. The angle of elevation of the top of a
of the top of the flagstaff from P is 45º. tower from the bottom of a building is 60º
Find the length of the flagstaff. and from top of the building is 45º. If
Take 3 = 1.732 height of tht tower is 120 m, the height
Hkwfe ij ,d fcanq
P ls] 10 ehVj Å¡ph bekjr ds of the building is :
'kh"kZ dk mÂ;u dks.k
30º gSA bekjr ds 'kh"kZ ij ,d bekjr ds vk/kj ls ,d ehukj ds 'kh"kZ dk mÂ;u
,d >aMk iQgjk;k tkrk gS vkSj fcUnq
P ls èotnaM dks.k60º gS vkSj bekjr ds 'kh"kZ ls gSA ;fn
45º
A

ds 'kh"kZ dk mÂ;u dks.k

45º gSA èotnaM dh yackbZ bl ehukj dh ÅapkbZ 120 ehVj gS] rks bekjr
ÅapkbZ gS%
Kkr dhft,A  3 = 1.732
(a) 40  3  1 m
(a) 10  3 + 2 m
(b) 40 3  3  1 m
(b) 10  3 +1
(c) 80  3  1 m
(c) 10 3 m
(d) 40 3 m
(d) 7.32 m
62. There are two temples, one on each bank 64. A vertical pole and a vertical tower are
of a river, just opposite to each other. One on the same level of ground in such a
temple is 54 m high. From the top of this way that from the top of the pole, the
temple, the angles of depression of the angle of elevation of the top of the tower
top and the foot of the other temples are is 60° and the angle of depression of the
30º and 60º respectively. The length of the bottom of the tower is 30°. If the height
temple is : of the tower is 76 m, then find the
,d unh ds çR;sd fdukjs ij] ,d nwljs ds Bhd height (in m) of the pole.
lkeus nks eafnj gSaA ,d eafnj 54 ehVj Åapk gSA bl ,d mèokZ/j [kaHkk vkSj ,d mèokZ/j ehukj le
tehu ij bl izdkj fLFkr gSa fd [kaHks ds 'kh"
eafnj ds 'kh"kZ ls vU; eafnjksa ds 'kh"kZ vkSj ikn ds
voueu dks.k Øe'k%30º vkSj 60º gSaA nwljs eafnj ehukj ls f'k[kj dk mUu;u dks.k
60° rFkk ehukj ds
dh yackbZ gS% ry dk voueu dks.k 30° gSA ;fn ehukj dh Å¡pk
(a) 18 m 76 ehVj gS] rks [kaHks ds ÅapkbZ (ehVj esa) K
(b) 36 m SSC CGL Pre (2021)
(c) 36 3 m (a) 38

r
(d) 18 3 m (b) 19 3

si
63. From the top of a cliff 200 m high, the (c) 19
angles of depression of the top and (d) 57
bottom of a tower are observed to be 30º

an by
and 45º, respectively. What is the height
of the tower?
65. There are two temples, one on each bank
of river just opposite to each other. From

n
the top of a temple the angle of elevation
200 ehVj Å¡ph ,d pV~Vku dh pksVh ls] ,d ehukj of the top of the other temple is 60º and
ds 'kh"kZ vkSj ikn ds voueu dks.k Øe'k%
30º vkSj the angle of depression of the foot of the

ja
R s
45º gSaA ehukj dh ÅapkbZ D;k gS\ other temple is 30º. Then what is the
(a) 400 m ratio of the heights of the two temples?
a th

,d unh ds çR;sd fdukjs ij] ,d nwljs ds Bhd

(b) 400 3 m
lkeus nks eafnj gSaA ,d eafnj ds 'kh"kZ ls vU
(c) 300 3 ds 'kh"kZ dk mÂ;u dks.k
60º vkSj ikn dk voueu
ty a

(d) None of these dks.k30º gSaA nksuksa eafnjksa dh Å¡pkbZ;k

D;k gS\
di M

Type-9 (a) 1 : 3
In this type, we will study the questions in (b) 2 : 3
which figure will be formed as shown below :
(c) 1 : 4
bl izdkj esa ge ,sls iz'uksa dk vè;;u djsaxs ftlesa fp=k
(d) 3:4
fuEufyf[kr izdkj ls curs gSaA
66. The angle of depression of top of a tower
A from the top of the mountain is 60º and
the angle of depression of the foot of the
mountain from the top of the tower is
C 30º. If the height of the tower is 36 m,
E the height of the mountain is :
A

igkM+ dh pksVh ls ,d ehukj ds 'kh"kZ dk vo

dks.k60º gS vkSj ehukj ds 'kh"kZ ls ioZr ds ikn
voueu dks.k 30º gSA ;fn ehukj dh ÅapkbZ
B D ehVj gS] rks igkM+ dh ÅapkbZ gS%
We can solve these types of questions by (a) 72 m
taking the two right-angled triangles AEC and (b) 144 m
CBD.
(c) 108 m
bl izdkj ds iz'uksa dks ge nks ledks.k f=kHkqtksa
AEC vkSj
(d) 72 3 m
CBD dks ysdj gy djrs gSaA
 Miscellaneous Questions ,d O;fDr ,d ekWy ds ikl lM+d ij [kM+k FkkA
ekWy ls 1425 ehVj nwj Fkk vkSj lM+d ls e
67. From the top of a 120 m high tower, the 'kh"kZ dks bl rjg ls ns[kus esa l{ke Fkk fd isM
angle of depression of the top of the pole
'kh"kZ] tks mlds vkSj ekWy ds chp esa gS]
is 45º and the angle of depression of the
'kh"kZ ds lkFk n`f"V dh js[kk esa Fkk vmu lc
3
foot of the pole is , such that tan = .
2
dh Å¡pkbZ 10 ehVj gS vkSj ;g mlls 30 ehVj
What is the height of the pole? gSA ekWy fdruk yack (ehVj esa) gS\
120 ehVj Å¡ph ehukj ds 'kh"kZ ls fdlh [kaHks ds SSC CPO 23/11/2020 (Morning)
'kh"kZ dk voueu dks.k
45º vkSj [kaHks ds ry dk (a) 475 (b) 300
3 (c) 425 (d) 525
voueu dks.k  bl izdkj gS fd tan = gSA 71. At a point on level grounde, the angle of
2
bl [kaHks dh Å¡pkbZ D;k gS\ elevation of a vertical tower is found to
SSC CPO 13 March 2019 (Eveing) 5
be such that its tangent is . On
(a) 60 m (b) 75 m 12
(c) 80 m (d) 40 m walking 192 m towards tower, the tangent
68. From the top of a 10 m high building, the 3

r
of angle of elevation is . Find height of
angle of elevation of the top of a tower is 4
tower?

si
60º and the angle of depression of the
foot of the tower is , such that tan = lery tehu ij fLFkr ,d fcUnq ij ,d yEcor
ehukj dk mÂ;u dks.k bl izdkj gS fd mldk
2
3
an by
. What is the height of the tower to
nearest metres?
5
tanq,

gSA ehukj dh vksj 192 ehVj pyus ij mÂ;

n
12
10 ehVj Å¡ph bekjr ls fdlh ehukj ds 'kh"kZ dk 3
dks.k dktanq, gks tkrk gSA ehukj dh Å¡p
mÂ; u d ks.k 60º gS RkFkk ehukuj ds ry dkvoueu 4

ja Kkr dhft,A
R s
2
dks.k bl izdkj gS fd tan = gSA fudVre (a) 300 (b) 200
3
a th

ehVj rd ehukj dh Å¡pkbZ Kkr djsaA (c) 180 (d) 100

SSC CPO 13 March 2019 (Morning) 72. Each side of a square subtends an angle
(a) 34 m (b) 35 m of 60º at the top of a tower h meter high
(c) 36 m (d) 33 m standing in the center of square. If a is
ty a

69. From the top of a hill 240 m high the the length of each side square then
angle of depression of the top and of the ,d oxZ dh izR;sd Hkqtk] oxZ ds dsUnz esa fL
di M

bottom of a pole are 30º and 60º, h ehVj Å¡ps ehukj ds f'k[kj ij
60º dk dks.k cukrh
respectively. the difference (ihn m)
gSA ;fn oxZ dh izR;sd Hkqtk dh yEckbZ
a gS] rks
between the height of the pole and its
distance from the hill is : (a) 2h2 = a2 (b) 2a2 = h2
2 2
(d) 2h2 = 3a2
,d igkM+h ds 'kh"kZ tks 240 ehVj Å¡pk gS ls ,d (c) 3a = 2h
73. A bird is sitting on the top a vertical pole
[kaHks ds Åij vkSj uhps ds ry ij dks.k Øe'k%
30º 20 m high and its elevation from a point
vkSj60º gSA [kaHks dh Å¡pkbZ vkSj igkM+h ls mldh O on the ground is 45º. It flies off
nwjh ds chp dk varj (ehVj esa) gS % horizontally straight away from point O.
SSC CGL 2019 Tier-II (16/11/9) After one second, the elevation of the bird
from O is reduced to 30º. Then the speed
(a) 80 2 – 3  (b) 120  3 – 1 (m/s) of the bird is :
A

(c) 120 2 – 3  (d) 80  3 – 1 ,d 20 ehVj Å¡ps yEcor [kaHks ds f'k[kj ij ,d

70. A person was standing on a road near a fpfM+;k cSBh gS vkSj tehu ij ,d fcUnq
O ls bldk
mall. He was 1425 m away from the mall mÂ;u dks.k45º gSA ;g fcUnq O ls {kSfrt ds lekarj
and able to see the top of the mall from fn'kk esa lh/s mM+rh gSA ,d lsdsaM ds ckn
the road in such a way that the top of the
dk mÂ;u dks.k fcUnq O ls 30º esa cny tkrk gSA
tree, which is in between him and the
mall, was exactly in the line of sight with
rks fpfM+;k dh xfr (eh@ls-) esa gS %
the top of the mall. The tree heigvht is (a) 40  2 – 1 (b) 40  3 – 2 
10 m and it is 30 m away from him. How
tall (in m) is the mall? (c) 20 2 (d) 20  3 – 1
74. A man standing between two vertical 77. The angle of elevation of a aeroplane from
posts finds that the angle subtended at a point on the ground is 60º. After flying
his eyes by the tops of the posts is a right for 30 seconds, the angle of elevation
angle. If the heights of the two posts are changes to 30º. If aeroplane is flying at a
two times and four times the height of height of 4500 m, then what is speed
the man and the distance between them (m/s) of aeroplane?
is equal to the length of the longer post, tehu ij fLFkr ,d fcUnq ls ,d gokbZ tgkt dk
then ratio of the distance of the man from mÂ;u dks.k60º gSA 30 lsdsaM rd mM+us d
the shorter and the longer post is : bldk mÂ;u dks.k30º esa cny tkrk gSA ;fn gokb
,d O;fDr nks yEcor [kaHkksa ds chp esa [kM+k gS] ogtgkt 4500 ehVj dh Å¡pkbZ ij mM+ jgk gS rks
ikrk gS fd [kaHkksa ds f'k[kjksa ds }kjk mldh vk¡[kksa
tgkt dh xfr (eh@ls) esa D;k gS\
ij cuk;k x;k dks.k ledks.k gSA ;fn nks [kaHkksa dh (a) 50 3
Å¡pkbZ] O;fDr dh Å¡pkbZ dh nksxquk vkSj pkSxquk gS
(b)
vkSj ;fn muds chp nwjh yEcs [kaHks dh yEckbZ ds 100 3
leku gS] rks O;fDr vkSj NksVs [kaHks rFkk yacs [kaHks dh 3
(c) 200
nwjh dk vuqikr D;k gksxk\

r
(d) 300 3
(a) 3 : 1 78. A ballon leaves from a point P rises aat a

si
(b) 2 : 3 uniform speed. After 6 minutes, an
(c) 3 : 2 observer situated at a distance of 450 3

75.
(d) 1 : 4
an by
The height of a tower is h and angle of
meters from point P observes that angle
of elevation of ballon is 60º. Assume that

n
elevation of the top of tower is a . On point of observation and point P are on
h same level. What is speed of Ballon? (m/s)
moving a distance towards the tower,

ja 2 ,d xqCckjk ,d fcUnq ls ,d leku xfr ls NksM+

R s
the angle of elevation becomes b. What is
the value of (cota – cotb)? tkrk gSA 6 feuV ds ckn ,d ifjn'kZd] tks fcUnq
P
a th

,d ehukj dh Å¡pkbZh vkSj ehukj ds f'k[kj dk ls 450 3 ehVj dh nwjh ij [kM+k gS] og xqCC
h mÂ;u dks.k60º ikrk gSA voyksdu fcUnq PrFk
mÂ;u dks.ka gSA ehukj dh vksj nwjh pyus ij fcUnq ,d gh lery ij fLFkr gSA xqCCkkjs dh
2
ty a

mÂ;u dks.kb gks tkrk gSA

(cota – cotb) dk eku (eh@ls) D;k gS\
D;k gS\ (a) 4.25
di M

(b) 3.75
1
(a) (c) 4.5
2
(d) 3.25
2 79. A navy captain going away from a light
(b)
3 house at the speed of 4  3 – 1 m/s. He
(c) 1 observes that it takes him 1 min to
(d) 2 change the angle of elevation of the top
76. A hydrogen filled ballon ascending at rate of lighthouse from 60º to 45º. What is
of 19 km/h was drifted by wind. Its angle height of lighthouse?
of elevation at 10º and 15º minutes were ,d usoh dIrku 4  3 – 1 eh@ls- dh xfr ls ,d
found to be 60º and 45º respectively. The
izdk'k LraHk ls nwj tkrk gSA og ;g ikrk gS
A

wind spedd in (whole numbers) during last

5 minutes approximately is equal to : mldks izdk'k LraHk dk mÂ;u dks.k
60º ls 45º esa
,d gkbMªkstu ls Hkjk gqvk xqCckjk 18 fdeh@?kaVk cnyus
dh esa 1 feuV dk le; yxrk gSA izdk'k Lra
nj ls mBrk gqvk gok ds }kjk eksM+ fy;k x;kA 10osa dh Å¡pkbZ Kkr dhft,A
vkSj 15osa feuV ij blds mÂ;u dks.k Øe'k%
60º (a) 240 3
vkSj45º ik, tkrs gSaA gok dh xfr (iw.kZ la[;k esa) (b) 480  3 – 1
vafre 5 feuV ds nkSjku yxHkx D;k gksxh\
(c) 360 3
(a) 7 (b) 2.6
(c) 11 (d) 33 (d) 280 2
80. An aeroplane is flying horizontally at a 81. Two trees are standing along the opposite
height of 18 km above ground. The angle sides of road. Distance between two trees
of elevation of plane from point X is 60º is 400 m. There is a point onthe road
and after 20 seconds, its angle of between the trees. The angle of
elevation from X becomes 30º. If point X depression of the point from top of trees
is on ground, then what is speed (km/h) are 45º and 60º. If the height of the tree
of aeroplane? which makes 45º is 200 m, then what will
,d gokbZ tgkt tehu ls 1-8 fdeh dh Å¡pkbZ ij be the height of other tree?
f{kfrt ds lekarj fn'kk esa mM+ jgk gSA tgkt dk nks o`{k ,d lM+d dh foijhr fn'kkvksa esa [kM
fcUnq X ls mÂ;u dks.k60º gS rFkk 20 lsds.M ds nksuksa o`{kksa ds chp dh nwjh 400 ehVj g
ckn bldk mÂ;u dks.k fcUnq X ls 30º gks tkrk gSA chp lM+d ij ,d fcUnq fLFkr gSA o`{kksa ds f'
;fn fcUnqX tehu ij fLFkr gS] rks gokbZ tgkt dh fcUnq ds voueu dks.k Øe'k%
45º vkSj60º gSA ;fn
xfr (fdeh@?kaVk) D;k gS\ ml o`{k dh Å¡pkbZ 200 ehVj gS] 45ºtksdk dks.k
(a) 216 3
cukrk gS] rc nwljs o`{k dh Å¡pkbZ D;k gksxh\
(a) 200

r
(b) 105 3
(b) 200 3

si
(c) 201 3
(c) 100 3
(d) 305 3 (d) 250

an by
n
ja
R s
Answer Key
a th

1.(c) 2.(a) 3.(c) 4.(d) 5.(b) 6.(b) 7.(b) 8.(a) 9.(c) 10.(c)
ty a

11.(c) 12.(a) 13.(c) 14.(b) 15.(d) 16.(a) 17.(d) 18.(d) 19.(a) 20.(d)
di M

21.(d) 22.(b) 23.(d) 24.(b) 25.(a) 26.(c) 27.(c) 28.(d) 29.(b) 30.(a)

31.(b) 32.(a) 33.(b) 34.(d) 35.(d) 36.(b) 37.(d) 38.(a) 39.(c) 40.(a)

41.(a) 42.(d) 43. (c) 44. (b) 45.(a) 46.(b) 47.(a) 48.(c) 49.(a) 50.(a)

51.(a) 52.(d) 53.(c) 54.(a) 55.(a) 56.(c) 57.(d) 58.(b) 59.(d) 60.(b)

61.(b) 62.(b) 63.(d) 64.(c) 65.(c) 66.(b) 67.(d) 68.(c) 69.(a) 70.(a)
A

71.(c) 72.(a) 73.(d) 74.(a) 75.(a) 76.(d) 77.(b) 78.(b) 79.(a) 80.(a)

81.(b)
 Height & Distance / Å¡pkbZ vkSj nwj
( Practice Sheet With Solution)
1. Two ships are sailing in the sea on the two ,d ehukj ds 'kh"kZ vkSj ry dks 100 ehVj Å¡ph ig
sides of a lighthouse. The angle of elevation pksVh ls 30
º vkSj 60º ds voueu dks.kksa ij ns•k x
of the top of the lighthouse is observed from
ehukj dh ÅapkbZ Kkr dhft;s\
the ships are 30° and 45° respectively. If the
(a) 42.2 mts (b) 33.45 mts
lighthouse is 100 m high, the distance between
(c) 66.6 mts (d) 58.78 mts
the two ships is:
6. A flagstaff 17.5 m high casts a shadow of lengt
,d çdk'k LraHk ds nksuksa vksj leqæ esa nks tgkt py jgs40.25 m. What will be the height of a building
gSaA tgktksa ls çdk'k LraHk ds 'kh"kZ dk mUu;u dks.k Øe'k%
which casts a shadow of length 28.75 m unde
30º vkSj 45º ns•k tkrk gSA ;fn ykbVgkml 100 ehVj similar conditions ?

r
Å¡pk gS] rks nksuksa tgktksa ds chp dh nwjh gS% 17-5 ehVj Å¡ps ,d èotnaM dh Nk;k 40-25 ehV

si
(a) 150 m (b) 200 m gSA ,d Hkou dh Å¡pkbZ fdruh gksxh] ftldh Nk
(c) 273 m (d) 400 m ifjfLFkfr;ksa esa 28-75 ehVj yach gksrh gS\ \
2. an by
An observer 1.6 m tall is 203 m away from a
tower. The angle of elevation from his eye to
(a) 14 cm (b) 13.5 cm

n
(c) 12.5 cm (d) 11.4 cm
the top of the tower is 30°. The height of the 7. A tower is broken at a point P above the ground
tower is: The top of the tower makes an angle 60° wit
ja
1-6 ehVj yack ,d i;Zos{kd ,d Vkoj ls 20 3 ehVj nwj
R s
the ground at Q. From another point R on th
gSA mldh vk¡• ls ehukj ds f'k•j dk mUu;u dks.kº 30 opposite side of Q angle of elevation of poin
a th

gSA Vkoj dh ÅapkbZ gS% P is 30°. If QR = 180 m, then what is the tota
(a) 21.6 m (b) 23.2 m height (in metres) of the tower?
(c) 24.72 m (d) None of these tehu ds Åij ,d fcanq P ij ,d Vkoj VwVk gqvk
VkWoj dk 'kh"kZ
Q ij tehu ds lkFk 60 º dk dks.k cuk
ty a

3. From a point P on a level ground, the angle of

elevation of the top tower is 30°. If the tower gSAQ ds foijhr fn'kk esa ,d vU; fcanqR ls fcanq
Pd
is 100 m high, the distance of point P from mUu;u dks.k º30gSA ;fnQR = 180 ehVj gS] rks V
di M

the foot of the tower is:

dh dqy ÅapkbZ (ehVj esa) fdruh gS\
lery Hkwfe ij fLFkr ,d fcanqP ls] 'kh"kZ ehukj
dk (a) 97 (b) 603
mUu;u dks.k º30gSA ;fn Vkoj 100 ehVj Åapk gS] rks (c) 45(3+4) (d) 45(3 + 2)
Vkoj ds vk/kj ls fcanq
P dh nwjh gS% 8. A ladder is resting against a vertical wall an
(a) 161 m (b) 170 m its bottom is 2.5 m away from the wall. If i
(c) 173 m (d) 200 m slips 0.8 m down the wall, then its bottom
4. A vertical toy 16 cm long casts a shadow 8 will move away from the wall by 1.4 m. Wha
cm long on the ground. At the same time a is the length of the ladder?
pole casts a shadow 48 cm long on the ground. ,d lh<+h ,d •M+h nhokj ds lgkjs fVdh gS vkS
Then find the height of the pole ? ry nhokj ls 2-5 ehVj dh nwjh ij gSA ;fn ;g nho
16 lsaVhehVj yack ,d ÅèokZ/j f•ykSuk tehu ij 8 0-8 ehVj uhps f•ldrh gS] rks bldk ry nhokj ls
A

lsaVhehVj yach Nk;k cukrk gSA mlh le; ,d •aHks dh ehVj nwj pyk tk,xkA lh<+h dh yackbZ fdruh g
tehu ij 48 lseh yach Nk;k iM+rh gSA rks •aHks dh ÅapkbZ
(a) 6 m (b) 6.5 m
Kkr djsa\ (c) 6.9 m (d) 7 m
(a) 1080 cm (b) 96 cm 9. The tops of two poles of height 60 metres an
(c) 108 cm (d) 118 cm 35 metres are connected by a rope. If the rop
5. The top and bottom of a tower were seen to makes an angle with the horizontal whos
be at angles of depression 30° and 60° from 5
tangent is metres, then what is the distanc
the top of a hill of height 100 m. Find the 9
height of the tower ? (in metres) between the two poles?

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 1

 60 ehVj vkSj 35 ehVj Å¡ps nks •aHkksa ds 'kh"kZ ,d jLlh ls
tqM+s gq, gSaA ;fn jLlh {kSfrt ls ,d dks.k cukrh gS ftldh against a wall is 60º and the foot of the ladde
is 4.6 m away from the wall. The length o
5
Li'kZ js•k ehVj gS] rks nksuksa •aHkksa ds chp dh nwjh
the ladder is
9
,d nhokj ls yxh lh<+h dk mUu;u dks.ko gS
60 vkSj l
(ehVj esa) D;k gS\
dk fupyk fljk nhokj ls 4-6 ehVj nwj gSA lh
(a) 42 (b) 49
(c) 36 (d) 45 yEckbZ gS
10. The height of a tower is 300 meters. When its (a) 6 m (b) 7 m
top is seen from top of another tower,then (c) 9.2 m (d) 7.5 m
the angle of depression is 60°. The horizontal 15. Two trees are standing along the opposite side
distance between the bases of the two towers of a road. Distance between the two trees i
is 120 metres. What is the height (in metres) 400 metres. There is a point on the roa
of the small tower? between the trees. The angle of depression
,d ehukj dh Å¡pkbZ 300 ehVj gSA tc bldk 'kh"kZ fdlh of the point from the top of the trees are 45
vU; ehukj ds 'kh"kZ ls ns•k tkrk gS] rks voueu dks.k
o
60 and 60º. If the height of the tree which make
gksrk gSA nksuksa ehukjksa ds vk/kjksa ds chp dh {kSfrt 45º
nwjhangle
120 is 200 metres, then what will b

r
ehVj gSA NksVs VkWoj dh ÅapkbZ (ehVj esa) fdruh gS\ the height (in metres) of the other tree?

si
(a) 88.24 (b) 106.71 ,d lM+d ds foijhr fn'kk esa nks isM+ •M+s gS
(c) 92.4 (d) 112.64 ds chp dh nwjh 400 ehVj gSA isM+ksa ds chp l
11. an by
The angles of elevation of the top of a tree
220 meters high from two points lie on the fcanq gSA isM+ksa ds 'kh"kZ ls fcanq ds voue
o
vk

n
same plane are 30° and 45°. What is the distance 60 gSaA ;fn 45
o o
dks.k cukus okys isM+ dh Åa
(in metres) between the two points? ehVj gS] rks nwljs isM+ dh ÅapkbZ (ehVj esa
ja
,d gh ry ij fLFkr nks fcanqvksa ls 220 ehVj Å¡ps ,d isM+ (a) 200
R s
(b) 2003
ds 'kh"kZ ds mUu;u dks.k ° vkSj
30 45° gSaA nks fcanqvksa ds(c) 3003 (d) 265
a th

chp dh nwjh (ehVj esa) fdruh gS\ 16. A boat is moving away from an observatio
(a) 190.22 (b) 140 tower. It makes an angle of depression of 60
(c) 150 (d) 161.04 with an observer's eye when at a distance o
12. On a ground , there is a vertical tower with a 50m from the tower. After 8 sec., the angle o
ty a

flagpole on its top . At a point 9 m away from depression becomes 30°. By assuming that i
the foot of the tower , the angles of elevation of is running in still water, the approximate spee
di M

the top and bottom of the flagpole are 60° and of the boat is
30° respectively . The height of the flagpole is
,d uko voyksdu Vkoj ls nwj tk jgh gSA ;g Vkoj
,d tehu ij ,d ÅèokZ/j ehukj gS ftlds 'kh"kZ ij ,d
èot LraHk gSA ehukj ds ikn ls 9 ehVj dh nwjh ij fLFkr ehVj dh nwjh ij i;Zos{kd dh vka• ds lkFk 60
° d

,d fcanq ij èotLraHk ds 'kh"kZ vkSj ry ds mUu;u dks.k voueu dks.k cukrk gSA 8 lsdaM ds ckn voueu
Øe'k% 60 ° vkSj 30° gSaA èot LraHk dh ÅapkbZ gS 30o gks tkrk gSA ;g ekudj fd ;g 'kkar ty esa py j
(a) 56 m (b) 63 m gS] uko dh vuqekfur xfr gS
(c) 65m (d) 75 m (a) 37 km/hr (b) 39 km/hr
13. A ladder 13 m long reaches a window which is (c) 45 km/hr (d) 56 km/hr
12 m above the ground on side of a street. 17. A and B standing on same side of a wall an
Keeping its foot at the same point, the ladder observe that the angle of elevation to the to
A

is turned to the other side of the street to of the wall are 45º and 60º respectively. If th
each a window 5m high, then the width of height of the wall is 50 meter,the distanc
the street is :
between A and B.
13 ehVj yach ,d lh<+h ,d f•M+dh rd igq¡prh gS tks
A vkSjB ,d nhokj ds ,d gh vksj •M+s gksdj ns•r
lM+d ds fdukjs tehu ls 12 ehVj Åij gSA vius iSj dks
fd nhokj ds 'kh"kZ dk mUu;u dks.k Øe'k%
º vkSj
456
mlh fcanq ij j•rs gq,] lh<+h dks lM+d ds nwljh vksj 5
ehVj Å¡ph f•M+dh dh vksj eksM+ fn;k tkrk gS] rks lM+d gSA ;fn nhokj dh ÅapkbZ 50 ehVj gS]
A vkSj
rksB d
dh pkSM+kbZ gS% chp dh nwjhA
(a) 17 m (b) 19 m (a) 21.1 (b) 22.1
(c) 15.5 m (d) None (c) 13.9 (d) 14.65

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 2

 a point in the ground is 60º. After flying for 5 point on the ground is 60º. After flying for 4
seconds the angle of elevation drops to 30º. seconds, the angle of elevation changes to 30
If the drone is flying at the height of 10003 If the aeroplane is flying at a height of 6000m
m the distance travelled by the drone is. then what is the speed (in m/s) of aeroplane
tehu ij ,d fcanq ls mM+us okys Mªksu dk mUu;u dks.ktehu ij ,d fcanq ls ,d gokbZ tgkt dk mUu;u dks
60º gSA 5 lsdaM rd mM+ku Hkjus ds ckn mUu;u º dks.k 30
60º gSA 40 lsdaM rd mM+ku Hkjus ds ckn] m
rd fxj tkrk gSA ;fn Mªksu 1000
 3 ehVj dh mM+ku Hkj 30º esa cny tkrk gSA ;fn gokbZ tgkt 6000 ehV
jgk gS rks Mªksu }kjk r; dh xbZ nwjh gSA ÅapkbZ ij mM+ jgk gS] rks gokbZ tgkt dh xfr (e
(a) 1200m (b) 1000m
esa) D;k gS\
(c) 2000m (d) None
(a) 503 (b) 1003
19. The angle of elevation of a pole from the two
points that are 75m and 48m away from its (c) 2003 (d) 3003
base are  and  respectively. If = 15º and  24. From the top of 75m high tower, the angle o
= 75º , then the height of the tower is: depression of two points A and B on opposit
side of the base of the tower on level groun
,d •aHks ds vk/kj ls 75 ehVj vkSj 48 ehVj dh nwjh ij
fLFkr nks fcanqvksa ls mUu;u dks.k vkSj
Øe'k%
 gSaA ;fn 3 5

r
is  and  , such that tan=and tan=
= 15º vkSj= 75º gks] rks ehukj dh ÅapkbZ gS% 4 8

si
(a) 50m (b) 60m What is distance between A and B.
(c) 63m (d) 65m 75 ehVj Å¡ps VkWoj ds 'kh"kZ ls] lery tehu ij V
20. an by
From the top of a tower 60 meter high the
angle of depression of the top and bottom of a
vk/kj ds foijhr fn'kk esa nks fcanqvksa
A vkSj B d

n
pole are observed to be 45° and 60° respectively. 3
If the pole and tower stand on the same plane,
voueu dks.k  vkSj gS] tSls fd tan= vk
4

ja
the height of the pole in meters is
R s
5
60 ehVj Å¡ps ,d ehukj ds 'kh"kZ ls ,d •aHks ds 'kh"kZ vkSj
tan= . A vkSjB ds chp dh nwjh D;k gS\
8
ry dk voueu dks.k Øe'k% 45° vkSj 60° ns•k tkrk gSA
a th

(a) 220m (b) 170m

;fn •aHkk vkSj ehukj ,d gh ry ij •M+s gksa] rks •aHks dh(c) 185m (d) 190m
ÅapkbZ ehVj esa gS 25. The angle of elevation of a tower at a poin
(a) 60(3 – 1) (b) 20(3 +3)
ty a

(c) 603 (d) 20(3 + 1) 4

21. The angles of elevation of the top of a tower 90 m from it is cot–1   .Then the height o
5
di M

from two points at a distance of 9m and 16m the tower is

from the base of the tower and in the same
,d ehukj dk mlls 90 ehVj dh nwjh ij mUu;u d
straight line with it are complementary. The
height of the tower is 4
cot   gSA rks ehukj dh ÅapkbZ gS
–1
,d ehukj ds vk/kj ls vkSj mlh lh/h js•k esa 9 eh vkSj 5
16 eh dh nwjh ij fLFkr nks fcanqvksa ls ehukj ds 'kh"kZ(a)ds45 (b) 90
mUu;u dks.k iwjd gksrs gSaA ehukj dh Å¡pkbZ gS (c) 112.5 (d) 150
(a) 9 m (b) 11 m 26. The angles of elevation of the tops of two vertica
(c) 12 m (d) 10 m towers as seen from the middle point of th
22. The angle of elevation of an aeroplane as lines joining the foot of the towers are 45° &
observed from a point 30 m above the 60°.The ratio of the height of the towers is
transparent water-surface of a lake is 30° and Vkojksa ds vk/kj dks feykus okyh js•k ds eè;
A

the angle of depression of the image of the ns•s tkus okys nks ÅèokZ/j Vkojksa ds 'kh"kZ
aeroplane in the water of the lake is 60°. The
height of the aeroplane from the water-surface
45° vkSj 60° gSaA Vkojksa dh ÅapkbZ dk vuq
of the lake is (a) 3 : 2 (b) 1 : 3
(c) 2 :3 (d) 2 : 1
,d >hy ds ikjn'khZ ty&lrg ls 30 ehVj Åij ,d fcanq
27. A man is standing on the deck of a ship, whic
ls ns•us ij ,d gokbZ tgkt dk mUu;u dks.k 30° gS vkSj is 10m above water level. He observes the angl
>hy ds ikuh esa gokbZ tgkt ds çfrfcEc dk voueu of elevation of the top of a light house as 60
dks.k 60° gSA >hy dh ty&lrg ls ok;q;ku dh Å¡pkbZ gS and the angle of depression of the base o
(a) 60 m (b) 42 m lighthouse as 30º. Find the height of the ligh
(c) 65 m (d) 75 m house.

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 3

 ,d vkneh ,d tgkt ds Msd ij •M+k gS] tks ty Lrj ls 8000 7500
10 ehVj Åij gSA og ,d ykbV gkml ds 'kh"kZ dk mUu;u (a) 3 m (b)
7
m
dks.k 60° vkSj ykbVgkml ds vk/kj dk voueu dks.k
6600
30° ns•rk gSA ykbV gkml dh ÅapkbZ Kkr dhft,A (c) m (d) 1200m
7
(a) 30m (b) 40m
(c) 45m (d) 38m 30. A car is moving at uniform speed towards
28. From the top of a building 60m high, the angle tower. It takes 15 minutes for the angle o
of elevation and depression of the top and the depression from the top of tower to the car t
foot of another building are  and  change from 30º to 60º. What time after this
respectively. Find the height of the second the car will reach the base of the tower?
building. ,d dkj ,d leku xfr ls ,d Vkoj dh vksj tk jgh gSA
60 ehVj Å¡ps ,d Hkou ds 'kh"kZ ls] ,d vU; Hkou ds VkWoj ds f'k•j ls dkj ds voueu dks.k dks °30 ls 60
'kh"kZ vkSj ikn ds mUu;u vkSj voueu dks.kØe'k%
vkSj esa cnyus esa 15 feuV dk le; yxrk gSA blds
 gSaA nwljs Hkou dh Å¡pkbZ Kkr dhft,A le; ckn dkj Vkoj ds vk/kj ij igqapsxh\
(a) 60(1 + tan tan) (a) 6 min (b) 6.5 min
(b) 60(1 + cottan) (c) 7 min (d) 7.5 min

r
(c) 60(1 + tan cot) 31. A man is watching from the top of a tower,

si
(d) 60(1 – tancot) boat speeding away from the tower. The angl
29. An aeroplane, when 4000m high from the of depression from the top of the tower t

an by
ground, pass vertically above another aeroplane
at an instance when the angles of elevation
the boat is 60º when the boat is 80m from th
tower. After 10 seconds, the angle of depressio
becomes 30º. What is the speed of the boat

n
of the two aeroplanes from the same point on
the ground are 60º and 30º respectively. Find (Assume that the boat is running in still water
the vertical distance between the two aero ,d vkneh ,d ehukj ds Åij ls ns• jgk gS] ,d uko
ja ehukj ls nwj tk jgh gSA tc uko Vkoj ls 80 ehVj d
R s
planes.
,d gokbZ tgkt] tc tehu ls 4000 ehVj Å¡pk gksrk gS] ij gS rks Vkoj ds 'kh"kZ ls uko dk voueu dks.k
° gSA
60
a th

,d nwljs gokbZ tgkt ds Åij ls yacor :i ls xqtjrk gS] lsdaM ds ckn voueu dks.k 30 ° gks tkrk gSA uko d
tc tehu ij ,d gh fcanq ls nks gokbZ tgkt ds mUu;u D;k gS\ (eku yhft, fd uko 'kkar ty esa py jgh gS
dks.k Øe'k% 60 ° vkSj 30° gksrs gSaA nks ok;q;kuksa ds chp
(a) 20 m/sec (b) 10 m/sec
ty a

dh ÅèokZ/j nwjh Kkr dhft,A (c) 16 m/sec (d) 18 m/sec

di M

ANSWER KEY
1.(c) 2.(a) 3.(c) 4.(b) 5.(c) 6.(c) 7.(d) 8.(b) 9.(d) 10.(c)

11.(d) 12.(b) 13.(a) 14.(c) 15.(b) 16.(c) 17.(a) 18.(c) 19.(b) 20.(b)

21.(c) 22.(a) 23.(b) 24.(a) 25.(c) 26.(b) 27.(b) 28.(c) 29.(a) 30.(d)

31.(c)
A

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 4

 SOLUTION
1. (c) 4. (b)
A At the same time, ratio of height and shadow
are same.

60° 45°
16 h
=
8 48
h = 16 × 6
h = 96 cm.
5. (c)
30° 45°
A
B D C
Let AD = h = 100 m. 30°
AD = CD = 1 unit  100 m. 60°

BD = 3 unit  100 3 m. 30°

r
E D

BD + CD = 100 3  100

si
= 100( 3  1)

2.
an by
= 100 (1.73 + 1) = 100 × 2.73 = 273 m.
(a)
B
AB = 100 m.
60°
C

n
A AB = 3 unit  100 m.

ja 100
R s
60° BC = 1 unit  m.
3
a th

Now, in AED,
D 30°
E 100
20 3 AE = 1 unit  m.
3
ty a

1.6 m 1.6 m
100 200
B CD = 100 – = = 66.6 m.
di M

C 3 3
20 3
6. (c)
Let AE = h Ratio will be same.
CD = BE = 1.6 m.
17.5 H
CB = DE = 20 3 m. =
40.25 28.75
DE = 3 unit  20 3 m. 70 H
=
AE = 1 unit  20 m. 161 28.75
AB = 20 + 1.6 = 21.6 m.
3. (c) 287.5
H=
23
R
H = 12.5 cm.
A

7. (d)
60°
P

100 m.
60° 30°

30°
Q P
RQ = 1 unit  100 m.
PQ = 3 unit  100 3 m. 30° 60°

= 100 × 1.73 = 173 m. R O Q

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 5

 In OPR, OR = ( 3 unit)  3 = 3 unit
1 unit  5 m.
OP = (1 unit) × 3 = 3 unit AO = BQ = 9 unit  45 m.
10. (c)
In OQP, OP = 3 unit
A
OQ = 1 unit and PQ = 2 unit 60°
RQ = 3 unit + 1 unit = 4 unit  180 m. 30°
60°
1 unit  45 m. D E
OP  3 unit  45 3 m.
PQ  2 unit  45 × 2 = 90 m. 300m
Height of tower = OP + PQ
= 45 3  90 = 45 ( 3  2) m.
C B
8. (b) 120m
A
AB = 300 m,
BC = 120 m. = DE

r
DE = 1 unit  120 m.
E

si
AE = 3 unit  120 3 m.
= 120 × 1.73
an by = 207.6 m.
CD = BE = Height of small tower = 300 – 207.

n
D = 92.4 m.
B
1.4 m. C 2.5 m. 11. (d)
Let AC = DE = Y m.
ja A
R s
AE = 0.8 m.
Let BE = H m.
a th

H² + (3.9)² = Y2 (in BDE) 45°

(H + 0.8)² + (2.5)² = Y² (in ABC) 220 m.

(H + 0.8)² + (2.5)² = H² + (3.9)²
H² + (0.8)² + 1.6 H + (2.5)² = H² + (3.9)²
ty a

1.6 H = 8.32 30° 45°

D B
C
di M

8.32
H= = 5.2
1.6 In ABC,
AB = 5.2 + 0.8 AB = 1 unit  220 m.
= 6 m. BC = 1 unit
In ABD
Length of the ladder = Y = (6)²  (2.5)²
AB = 1 unit
Y = 6.5 m. BD = 3 unit
Y = 6.5 m. BD – BC = 3 unit – 1unit
9. (d)
P = ( 3 – 1) unit = ( 3 – 1)  220
= (0.732 × 220) = 161.04
12. (b)
A

A O
60m A

35 m. 2unit
3unit
B
B Q
AB = OQ = 35 m. 1unit
PO = PQ – OQ = 60 – 35 = 25 m. 30° 60°
D C
5 PO 9 m.
tan = = 3unit
9 AO

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 6

 In ABC,
9 BC = BD – DC
AB (2 unit) = 2 = 400 – 200
3
BC = 200 m.
18 BC = 1 unit  200 m.
=  3 = 63
3
AB = 3 unit  200 3 m.
13. (a) 16. (c)
A A

E 13 m. 30°
12 m. 3unit
13 m.
5 m.

D B

r
C
30° 60°

si
BC = 5 m (Pythagoras in ABC) D B
CD = 12 m (Pythagoras in EDC) C 50 m.
BD = CD + BC
= 12 + 5
an by 1 unit = 50
3unit
1unit

n
= 17 m.
So, DC 2 unit = 100
14. (c)

ja A
100
R s
Speed = m/s
8
a th

30° 100 18
× = 45 km/h
8 5
17. (a)
D
ty a
di M

60°
C B
4.6 m.
BC = 1 unit  4.6 m.
AC = 2 unit  4.6 × 2 = 9.2 m.
Length of the ladder = 9.2 m. 50 m.
15. (b)
A
60°
30° 45° 60°
E A
45° B C
A

In BDC,
1 1
BC = 1 unit × = unit
45° 60° 3 3
D B
C 1
DC = 3 unit × = 1unit
BD = 400 m. 3
DE = 200 m. In ADC, 
DC = 200 m.
DC = 1 unit 50 m.
In EDC,
AC = 1 unit
DE = CD = 200 m.

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 7

 1
AB = AC – BC = 1 unit – unit = 60 + 203 = 20(3 +3) m
3 21. (c)
 3 – 1  3 – 1 H = 9  16
=  unit =    50 H=3×4
 3   3 
H = 12 m.
1.732 – 1 22. (a)
=  50 = 21.1 m.
1.732 A
18. (c)
C D
B 30°
C
30 m. 60°
G D
10003
Water surface
1000 3
30 m.

r
60° 30° H
I

si
A 1000 B E
3000 60°
an by
BE = 3000 – 1000 = 2000
19. (b) Let, AC = H
F E

n
A Then HE = H
HE = FI = H

ja 1 H
R s
H =
In ABC, tan 30° = 
BC 3 BC
a th

BC = H 3 = EF
In BFE,
60  H 60  H
ty a

tan 60 =  3 =

B H 3 H 3
D C 48 m.
di M

3 H – H = 60
75 m.
 = 15°,  = 75° H = 30 m.
 +  = 90 °  complementary Angles Height of the aeroplane form the water – surfac
Then, of the lake = 30 + H = 30 + 30 = 60m.
23. (b)
H = 48  75
H = 8  6  15  5 B C
H=2×2×3×5
H = 60 m. 60°
30°
20. (b)
E

6000 6000
203
A

60° 30°
45º A 2000 3 E D
C D 60003
60º 203
Distance cover in 40 second = DE
60 60 = 60003 – 20003 = 40003
So,
60º
4000 3
Speed = = 1003
B 203 A 40

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 8

 D A

30°
5×3 3unit
60°
5 × 3 75 m. B E
30° 1unit
60°
10 m. 10 m.
A 4×5 C 8×3 B
44
30°
15 unit = 75 C D
1 unit = 5 In BCD
44 unit = 220m BC = 1 unit  10 m.
25. (c)
A CD = 3 unit  10 3 m.

r
In ABE,

si
BE = 1 unit  10 3 m.
AE = 3 unit  10 3  3 = 10 × 3 = 30 m.
an by Height of the light house = 30 + 10 = 40 m.

n
4  28. (c)
cot –1  
5  A
B C
ja 90 m.
R s
4 h
 = cot –1  
a th

5
E D
4 90
cot = =
5 AC
ty a

450 60 m
AC =
4
di M

AC = 112.5 m. B C
26. (b) CD = BE = 60 m
A E Let, AE = h
h
In ADE, tan =
ED
45° 30°
h
H1 H2 ED = .......... (1
tan
Also, ED = BC

45° 60° 60
D In BDC, tan =
B ED
A

C
BC = CD (Given) 60 (Using (I))
tan =
Also, BC = AB = H, h
In CDE, tan
DE 60 tan
tan 60 = h= ..........(2
CD tan
H AB = 60 + h
3= 2
H1 60 tan
= 60  (Using (II))
H1 : H2 = 1 : 3 tan

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 9

  tan A
= 60 1  = 60 (1 + tan  cot )
 tan
29. (a)
A

3 unit
B

1 unit
60º
30º
C D 30° 60°
3unit D
B C 80 m.
Given, AC (3 unit) = 4000
8000 10 Seconds
So, AB (2 unit) = m In ACD,
3
30. (d)

r
AD = 3 unit  80 3 m.
D

si
In ABD,
30°
AD = 1 unit  80 3 m.
an by 60° 30°
3unit BD  3 unit  80 3  3 m.

n
= 240 m.
BC = BD – DC
ja
R s
30° 60° = 240 m. – 80 m.
= 160 m.
a th

A B 1 unit C

15 min. 160
Speed = = 16 m/s
3unit 10
ty a

2 unit = 15minute
di M

1unit = 7.5 minute

A

ditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs 1

 Mensuration-2D

MENSURATION -2D/f}foeh; {ks=kfefr

[CLASSROOM SHEET]

What is Mensurtion?/{ks=kfefr D;k gS\ Important Unit Coversion

The literal meaning of the word Mensuration 1 Kilometer (km) = 10 Hectometer
is 'to measure'. It is a branch of mathematics 1 Hectometer = 10 Decametre
that deals withthe measurement of 1 Decametre = 10 Metre (m)
perimeter, area and volume of the different
1 metre = 10 Decimetre (dm)
geometrical figures.
1 Decimetre = 10 centimetre (cm)
{ks=kfefr 'kCn dk 'kkfCnd vFkZ gS ^ekiuk*A ;g xf.kr
1 centimetre = 10 milimetre (mm)
dh ,d 'kk•k gS tks fofHkUu T;kferh; vkÑfr;ksa dh

r
1 Mile = 1760 Yard
ifjf/] {ks=kiQy vkSj vk;ru ds eki ls lacaf/r gSA

si
1 Yard = 3 Feet
The mensuration is divided in the following
1 Feet = 12 Inch
two parts :
an by
{ks=kfefr dks fuEufyf[kr nks Hkkxksa esa foHkkftr1fd;k
Inch = 2.54 Centimetre (cm)

n
x;k gS % 1 Feet = 0.3048 Metre
ja
(i) Two-dimension mensuration 5 Mile = 8 Kilometre (km)
R s

f}foeh; {ks=kfefr 1 Hectare = 10000 metre2 (m2)

a th

(ii) Three-dimension mensuration 1 Cubic metre (m3) = 1000 Litre

1 Litre = 1000 cubic cm (cm3)
f=kfoeh; {ks=kfefr
What is Area?/{ks=kiQy D;k gS
ty a

In two-dimension mensuration we will study

the two-dimension figures (plane figures), The area can be defined as the space
di M

like triangle, quadrilateral, polygon, circle occupied by a flat shape or the surface of
etc. an object.
f}foeh; {ks=kfefr esa ge f}foeh; vkÑfr;ksa (lery {ks=kiQy dks fdlh ,d piVs vkdkj ;k fdlh o
vkÑfr) tSls&f=kHkqt] prqHkqZt] cgqHkqt] o`Ùk vkfndh dk
lrg }kjk ?ksjs x;s LFkku ds :i esa ifjH
vè;;u djrs gSaA fd;k tk ldrk gSA
In three-dimension mensuration we will The area of a figure is the number of unit
study the three-dimension figures like cube, squares that cover the surface of a closed
figure. Area is measured in square units such
cuboid, cylinder, cone, frustum, sphere,
as square centimeters, square meter, etc.
hemisphere, Prism, Pyramid etc.
fdlh vkÑfr dk {ks=kiQy mldh can lrg }kjk
A

f=kfoeh; {ks=kfefr esa ge f=kfoeh; vkÑfr;ksa tSls&

x, oxZ ek=kdksa dh la[;k gSA {ks=kiQy dk
?ku] ?kukHk] csyu] 'kadq] fNÂd] xksyk] v¼Zxksyk]
esa ekik tkrk gS tSls dh oxZ lsaVhehVj]
fçTe] fijkfeM vkfn dk vè;;u djrs gSaA vkfnA
In mensuration, Pythagorean triplets and
divisibility tricks will be used.
 Mensuration-2D

Perimeter/ifjeki (b) If the area of ABC is , then

A
Perimeter can be defined as the path or
the boundary that surrounds a figure . It 
can also be defined as the length of the
4 R
outline of a shape. P
ifjeki dks fdlh vkÑfr dks ?ksjus okys iFk ;k 
 
4
ifjlhek ds :i esa ifjHkkf"kr fd;k tk ldrk gSA 4 4
B C
bls fdlh vkÑfr dh ifjlhek dh yackbZ ds :i esa Q
Hkh ifjHkkf"kr fd;k tk ldrk gSA (c) Let a triangle PQR is formed by joining
the mid-points of the sides of ABC, then
again a XYZ is formed by joining the mid-
points fo the sides of PQR, if this process
continue till infinite, then
eku yhft, fd ABC dh Hkqtkvksa ds eè;&f

r
dks feykus ls ,d f=kHkqt
PQR curk gS] fiQjPQR
dh Hkqtkvksa ds eè;&fcanqvksa dks feykdj
XYZ,

si
Mensuration 2D
an by curk gS] ;fn ;g çfØ;k vuar rd pyrh jgs] rks
In this section, we will study the (i) The area of all triangles/lHkh f=kHkq
{ks=kiQy

n
measurements of perimeter and area of
figures which lie on a plane.
ja 4
= × Area of ABC
bl Hkkx esa ge ,d gh lery esa fLFkr vkÑfr;ksa 3
R s
ds ifjeki ,oa {ks=kiQy dh eki dk vè;;u djsaxsA (ii) The perimeter of all triangles/lHkh f=kH
a th

dk ifjeki
Triangle
= 2 × Perimeter of ABC
 The perimeter and the area of a triangle A
ty a

made by joining the mid-points of the sides

will be half of original perimeter and one-
di M

fourth of the original area respectively. P Z R

Hkqtkvksa ds eè;&fcanqvksa dks feykus ls cus f=kHkqt
X Y
dk ifjeki vkSj {ks=kiQy Øe'k% ewy ifjeki dk vkèkkj
vkSj ewy {ks=kiQy dk ,d&pkSFkkbZ gksxkA B C
Q
If P, Q and R be the mid-point of AB, BC 1. Consider an equilateral triangle of a side
and AC, respectively, then of unit length. A new equilateral triangle
;fn P, Q vkSj R Øe'k% Hkqtkvksa
AB, BC vkSj is formed by joining the mid-points of one,
then a third equilateral triangle is formed
AC ds eè; fcUnq gks] rks
by joining the mid-point of seconds. The
A process is continued. The perimeter of all
A

triangle, thus formed is:

P R bdkbZ yackbZ dh Hkqtk ds ,d leckgq f=

eè;&fcanqvksa dks feykdj ,d u;k leckgq f=kH
B
gS] fiQj nwljs ds eè;&fcanqvksa dks feykdj
C
Q leckgq f=kHkqt curk gSA çfØ;k tkjh jgrh gSA
 Mensuration-2D

 If the lengths of the perpendiculars drawn  ABC is an equilateral triangle (leckgq f=kH
from a point inside an equilateral triangle & DE || BC, then ADE is also equilateral.
to its sides are P1, P2 and P3, then A

;fn fdlh leckgq f=kHkqt ds vanj fdlh fcUnq ls

bldh Hkqtkvksa ij Mkys x, yacksa dhP1,
yackbZ
P2
D E
vkSjP3 gks] rks

3a
(a) P1 + P2 + P3 = = h (height) B C
2
3. ABC is an equilateral triangle, P and Q are
(b) Area of the equilateral triangle two points on AB and AC respectively such
that PQ ||BC. If PQ = 5 cm, then area of
(P1 + P2 +P3 )2 APQ is:
=
3
ABC ,d leckgq f=kHkqt gSA Pfcanq
vkSjQ Øe'k%
2. If the length of the three perpendiculars AB rFkkAC ij bl rjg fLFkr gSa fd PQ||BC,

r
from a point in the interior of an APQ dk {ks=kiQy Kkr djsaA

si
equilateral triangle to the sides are 4 cm,
5 cm and 6 cm, then find the area of the
an by 25 25
(a) cm2 (b) cm2
triangle. 4 3
;fn fdlh leckgq f=kHkqt ds vUnj fLFkr fdlh fcanq

n
25 3
ls f=kHkqt dh Hkqtkvksa ij •haps x;s yEc dh yEckb;ka (c) cm2 (d) 25 3 cm2
ja 4
4 lseh] 5 lseh vkSj 6 lseh gks] rks f=kHkqt dk {ks=kiQy
R s
4. ABC is an equilateral triangle (leckgq f=kH
Kkr djsaA
DE || BC & MN || DE.
a th

(a) 50 3 cm 2
(b) 75 3 cm 2 A

(c) 225 cm2 (d) 100 cm2

M N
ty a

 In equilateral 
D E
If r = inradius (var%f=kT;k)
di M

R = circumradius (ifjf=kT;k) B C

S = Side of equilateral triangle DE 2 MN 3

If = , = , then
BC 3 DE 4
(leckgq f=kHkqt dh Hkqtk,¡)
Area of AMN
=?
Area of ABC
(a) 1/4 (b) 2/3
(c) 1/2 (d) 3/4
5. The difference between the area of the
circumscribed circle and the area of the
A

inscribed circle of an equilateral triangle

is 2156 sq. cm. What is the area of the
r equilateral triangle?
(i) Find :
R fdlh leckgq f=kHkqt ds ifjo`Ùk vkSj var%
{ks=kiQyksa dk varj 2156 oxZ lseh gSA le
Area of incircle
dk {ks=kiQy D;k gS\
 Mensuration-2D

6. Find the area of the hexagon formed after ;fn ,d lef}ckgq f=kHkqt dh çR;sd leku H
cutting the corners of the sides of an
dh yEckbZ 'a' bdkbZ gS vkSj rhljh Hkqtk dh
equilateral triangle of side 9 cm:
'b' bdkbZ gS] rks mldk {ks=kiQy D;k gksx
9 lseh Hkqtk ds ,d leckgq f=kHkqt dh Hkqtkvksa dks
fdukjs ls dkVus ij cus "kVHkqt dk {ks=kiQy Kkr a 4b 2  a 2
(a) sq. units
dhft;s\ 4

2 3 29 3 a
(a) (b) (b) 2a 2  b 2 sq. units
2 2 2

b
27 3 20 3 (c) 4a 2  b 2 sq. units
(c) (d) 4
2 2
7. In the figure above, ABCD is a rectangle b
(d) a 2  2b 2 sq. units
and triangle AFE and triangle EFC are 2
equilateral triangles. If the area of triangle 10. Find the area of an isosceles triangle whose

r
BEC is 8 3 cm2, what is the area of the sides are 8 cm, 5 cm and 5 cm.
complete rectangle? lef}ckgq f=kHkqt dk {ks=kiQy Kkr dhft;

si
fn, x, fp=k esa] ABCD ,d vk;r vkSj f=kHkqt
an by Hkqtk,a dh yEckbZ Øe'k% 8 lseh] 5 lseh
AFE vkSj f=kHkqt lseh gSA
EFC leckgq f=kHkqt gSaA ;fn f=kHkqt
BEC dk {ks=kiQy (a) 12 cm2 (b) 15 cm2
8 3 oxZ lseh gS rks laiw.kZ vk;r

n
(c) 18 cm2 (d) 20 cm2
dk {ks=kiQy D;k gS\ ja 11. The Altitude drawn to the base of an
E
R s
A B isosceles is
95 cm and the perimeter is
38 cm. Find the area of the isosceles
a th

triangle.
,d lef}ckgq ds vk/kj ij Mkyk tkus okyk yE
95 lseh gS vkSj vkSj bldk ifjeki 38 lseh
ty a

lef}ckgq f=kHkqt dk {ks=kiQy Kkr dhft,A

D
di M

F C (a) 695 cm2 (b) 1295 cm2

(a) 123 (b) 36 (c) 1495 cm2 (d) 795 cm2
(c) 483 (d) None of these 12. ABC is an isosceles right triangle and AC
8. The ratio of the length of each equal side is its hypotenuse. The area of the square
and the third side of an isosceles triangle drawn on hypotenuse as side is 128 cm2.
is 3:5. If the area of the triangle is What is the sum of areas of equilateral
2
30 11 cm then the length of the third triangles drawn on AB and BC as sides?
side (in cm) is: ABC ,d lef}ckgq ledks.k f=kHkqt gS vkSAC
,d lef}ckgq f=kHkqt dh izR;sd leku Hkqtk dh yackbZbldk fod.kZ gSA fod.kZ dks Hkqtk ekudj
vkSj rhljh Hkqtk dh yackbZ dk vuqikr
3 : 5 gSA ;fn
cuk, x, oxZ dk {ks=kiQy 128 oxZ lsa-eh-
AB
f=kHkqt dk {ks=kiQy
30 11 lseh gS] rks rhljh Hkqtk
2
vkSjBC Hkqtkvksa ij cuk, x, leckgq f=kHk
A

dh yackbZ (lseh esa) D;k gksxh\ {ks=kiQyksa dk ;ksx D;k gS\

SSC CGL TIER I 18/07/2023 (Shift-02)
(a) 10 6 (b) 5 6 (a) 32 2 cm2 (b) 32 3 cm2
(c) 32 cm2 (d) 64 cm2
(c) 13 6 (d) 11 6
13. If the perimeter of an isosceles right
9. If for an isosceles triangle the length of
 Mensuration-2D

,d lef¼ckgq f=kHkqt ds vk/kj ij Mkys x, y

;fn lef}ckgq ledks.k f=kHkqt dh ifjf/
8  
2 +1
dh yEckbZ 8 lseh rFkk f=kHkqt dk ifjeki 6
lseh gS] rks f=kHkqt ds fod.kZ dh yackbZ gS gSA f=kHkqt dk {ks=kiQy (lseh2 esa) gksx
SSC CGL TIER - II 18/11/2022 (c) 240 (d) 180
(a) 10 cm (b) 8 cm (c) 360 (d) 120
(c) 24 cm (d) 12 cm 18. The perimeter of an isosceles triangle is 91
14. The ratio of length of each equal side and cm. If one of the equal sides measures 28
the third side of an isosceles triangle is cm, then what is the value of the other non-
equal side?
3 : 4. If the area of the triangle is 18 5 ,d lef}ckgq f=kHkqt dk ifjeki 91 lseh gSA ;fn
square units, the third side is: Hkqtkvksa esa ls ,d dk eki 28 lseh gS] rks vlek
fdlh lef}ckgq f=kHkqt dh cjkcj Hkqtk esa ls ,d dk eku D;k gS\
rFkk vleku Hkqtk dk vuqikr 3 % 4 gSA ;fn f=kHkqt SSC CHSL 17/03/2023 (Shift-03)
(a) 56 cm (b) 42 cm
dk {ks=kiQy18 5 oxZ bdkbZ gks] rc rhljh Hkqtk
(c) 14 cm (d) 35 cm
Kkr djsa  Side of Largest Square inside a Triangle

r
(a) 16 units (b) 510 units whose base is ‘B’ and height is ‘H’

si
(c) 82 units (d) 12 units fdlh f=kHkqt ds vanj lcls cM+s oxZ dh Hkqt
15. The perimeter of an isosceles, right-angled
an by B×H
triangle is 2p unit. The area of the same vk/kj B vkSj Å¡pkbZ
H gks=
B+H
triangle is-

n
,d ledks.kh; lef¼ckgq f=kHkqt dk ifjeki2p
bdkbZ gS bldk {ks=kiQy gksxk&
ja
R s
(c) 3 – 2 2  p 2
sq.unit
a th

(d) 2  2 2  p 2
sq.unit
19. In a right angled triangle ABC, AB = 12
cm and AC = 15 cm. A square is inscribed
(c) 2 – 2 2  p 2
sq.unit
ty a

in a triangle. One of the vertices of square

coincides with vertex of triangle. What is
(d) 3 – 2  p2
sq.unit
di M

the maximum possible area (in cm2) of the

16. In a triangle ABC, AB = BC and the square?
,d ledks.k f=kHkqtABC esaAB = 12 lseh rFk

perimeter of  ABC is 8 2 + 2 cm. If the 
AC ¾ 15 lseh gSA f=kHkqt ds Hkhrj ,d o
length of BC is 2 times the length of gqvk gSA oxZ ds 'kh"kks± esa ls ,d f=kHk
AB, then find the area of  ABC. Li'kZ djrk gSA oxZ dk vf/dre laHko {ks=kiQ
,d f=kHkqtABC esaAB = BC vkSj  ABC dh esa) D;k gS\
SSC CGL TIER - II 18/02/2018
ifjf/ 8  2 + 2  lseh gSA ;fnBC dh yackbZ
AB
1296
(a) (b) 25
dh yackbZ ls 2 xquk gS] rks
 ABC dk {ks=kiQy 49
A

1225 1225
Kkr dhft,A (c) (d)
36 64
SSC CGL TIER - II 18/11/2020 20. The area of largest square which is
(a) 28 cm² (b) 36 cm² inscribed in a triangle whose sides PQ, QR
(c) 32 cm² (d) 16 cm² and PR are 6 units, 8 unit and 10 unit
17. The altitude drawn to the base of an a
respectively is in the form of square
 Mensuration-2D

ml lcls cM+s oxZ tks 6 ;wfuV dh Hkqtk

PQ, 8 (a) Area of ABC
;wfUkV dh Hkqtk
QR rFkk 10 ;wfuV dh Hkqtk
PR (b) 2 times the area of ABC

a (c) Area of semi-circle ABC

okys fdlh f=kHkqt ds Hkhrj cuk gS] dk {ks=kiQy
b (d) None of the above
o x Z ; w fu V g S ] t gak ar Fk k b i w . k k Z a d g24.
S A In the given figure, 3 semicircles are drawn
(a + b) dk eku Kkr dhft,A on three sides or triangle ABC. AB = 21
cm, AC = 28 cm and BC = 35 cm. What is
(a) 225 (b) 625
the area (in cm2) of the shaded part?
(c) 445 (d) 289
21. An equilateral triangle of side 12 cm is
nh x;h vkÑfr esa] f=kHkqt
ABC dh rhuks Hkqt
drawn. What is the area (in cm2) of the largest ij 3 v¼Zo`Ùk cuk;s x, gSaA
AB = 21 cm, AC =
square which can be drawn inside it? 28 cm rFkkBC = 35 cm gSA Nk;kafdr Hk
12 lseh Hkqtk okyk ,d leckgq f=kHkqt cuk;k x;kA {ks=kiQy (lseh
2
) esa D;k gS\
blesa cuk;s tk ldus okys lcls cM+s oxZ dk {ks=kiQy

r
A
(lseh2 esa) D;k gS\

si
SSC CGL TIER - II 09/03/2018
(a) 1512 – 8643 an by(b) 3024 – 17283
(c) 3024 + 17283 (d) 1512 + 8643
22. Find the area of shaded region, where BC B C

n
= 8 cm, AB = 6 cm and AC = 10 cm. (a) 588
Nk;kafdr Hkkx dk {ks=kiQy Kkr djsa]
BC tgka
ja = (b) 324
R s
8 cm, AB = 6 cm and AC = 10 cm. (c) 294
a th

C (d) 286
25. In the given figure, ABC is a right angled
triangle, right angled at B. BC = 21 cm
ty a

and AB = 28 cm. Width AC as diameter

A of a semi-circle and width BC as radius a
B
di M

quarter circle are drawn. What is the area

of the shaded portion?
(a) 36 cm² (b) 25 cm² fn;s x, fp=k esa]ABC, B ij ledks.k f=kHkqt
(c) 48 cm² (d) 12.5 cm² AB = 21 lseh rFkkAB = 28 lseh gSA AC dk
23. In the given figure, ABC is a right angled O;kl ekudj ,d v¼Zo`Ùk rFkkBC dks f=kT;k eku
triangle, right angled at A. Semi-circles are ,d o`Ùk[k.M [khpk x;k gSA rc Nk;kafdr H
drawn on the sides AB, BC and AC. Then,
{ks=kiQy D;k gksxk\
the area of shaded portion is equal to
which one of the following? A
fn;s x, fp=k esaABC, A ij ledks.k f=kHkqt gSA
HkqtkAB, BC rFkk AC ij v¼o`Ùk cuk;s x, gSaA
A

rks Nk;kafdr Hkkx dk {ks=kiQy fuEu esa ls fdlds28

cjkcj gksxk\

A B
21 C
(a) 425 cm²
 Mensuration-2D

uksV % 30. The perimeter of a triangle is 30 cm and

its area is 30 cm². If the largest side
C
measures 13m, what is the length of the
smallest side of the triangle?
,d f=kHkqt dk ifjeki 30 lseh rFkk bldk {ks=
H 30 lseh2 gSA ;fn lcls cM+h Hkqtk dh yE
eh gS rkss lcls NksVh Hkqtk dh yEckbZ Kk
UPSSSC Revenue Lekhpal
(c) 3 cm (d) 4 cm
(c) 5 cm (d) 6 cm
B A 31. The in-radius and circumradius of a right-
H2 angled triangle is 3 cm and 12.5 cm,
Area of right angle triangle = sin2 respectively. The area of the triangle is:
4
Where, H  Hypotenuse (d.kZ) and,   fdlh ledks.k f=kHkqt dh vUr% f=kT;k vkSj ifjf=k
one of the acute angle of right angle 3cm vkSj12.5cm gSaA f=kHkqt dk {ks=kiQy

r
triangle. SSC CHSL 05/08/2021 (Shift- 03)

si
ledks.k f=kHkqt dk dksbZ ,d U;wudks.k gSA (a) 84 cm² (b) 88 cm²
26. One of the angles of a right-angled triangle
an by (c) 48 cm² (d) 64 cm²
is 15o and the hypotenuse is 1 m. The area 32. The area of the largest triangle that can
of the triangle (in sq. cm.) is be inscribed in a semicircle of radius 4 cm

n
in square centimeters
ledks.k f=kHkqt ds dks.kksa esa ls ,d dks.k
15° rFkk
lcls cM+k f=kHkqt dk {ks=kiQy Kkr dhft
f=kHkqt dk d.kZ 1 eh gSA {ks=kiQy fdruk gksxk
ja lseh ds f=kT;k okys v/Zo`Ùk esa vafdr fd;k
R s
(lseh2 esa)
SSC CHSL 15/10/2020 (Shift- 03)
a th

(a) 1220 (b) 1250

(a) 16 cm² (b) 14 cm²
(c) 1200 (d) 1215
27. If hypotenuse of a right angle D is 10 cm. (c) 12 cm² (d) 18 cm²
What can be its maximum area? 33. The area of the largest triangle that can
ty a

;fn fdlh ledks.k f=kHkqt dk d.kZ 10 lseh gS] rks be inscribed in a semi-circle of radius 6 cm
is:
mldk vf/dre {ks=kiQy D;k gks ldrk gS\
di M

(a) 36 cm² (b) 25 cm² 6 cm f=kT;k okys ,d v/Z&o`Ùk esa cu ldu

(c) 16 cm² (d) 30 cm² lcls cM+s f=kHkqt dk {ks=kiQy fdruk gksx
28. The area of triangle is 15 sq cm and the SSC CHSL 11/08/2021 (Shift- 03)
radius of its incircle is 3 cm. Its perimeter (a) 35 cm² (b) 34 cm²
is equal to: (c) 38 cm² (d) 36 cm²
f=kHkqt dk {ks=kiQy 15 oxZ lseh gS vkSj blds var%o`Ùk
dh f=kT;k 3 lseh gSA bldh ifjf/ fdruh gS% Circle (o`Ùk)
SSC CGL 04/06/2019 (Shift- 03)
 Area of quadrant of circle
(a) 12 cm (b) 20 cm
1
(c) 5 cm (d) 10 cm o`Ùk ds prqFkk±'k dk {ks=kiQy
= r 2
4
A

29. The sides of a triangle are 56 cm, 90 cm  Circumference of quadrant of circle

and 106 cm. The circumference of its
r
circumcircle is : o`Ùk ds prqFkk±'k dk ifjf/
= + 2r
2
,d f=kHkqt dh Hkqtk,¡ 56 lseh] 90 lseh vkSj 106
lseh gSaA blds ifjo`Ùk dh ifjf/ Kkr djsaA
SSC CGL TIER II 12/09/2019 r
 Mensuration-2D

36. If length of the arc = 6 cm and radius of

r 2 
 Area of sector/f=kT;[kaM dk {ks=kiQy
= circle = 5 cm. Find area of sector of a
360º
circle.
;fn pki dh yEckbZ = 6 lseh vkSj f=kT;k
= 5
lseh gksA o`Ùk ds f=kT;•aM dk {ks=kiQy
(a) 15 cm² (b) 20 cm²
(c) 30 cm² (d) 24 cm²
A B  Area of segment (o`Ùk[kaM dk {ks=ki

34. The area of a sector of a circle with central

angle 60° is A. The circumference of the
circle is c. Then A is equal to:
dsaæh; dks.k 60° okys ,d o`Ùk ds o`Ùk•.M
c
dk {ks=kiQy A gSA o`Ùk dh ifjfèkc gS] rks A fuEu A B
esa ls fdlds cjkcj gksxk\

r
Area of segment = Area of AOB
c2 c2 o`Ùk[kaM

si
(a) (b) dk {ks=kiQy
= f=kHkqt
AOB dk {ks=kiQy
6 18
r 2  1 2
(c)
2
c
an by (d)
c2 =
360º 2
 r sin 
24 4 37. A sector of a circle of radius 10 cm is formed

n
2r at 60° angle at the centre. What will be its
 Length of the Arc AB (l ) = area (take  = 3.14)?
360º
ja
R s
pki AB dh yackbZ 10 cm f=kT;k ds ,d o`Ùk dk f=kT;k[kaM 60°
dsa
dks.k ij curk gSA bldk {ks=kiQy D;k (gksxk\
 = 3.14
a th

ekfu,)
SSC CGL TIER I 17/07/2023 (Shift-04)
(a) 52.33 cm² (b) 75.28 cm²
ty a

A B (c) 60.67 cm² (d) 55.00 cm²

Area of a sect or i s 1848 m 2 and the central
di M

38.
35. Find the length of the arc of the sector of angle of the sector is 270°. Find the radius
a circle of diameter 7 cm with a central of the circle. (Take  = 22/7 )
angle of 108°. [Use  = 22/7] fdlh o`Ùk ds ,d f=kT;•aM dk {ks=kiQy
1848 m2 g
7 cm O;kl okys ,d o`Ùk ds f=kT;•aM dh pki dh vkSj f=kT;•aM dk dsaæh; 270°
dks.kgSA o`Ùk dh
yackbZ Kkr dhft, ftldk dsaæh; dks.k
108° dk gSA Kkr dhft,A ¹ = 22/7 dk ç;ksx dhft,º
¹  = 22/7 dk ç;ksx dhft, SSC CGL TIER I 18/07/2023 (Shift-04
SSC CGL TIER I 19/07/2023 (Shift-01) (a) 784 m (b) 22 m
(c) 27 m (d) 28 m
(a) 6.6 cm (b) 5.6 cm
 Area enclosed by two concentric circle
(c) 13.2 cm (d) 11.2 cm
(nks ladsUnzh; o`Ùkksa }kjk ?ksjk x;k {ks
A

1
 Area of Sector/ f=kT;[kaM dk {ks=kiQy
= lr
2

r R

c
 Mensuration-2D

If R and r are radii of two concentric (a) 8800 (b) 8756

circles, then (c) 8558 (d) 8778
;fn R vkSjr nks ladsUnzh; o`Ùkksa dh f=kT;k,¡ gSa] rks
42. A race track is in the shape of a ring whose
Area enclosed by two circle = R2 – r2
inner and outer circumference is 440m and
nks o`Ùkksa }kjk ?ksjk x;k={ks=kiQy
R2 – r2) 506, respectively. What is the cost of
= (R + r)(R – r)
Width of path/jkLrs dh pkSM+kbZ =R–r  22 
levelling the track at Rs. 6/m² ?  π = 
outer circumference – inner circumference  7 
=
2
39. The inner circumference of a circular path ,d /kou iFk NYys ds vkdkj esa gS ftldh vkarf
enclosed between two concentric circles vkSj ckgjh ifjf/ Øe'k% 440 ehVj vkSj 506
is 264 m. The uniform width of the circular bl iFk dks 6 :i;s izfr oxZ ehVj dh nj ls lery
path is 3 m. What is the area (in m², to
the nearest whole number) of the path? djus dh ykxr fdruh vk,xh\
 22  SSC CGL 03/03/2020 (Shift- 02)
 Take   
7 

r
 (a) Rs. 18966 (b) Rs. 24832
nks ladsafær o`Ùkksa ds chp f?kjs ,d o`Ùkkdkj iFk(c)
dh Rs. 19866 (d) Rs. 29799

si
vkarfjd ifjf/ 264m gSA o`Ùkkdkj iFk dh ,dleku
43. The sum of the radii of two circles is
pkSM+kbZ
3m gSA iFk dk {ks=kiQym2 esa]
( fudVre
an by 22
286cm and the area between the
concentric circles is 50336cm2. What are
iw.kZ la[;k rd) D;k gS\
(π = ysa
)

n
7 the radii (in cm) of the two circles?
SSC CGL 20/04/2022 (Shift- 01)
ja  22 
(a) 696 (b) 948  Take  = 
R s
(c) 756 (d) 820  7 
40. The area of a circular path enclosed by two
nks o`Ùk dh f=kT;kvksa dk286cm
;ksx gS vk
a th

concentric circles is 3080 m². If the

difference between the radius of the outer ladsafær o`Ùkksa ds eè; dk {ks=kiQy
50336cm 2
edge and that of inner edge of the circular gSA nksuksa o`Ùkksa (cm
dh f=kT;k,a
esa
) fdruh&fdru
path is 10 m, what is the sum (in m) of
ty a

22
the two radii? (Take  = 22/7) gksaxh\= eku ysaA
7
nks ladsafær o`Ùkksa ls f?kjs ,d o`Ùkkdkj iFk dk {ks=kiQy
di M

3080 oxZ ehVj gSA ;fn ckgjh fdukjs dh f=kT;k SSC CHSL 15/04/2021 (Shift- 02)
vkSj o`Ùkkdkj iFk ds Hkhrjh fdukjs dh f=kT;k ds (a) 91 and 84 (b) 171 and 84
chp dk varj 10 ehVj gS] rks nksuksa f=kT;kvksa (c) dk 115 and 91 (d) 115 and 171
;ksx (ehVj esa) D;k gS\
Some Useful Results/dqN mi;ksxhifj.k
SSC CGL 23/08/2021 (Shift- 02)
(a) 70 (b) 112 (i) If two circles touch internally, then the
(c) 98 (d) 84 distance between their centres is equal to
the difference of their radii.
41. The perimeter of a circular lawn is 1232
m. There is 7 m wide path around the lawn. ;fn nks o`Ùk vkrafjd :i ls Li'kZ djsa] rks
The area (in m²) of the path is: dsUæksa ds chp dh nwjh mudh f=kT;kvks
A

22 cjkcj gksrh gSA

=
7
,d o`Ùkkdkj ykWu dk ifjeki 1232 ehVj gSA ykWu
ds pkjksa vkSj 7 ehVj pkSM+k ekxZ gSA ekxZ dk {ks=kiQy O O’

22
 Mensuration-2D

Radius of bigger circle/nh?kZ o`Ùk dh f=kT;k

=R 46. Two circles touch externally. The sum of
their areas is 130 sq. cm and the distance
Radius of smaller circle/y?kq o`Ùk dh f=kT;k
=r
between their centres is 14 cm. Find the
OO = R – r
radii of the circles.
44. Two circles touch each other internally,
if the sum of the areas of two circles is nks o`Ùk cká :i ls Li'kZ djrs gSaA muds
116 cm2 and the distance between their dk ;ksx 130 oxZ lseh gS vkSj muds dsUæk
centres is 6 cm. Find the radius of both dh nwjh 14 lseh gSA o`Ùkksa dh f=kT;k,¡ K
the circles. (a) 11,3 (b) 10,4
nks o`Ùk vUr% Li'kZ djrs gSa A buds {ks=kiQyksa dk
(c) 9,5 (d) 8,6
;ksxiQy 116
 oxZ lseh rFkk buds dsUæksa ds(iii)
chp Distance moved by a rotating wheel in one
dh nwjh 6 lseh gS A o`Ùkksa dh f=kT;k,¡ Kkr dhft;s
revolution is equal to the circumference
of the wheel.
,d ?kwers gq, ifg;s }kjk ,d pDdj esa r; d
C C' x;h nwjh o`Ùk dh ifjf/ ds cjkcj gSA

r
(iv) The number of revolutions completed by

si
a rotating wheel in one minute
an by ,d ?kwers gq, ifg;s }kjk ,d feuV esa iwjs fd
(a) 10,4 (b) 11,3 pDdjksa dh la[;k
(c) 8,6 (d) 9,5

n
45. A smaller circle touches a bigger circle Distance moved in one minute
internally and also passes through the
ja Circumference
center 'O' of the bigger circle. If the area 47. If a wheel has diameter 42 cm, then how
R s
of the smaller circle is 192 cm², the area far does the wheel go (in meters) in 12
a th

of the bigger circle (in cm²) is:

22
,d NksVk o`Ùk ,d cM+s o`Ùk dks vkarfjd :i ls revolutions?  π  

Li'kZ djrk gS vkSj cM+s o`Ùk ds 'O' dsaæ
ls Hkh  7 
xqtjrk gSA ;fn NksVs o`Ùk dk {ks=kiQy 192 ehVj2 ;fn ,d ifg;s dk O;kl 42 lseh gS] rks 12 pDdjk
ty a

esa ifg;k fdruh nwj (ehVj esa) r; djsxk\

gS] rks cM+s o`Ùk dk {ks=kiQy (ehVj2 esa) Kkr dhft,A
di M

SSC MTS 05/07/2022 (Shift- 03) CPO 2019 23/11/2020 (Shift- 03)
(a) 768 (b) 384 (a) 17.64 (b) 15.84
(c) 1024 (d) 720 (c) 23.27 (d) 21.45
(ii) If two circles touch externally, then the
48. An athlete runs 8 times around a circular
distance between their centres is equal to
field of radius 7 m in 3 minutes 40 seconds.
the sum of their radii.
;fn nks o`Ùk oká :i ls Li'kZ djsa] rks muds dsUæksa
His speed (in km/h) is: (Taken  =
22
)
ds chp dh nwjh mudh f=kT;kvksa ds ;ksx ds cjkcj 7
gksrh gSA ,d /kod 3 feuV 40 lsdaM eas 7 m f=kT;k o
,d o`Ùkkdkj eSnku ds
8 pDdj yxkrk gSA mldh p
A

22
O O’ (km/h esa
) fdruh gS\ (  = dk ç;ksx djs)
7
SSC CGL MAINS 03/02/2022

72 118
(a) (b)
25 25
Radius of bigger circle/nh?kZ o`Ùk dh f=kT;k
=R
 Mensuration-2D

49. The radius of a roller is 14 cm and its fn;s x;s fp=k eas] izR;sd ckgjh o`Ùk ftudh
length 20 cm. It takes 235 complete 'R' gS rks vkUrfjd o`Ùk dh f=kT;k gksxh&
revolutions to move once over to level a
playground. Find the area of the
playground.(Use =22/7)
,d jksyj dh f=kT;k 14 lseh vkSj bldh yackbZ 20
lseh gSA ,d [ksy ds eSnku dks lery djus dss
fy, blds 235 iw.kZ pDdj yxrs gSaA [ksy ds eSnku
dk {ks=kiQy Kkr dhft,A
SSC CHSL 09/06/2022 (Shift- 02) 2 1
R
(a) 4136 cm² (b) 4136 × 10³ cm²
(a)
 
2 +1 R (b)
2

(c) 41360 cm² (d) 4136 × 10² cm² (c)  2 – 1 R (d) 2R

50. The diameter of a wheel is 1.33 m. What 53. An equilateral triangle circumscribes all

r
distance (in m, to the nearest whole the circles, each with radius 10 cm. What
number) will it travel in 380 revolutions? is the perimeter of the equilateral triangle?

si
,d leckgq f=kHkqt lHkh o`+Ùkksa] ftudh
22
(Take  = ). lseh gS dks ifjxr djrk gSA leckgq f=kHkqt dk
7
an by gksxk\
fdlh ifg, dk O;kl 1.33 m gSA 380 pDdjksa esa]

n
;g ifg;k fdruh nwjh (m esa] fudVre iw.kZ la[;k
ja
R s
22
rd ) r; djsxk\ (  = ysa)A
7
a th

SSC CHSL 16/04/2021 (Shift- 01)

(a) 1856 (b) 1855
ty a

(c) 1685 (d) 1588 (a) 20 (2 + 3) cm (b) 30 (2 + 3) cm

51. The minute hand of a clock is 20 cm long. (c) 60 (2 + 3) cm (d) None of these
di M

Find the area on the face of the clock 54. In the figure given below, AB is line of
swept by the minute hand between 8 am length 2a, with M as mid-point. Semi-
and 8:45 am. circles are drawn on one side with AM, MB
,d ?kM+h dh feuV dh lqbZ 20 lseh yach gSA lqcgand AB as diameters. A circle with centre
8 am ls lqcg 8 % 45 am ds chp feuV dh lqbZ O and radius r is drawn such that this
}kjk ?kwes x, ?kM+h ds i`"B dk {ks=kiQy
r dhft,A Kk circle touches all the three semi-circles.
What is the value of r ?
SSC CGL 12/12/2022 (Shift- 01)
uhps fn;s x;s fp=k esa
AB Hkqt dh yEckbZ 2a gS
6600
(a) cm² M mldk eè; fcUnq gSAAB, MB, AB Hkqtkvks
7
O;kl eku dj v¼Zo`Ùk cuk;s tkrs gSaA O dsU
,d
6600
A

(b) cm² okyk rFkkr f=kT;k okyk o`Ùk bl izdkj cuk;k

9
fd og rhuksa v¼Zo`Ùkksa dks Li'kZ rdjrk dk
6600
(c) cm² eku Kkr djks&
14
6600
(d) cm² O
18
 Mensuration-2D

(a) 60 cm2 (b) 72 cm2

2a a
(a) (b) (c) 144 cm2 (d) 90 cm2
3 2
58. Find the area of the quadrilaterl formed
a a by joining the mid-points of the sides of
(c) (d)
3 4 the quadrilateral of area 160 cm2.
55. Three circles of diameter 10 cm each, are 160 oxZ lseh {ks=kiQy ds fdlh prqHkqZt d
bound together by a rubber band, as shown ds eè; fcUnqvksa dks feykus ls cuh prqHkqZt
in the figure. The length of the rubber band
Kkr dhft,A
(in cm) if it is stretched as shows, is
(a) 60 cm2 (b) 80 cm2
izR;sd 10 lseh f=kT;k rhu o`Ùk fdlh jcj cSaM }kjk
(c) 72 cm2 (d) 100 cm2
,d lkFk c/sa gq, gSa] tSlk fd uhps nh xbZ vkÑfr
59. Two sides of a plot measuring 32m and
eas fn[kk;k x;k gS f[kps gq, jcj cSaM dh yEckbZ
24 m and the angle between them is a
(lseh esa) gSA perfect right angle. The other two sides

r
measure 25m each and the other three
angles are not right angles. The area of

si
the plot in m² is-
,d eSnku dh nks Hkqtk,¡ Øe'k% 32 eh] 24
an by
vkSj muds chp dk dks.k
90º gSA vkSj ckdh

n
Hkqtkvksa dk 25
ekueh gS] ijUrq ckdh rhu d
ledks.k ugha gS ml eSnku dk {ks=kiQYk K
ja
R s
(a) 786 (b) 534
a th

(c) 696.5 (d) 684

(a) 30 (b) 30 + 10 60. Find the area of a circle whose radius is
(c) 10 (d) 60 + 20 equal to the side of a square whose
perimeter is 196 m.
ty a

56. Three circles of radius 7 cm are kept

touching each other. The string is tightly ml o`Ùk dk {ks=kiQy Kkr djsa ftldh f=kT;k m
Hkqtk ds cjkcj gS ftldk ifjeki 196 ehVj gSA
di M

tied around these three circles. What is

the length of the string? SSC CHSL 20/03/2023 (Shift-03)
7 ls-eh f=kT;k okys rhu o`Ùkksa dks vkil esa Li'kZ (a) 7457 m2 (b) 7546 m2
djrs gq, j[kk x;k gSA bu rhu o`Ùkksa ds pkjksa vksj ,d
(c) 6477 m2 (d) 8844 m2
/kxk dldj ck¡/k x;k gSA /kxs dh yEckbZ D;k gS\(a) A circle of radius r is inscribed in the
square, then radius
SSC CGL 08/12/2022 (Shift- 01)
oxZ ds vanjr f=kT;k dk dksbZ o`Ùk cuk gks
(a) 42 + 7p cm (b) 21p + 14 cm
Side
(c) 42 + 14p cm (d) 7 + 14p cm r=
2
Quadrilateral A D
A

57. In a quadrilateral ABCD, AC = 12 cm. If

length of the perpendiculars drawn from
B and D to line AC are 5 cm and 7 cm, r
the area of the quadrilateral ABCD is :
fdlh prqHkqZt
ABCD esa]
AC = 12 lseh gSA ;fn
B rFkk
 Mensuration-2D

61. Each edge of the following square is 20

cm long, and a circle is inscribed in the A B
square as shown. What is the area of the
shaded region? (Use  = 3.14).
fuEu oxZ dh izR;sd Hkqtk dh yEckbZ
20 cm gS]
vkSj oxZ ds vanj ,d o`Ùk gS tSlk fd fp=k esa
fn[kk;k x;k gSA Nk;kafdr ('ksM fd, x,) {ks=k
C D
dk {ks=kiQy D;k gksxk\
[  = 3.14 ysa
]
(d) As we have discussed in previous slide for
20 cm
circles, now we can generalise for square
also.
Side of smaller square 1
=
Side of bigger square 2
A B

r
si
an by
SSC CHSL 08/06/2022 (Shift- 2)

n
(a) 88 cm2 ja (b) 85 cm2
(c) 86 cm2 (d) 84 cm2
R s
(b) A circle of radius R circumsceribed the
square, then C D
a th

R f=kT;k dk dksbZ o`Ùk oxZ ds ifjr% cuk gks] rks

(e) (i) Find the area of the largest square that
can be drawn inside a circle of radius R.
ml lcls cM+s oxZ dk {ks=kiQy Kkr djsa R
ty a

R f=kT;k ds o`Ùk ds vUnj cuk;k tk ldrk gS

di M

A B

a 
O
Diameter = Diagonal
 2R = 2a
C D
a
R= (ii) If one of the diagonal or the perimeter
2
become x times then the area will become
Side
 Radius = x2 times or increases by (x2 – 1) times.
A

2
(c) If side of square = a ;fn ,d fod.kZ ;k ifjeki x xquk gks tk, rks {k
a x2 xquk gks tkrk gS(x;k2 – 1) xquk c<+ tkrk g
Radius of incircle (r) = (iii) For two squares/nks oxks± ds fy,
2
a (a) Ratio of sides = Ratio of diagonal
Radius of circumcircle (R) = = Ratio of perimeter
2
(b) Ratio of area = (Ratio of sides)2
 Mensuration-2D

(iv) Side of square is given as 'a' ml lcls cM+s oxZ dk {ks=kiQy Kkr djsaR
oxZ dh Hkqtk
'a' f=kT;k ds prqFkZ o`Ùk•aM esa cuk;k tk ld
1
Area of square = R2
2

Q
r
C B

O A P
3a 2
Ungrazed area = 63. In the given figure, a square ABCD is
14
62. If radius of all the circles is 14 cm, find inscribed in a quadrant APCQ. If AB = 16
the area of the shaded region cm, find the area of the shaded region (take
 = 3.14) correct to two placed of decimal.
;fn lHkh o`Ùkksa dh f=kT;k,¡ 14 lseh gS rks Nk;kafdr

r
Hkkx dk {ks=kiQy Kkr djsaA nh xbZ vkÑfr esa prqFkkZa'k
APCQ eas ,d oxZABCD
mRdh£.kr gSA AB;fn = 16 cm gS] rks Nk;ka

si
an by {ks=k dk {ks=kiQy Kkr dhft,A
( = 3.14 ysa
) (n'keyo
ds nks LFkkuksa rd lgh mÙkj nhft,)
Q

n
ja D C
R s
(a) 168 cm² (b) 616 cm²
a th

(c) 156 cm² (d) 256 cm²

(v) Total area of three equilateral triangle
inscribed in a semicircle of radus 'r' cm
rhu leckgq f=kHkqtksa dk {ks=kiQy'r'tks lseh-
fdlh
ty a

A B P
f=kT;k ds v/Z o`Ùk esa cus gq, gSA
SSC CGL 02/12/2022 (Shift- 04)
di M

r (a) 155.98 cm² (b) 179.68 cm²

(c) 163.85 cm² (d) 145.92 cm²
r
r r (viii) A square is drawn inside a quadrant of
radius r cm in such a way that 2 of its
r r
vertices are on the radii of the quadrant
3 3
Area = r2 and they are at equal distance from the
4 centre of circle and remaining 2 vertices
(vi) Find the area of the largest square that can
are on the arc of the quadrant. Find the
be drawn inside a semi-circle of radius R.
side of square in term of r.
ml lcls cM+s oxZ dk {ks=kiQy Kkr djsa R ftls
f=kT;k ds v/Z&o`Ùk ds vUnj cuk;k tk ldrk gSA r f=kT;k ds ,d prqFkkZa'k o`Ùk•aM e
A

4
bl çdkj •hapk tkrk gS fd blds 2 'kh"kZ prq
R2
Area of square =
5 dh ifjf/ ij gSa vkSj os o`Ùk ds dsaæ ls
nwjh ij gSa vkSj 'ks"k nks 'kh"kZ prqF
ij gSaA oxZ dh Hkqtk
r ds inksa esa Kkr dhft
 Mensuration-2D

Q XBD ,d o`Ùk dk prqFkkZa'k gS]XB tgka

= 20
C lseh] XA = AB = XC = CD, XA , AB , XC vkS
CD dks O;kl ekudj pkj v/Zo`Ùk •haps x, gSaA N
Hkkx dk {ks=kiQy Kkr djsaA

D B

P
O A

(ix). r  ( 2 – 1)R

125 125
(a) – 25 (b) + 25
2 2

r
175 175
– 25 + 25

si
(c) (d)
2 2
66. If side of square = 14 cm, find the area
an by of shaded region.
How to Calculate Area of Leaf ;fn oxZ dh Hkqtk = 14 lseh gks] rks Nk;k

n
 Let side of square be ‘a’.
ja Hkkx dk {ks=kiQy Kkr djsaA

3  = 4 a2
R s
Area of leaf = a – 2
2 2
a 
14 7
a th
ty a
di M

(a) 98 cm² (b) 84 cm²

(c) 70 cm² (d) None of these
64. ABCD is a square whose side is 14 cm, find 67. An equilateral triangle is made on the
the area of the shaded region. diagonal of a square. Then find the ratio
ABCD ,d oxZ gS ftldh Hkqtk 14 lseh gS] of their areas.
Nk;kafdr Hkkx dk {ks=kiQy Kkr djsaA ,d oxZ ds fod.kZ ij leckgq f=kHkqt cuk;k
gSA rks muds {ks=kiQy dk vuqikr gksxk&
(a) 3 : 2 (b) 2 : 3
(c) 2 : 3 (d) 1 : 2
68. Four equal sized maximum circular plates
are cut from a square paper sheet of area
A

784 sq. cm. The circumference of each

plate is-
(a) 112 cm² (b) 126 cm² ,d oxkZdj dkxt dh 'khV ftldk {ks=kiQYk
(c) 140 cm² (d) 84 cm² lseh2 gSA mlls ,d leku vf/dre vkdkj dh pkj
65. XBD is quadrant of a circle where, XB = o`Ùkh; IysVsa dkVh tkrh gSA izR;sd o`Ù
20 cm, XA = AB = XC = CD. Four semi-
ifjfèk Kkr djks&
 Mensuration-2D

69. A circle is inscribed in a square whose Then,

diagonal is 122 cm. An equilateral triangle A b D
is inscribed in that circle. The length of
the side of the triangle is- a d a
,d oxZ ftldk fod.kZ 122 lseh gS ds vUnj
,d o`Ùk cuk gSA ml o`Ùk eas ,d leckgq f=kHkqt B
b
C
cuk gqvk gSA rks f=kHkqt dh Hkqtk dh yEckbZ gksxh&
Area of parallelogram ABCD
(a) 43 cm (b) 83 cm
= 2 s (s  a )(s  b )(s  d )
(c) 63 cm (d) 113 cm
 When the opposite sides are parallel, the a b d
where s =
quadrilateral is called a parallelogram. 2
70. Side AB = 24 of a parallelogram ABCD is
;fn foijhr Hkqtk lekukarj gks] rks prqHkqZt] lekarj
24cm and side AD = 16 cm. The distance
prqHkqZt dgykrk gSA between AB and CD is 10 cm, then find
In Parallelogram ABCD, let side AB = a cm the distance between AD and BC.

r
and BC = b cm, then ,d lekUrj prqHkqZt ABCD dh HkqtkAB = 24

si
lekarj prqHkqZt
ABCD esa] ;fnAB = a lseh vkSj lseh rFkk Hkqtk
AD = 16 lseh gSA
AB rFkkCD ds
BC = b lseh gS] rks eèk dh nwjh10 lseh gS rc AD rFkkBC ds eè;
an by
b
dh nwjh Kkr djks&
(a) 16 cm (b) 18 cm

n
h2 (c) 15 cm (d) 26 cm
a h1 a  If each side of a parallelogram is equal,
ja
R s
the parallelogram is called a rhombus.
;fn fdlh lekarj prqHkqZt dh izR;sd Hkqtk
a th

b
gks] rks lekarj prqHkqZt leprqHkqZt dgykr
(i) AB = CD and BC = AD In rhombus ABCD,
(ii) Each diagonal AC or BD divides the A D
ty a

parallelogram in the conguent triangles. d1

çR;sd fod.kZAC vFkok BD lekarj prqHkqZt dks d2
di M

lok±xle f=kHkqtksa esa foHkkftr djrk gSA O

(iii) AC2 + BD2 = AB2 + BC2 + CD2 + AD2
= 2(AB2 + BC2) = 2(a2 + b2) B C
(iv) Perimeter = 2(a + b) a
(v) (a) Area = Base × Height Let the side BC = a, AC = d1 and BD = d2,
= a(side) × (distance between the side then
(i) AB= BC = CD = DA = a
and its parallel side)
(ii) Diagonals bisect each other at right angle.
= a × h1 = b × h2
fod.kZ ,d&nwljs dks ledks.k ij izfrPNsfnr djr
(b) Area of all triangle of same base and
between the parallel lines are the same. 1
(iii) Side = a = d12  d 22
A

2
leku vk/kj vkSj lekarj Hkqtkvksa ds eè; cus
izR;sd f=kHkqt dk {ks=kiQy leku gksrk gSA or, 4a2 = d12  d 22
(c) The length of one diagonal is d. (iv) Perimeter = 4a

b 1
A D (v) (a) Area = × d1 × d2
2
a a 1
 Mensuration-2D

(b) Area = Base × Height 75. Perimeter of a rhombus is 2p unit and sum
Area of AOB = Area of BOC = Area of the lengths of diagonals is m unit, then
of COD = Area of AOD the area of the rhombus is-
,d leprqHkZqt dh ifjeki2p ek=kd gSa] vkSj fo
A D dk ;ksx m ek=kd gSA mldk {ks=kiQy Kkr d
1 2
(a) m p sp. unit
4
O 1
mp2 sp. unit
(b)
4
1
B C (b)
4
m2 – p2  sp. unit
1 2
71. The sum of the squares of the sides of a (d)
4
p – m2  sp. unit
rhombus is 1600 cm². What is the side of
 If each angle of a quadrilateral is 90º and
the rhombus? length of the opposite sides are equal, then

r
,d leprqHkqZt dh Hkqtkvksa ds oxks±1600cm2
dk ;ksx it is called a rectangle.
gSA ml leprqHkqZt dh Hkqtk dk eki D;k gksxk\ ;fn fdlh prqHkqZt dk izR;sd dks.k90º gks vk

si
SSC CHSL 10/08/2021 (Shift- 02)
an by foijhr Hkqtkvksa dh yackbZ cjkcj gks rks
(a) 25 cm (b) 15 cm dgykrk gSA
(c) 20 cm (d) 10 cm  Let ABCD is a rectangle such that AB = a

n
72. Length of each side of a rhombus is 13 and BC = b, then
cm and one of the diagonal is 24 cm. What
ja ;fn ,d vk;r ABCD bl izdkj gS fd AB = a
vkSjBC = b gS] rks
R s
is the area (in cm²) of the rhombus?
A D
,d leprqHkqZt dh izR;sd Hkqtk dh yackbZ 13 lseh
a th

rFkk blds ,d fod.kZ dh yackbZ 24 lseh gSA leprqHkqZt O

dk {ks=kiQy (lseh2 esa) Kkr djsaA a
ty a

SSC CGL 13/08/2021 (Shift- 01)

(a) 240 (b) 60 B b C
di M

(c) 300 (d) 120 (i) AB = CD = a and BC = AD = b

73. The length of one side of a rhombus is 6.5 (ii) The diagonals bisect each other,
cm and its altitude is 10 cm. If the length (a) AC = BD = a 2  b2
of its one diagonal be 26 cm, the length
of the other diagonal will be- a2  b2
(b) AO = OC = OB =
,d leprqHkqZt dh Hkqtk dh yEckbZ 6-5 lseh (iii)
rFkkPerimeter = 2(length + breadth)
2
yEc 10 lseh gSA ;fn blds ,d fod.kZ dh yEckbZ = 2(a + b)
(iv) Area = Length × Breadth = ab
26 lseh gks rks blds nwljs fod.kZ dh yEckbZ gksxh&
76. The area of a square is 1156 cm2. What is
(a) 5 cm (b) 10 cm the respective ratio between the length and
(c) 6.5 cm (d) 26 cm the breadth of a rectangle whose length is
74. The measure of each of two opposite angles twice the side of the square and whose
A

of a rhombus is 60º and the measure of breadth is 14 cm less than the side of the
square?
one of its sides is 10 cm. The length of
its smaller diagonal is: ,d oxZ dk {ks=kiQy 1156 oxZ lseh gSA ml v
,d leprqHkqZt dh izR;sd Hkqtk dh yEckbZ 10 lserh yackbZ vkSj pkSM+kbZ ds chp Øe'k% vuqikr
rFkk ,d dks.k 60º gS rks blds NksVs fod.kZ dh yackbZ oxZ dh Hkqtk ls nksxuh gS vkSj ftldh
yEckbZ gkxh& dh Hkqtk ls 14 lseh- de gS\
(a) 10 cm (b) 103 cm SSC CHSL 20/03/2023 (Shift-04)
 Mensuration-2D

77. The breadth of a rectangular floor is 3/5 of (b) Radius of the circle circuscribed the
its length. If the area of the floor is 60 rectangle ABCD be R, then
metre2 then what is the difference between vk;r ABCD ds ifjr% cus o`Ùk dh f=kT R
the length and breadth of the floor?
gks] rks
,d vk;rkdkj iQ'kZ dh pkSM+kbZ mldh yackbZ dh 3@5 gSA
;fn iQ'kZ dk {ks=kiQy 60 oxZ ehVj gS] rks iQ'kZ dh yackbZ a2  b2
R=
vkSj pkSM+kbZ esa fdruk varj gS\ 2
SSC CHSL 15/03/2023 (Shift-02)
(a) 6 m (b) 7.5 m A D
(c) 5 m (d) 4 m O
78. The area of a triangular park with sides 88 m, a
165 m, and 187 m is equal to the area of a
b
rectangular plot whose sides are in the ratio 5 B C
: 3. What is the perimeter (in m) of the plot?
88 ehVj] 165 ehVj vkSj187 ehVj Hkqtkvksa okys ,d
(vii) If the length of the rectangle will become
f=kdks.kh; ikdZ dk {ks=kiQy ,d vk;rkdkj Hkw•aM ds

r
x times and breadth will become y times,
{ks=kiQy ds cjkcj gS ftldh Hkqtkvksa dk5vuqikr
:3 the area of the rectangle will become xy

si
gSA Hkw•aM dh ifjf/ (ehVj esa) D;k gS\ times.
ICAR Mains, 07/07/2023 (Shift-1) ;fn vk;r dh yackbZx xquh vkSj pkZM+kbZ
y xquh
(a) 352
an by (b) 384 tk,] rks vk;r dk {ks=kiQy
xy xquk gks tk,xkA
(c) 400 (d) 320

n
(viii) For the rectangle,
79. The sides of a rectangular garden are 176 Ratio of area = (ratio of length) : (ratio of
m and 56 m. Its area is equal to the area
ja
of a circular field. What will be the cost breadth)
R s
(in `) of fencing the circular field at the (ix) Path around or in a rectangle/ vk;r ds ifjr%
22
vFkok vk;r ds vanj iFk
a th

rate of ` 35 per m? (Use  = )

7
,d vk;rkdkj ckx dh Hkqtk,¡176 ehVj vkSj56 ehVj (a) Area of the path of uniform width d all
gSA bldk {ks=kiQy ,d o`Ùkkdkj eSnku ds {ks=kiQyaroundds outside the rectangle ABCD
vk;r ABCD ds ifjr% ,d leku pkSM+kbZd oky
ty a

cjkcj gSA o`Ùkkdkj eSnku ` 35esa

izfr ehVj dh nj ls
22 jkLrs dk {ks=kiQy
= 2d(l + b + 2d)
ckM+ yxkus dh ykxr ` esa)
( fdruh gksxh\ ( = dk
di M

7
P S
mi;ksx dhft,A)
A d D
ICAR Mains, 07/07/2023 (Shift-2)
(a) 12,100 (b) 11,000 d b d
(c) 13,320 (d) 12,210
l
(v) Area of AOB = Area of BOC
= Area of COD B d C
Q R
ab
= Area of DOA = (b) Area of the path of uniform width 'd' all
4
around inside the rectangle ABCD
(vi) (a) Radius of the maximum possible circle
vf/dre laHkkfor o`Ùk dh f=kT;k vk;r ABCD ds vanj ,d leku pkSM+kbZ
d oky
A

Breadth b jkLrs dk {ks=kiQy

= 2d(l + b – 2d)
= =
2 2 A D
A D d
P S
b d d
b
Q R
 Mensuration-2D

(c) Area of the path of uniform width d along 84. What is the maximum area of a rectangle,
the length and the breadth the perimeter of which is 18 cm?
yackbZ vkSj pkSM+kbZ ds vuqfn'k ,dleku
d pkSM+kbZ
ml vk;r dk vf/dre {ks=kiQy D;k gksxk ftld
ds iFk dk {ks=kiQy
= (l + b – d)d ifjeki 18 lseh gS\
A C
(a) 20.25 cm² (b) 20.00 cm²
(c) 19.75 cm² (d) 19.60 cm²
l
85. A rectangular carpet has an area of 120
m2 and a perimeter of 46 metre. The length
of its diagonal is
B D
b ,d vk;rkdkj dkjisV dk {ks=kiQy120 m² rFk
80. There is a rectangular garden of 240 metres
× 80 metres. A path of width 4 metre is
ifjeki 46 m gS blds fod.kZ dh yEckbZ gks
build outside the garden along its four sides. (a) 17 meter (b) 21 meter
What is the area of the path? (c) 13 meter (d) 23 meter
240 ehVj × 80 ehVj dk ,d vk;rkdkj cxhpk gSA

r
86. A path of uniform width runs round the
cxhps ds ckgj pkjksa vksj 4 ehVj pkSM+k ,d iFk cuk;kinside of a rectangular field 38 m long and

si
x;k gSA bl iFk dk {ks=kiQy D;k gS\ 32 m wide, If the path occupies 600 m2,
SSC CHSL 10/03/2023 (Shift-04)
an by then the width of the path is-
(a) 2826 m2 (b) 2542 m2
(c) 2916 m 2
(d) 2624 m2 ,d vk;rkdkj eSnku ftldh yEckbZ 38 eh rF

n
81. The length and breadth of rectangular field pkSM+kbZ 32 eh gS ds vUnj ,d leku pkSM
are in the ratio 7 : 4. A path 4 m wide
ja jkLrk gSA ;fn jkLrs dk {ks=kiQy2 600
gS rks
eh jk
running all around outside has an area of
dh pkSM+kkbZ gksxh&
R s
416 m2. The breadth (in m) of the field is :
vk;rkdkj •sr dh yackbZ vkSj pkSM+kbZ dk vuqikr
(a) 30 m (b) 5 m
a th

(c) 18.75 m
7 % 4 gSA 4 ehVj pkSM+k ,d iFk tks ckgj pkjksa (d) 10 m
87.eSnku
vksj cuk gS] dk {ks=kiQy 416 oxZ ehVj gSA A street of width 10 metres surrounds from
outside a rectangular garden whose
dh pkSM+kbZ (ehVj esa) gS %
ty a

measurement is 200 m × 180 m. The area

(a) 28 (b) 14
of the path (in square metres) is.
di M

(c) 15 (d) 16
82. A path of uniform width runs round the ,d vk;rkdkj eSnku ftldh eki 200 eh × 180
inside of rectangular field 38m long and eh gS ds ckgj pkjks vksj 10 eh pkSM+kbZ
32m wide. If the path occupies 600 m2, gsA ml xyh dk {ks=kiQy Kkr djksA
then the width of the path is :
(a) 8000 (b) 7000
38 ehVj yacs vkSj 32 ehVj pkSM+s vk;rkdkj eSnku
(c) 7500 (d) 8200
ds vanj ,d leku pkSM+kbZ dk iFk cuk gSA ;fn iFk
88. A took 15 sec. to cross a rectangular field
600 ehVj txg ?ksjrk gS] rks iFk dh pkSM+kbZ gS%diagonally walking at the ratio of
(a) 30 m (b) 5 cm 52 m/min and B took the same time to
(c) 18.75 m (d) 10 m cross the same field along its sides walking
83. A rectangular area of 6 sq. m is to be at the rate of 68 m/min. The area of the
A

painted on a 3m × 4m board leaving of

field is:
uniform width on all sides. What should
be the width of the border? ,d O;fDr A 52 eh@feuV dh xfr ls ,d vk;rkdkj
3 ehVj × 4 ehVj cksMZ ij lHkh rjiQ ,d leku eSnku dks fod.kkZorZ 15 lsd.M esa ikj d
vkSj
pkSM+kbZ NksM+dj 6 oxZ ehVj ds ,d vk;rkdkj {ks=k dksB 68 eh@feuV ls Hkqtkvksa ds vuqfn'k
leku
isaV fd;k tkuk gSA ckWMZj dh pkSM+kbZ fdruh gksuh le; esa ikj djrk gSA eSnku dk {ks=kiQy
pkfg,\
 Mensuration-2D

89. There is a rectangular tank of length 180 92. A rectangular park is 60 m long and 40
m and breadth 120 m in a circular field, m wide. There are two paths in the middle
If the area of the land portion of the field of the plot parallel to its sides. The width
is 40000 m2, what is the radius of the field? of path is 4 meter. These paths cuts to
each of the at right angle. Then find the
 22  cost of cementing the path at the rate of
 Take  = 
 7  7.50 Rs./m²?

,d vk;rkdkj VSad ftldh yEckbZ 180 eh rFkk ,d vk;rkdkj eSnku 60 eh yEck rFkk 40 eh pkSM
eSnku
pkSM+kbZ 120 eh gS] ,d o`Ùkkdkj eSnku eas fLFkfr gSAds chp eas yEckbZ rFkk pkSM+kbZ ds
;fn eSnku ds tehuh Hkkx dk {ks=kiQy 40000 eh gSa tks ,d&nwljs dks dsUnz ij ledks.k ij dk
2

jkLrs dh pkSM+kbZ 4 eh gS rks 7-50 2

dh#Ik;s@
nj ls
 22 
gS rks eSnku dh f=kT;k gksxh\
  = yhft,  jkLr dks lhesaV djokus dh dqy ykxr gksxh&
 7 
(a) Rs.25780 (b) Rs.2880
(a) 130 m (b) 135 m
(c) Rs.2650 (d) Rs.2000

r
(c) 140 m (d) 145 m
93. A playground is in the shape of rectangle.
90. The area of a square shaped field is 1764 A sum of Rs. 1000 was spent to make the

si
m². The breadth of a rectangular park is
an by ground usable at the rate of 25 paise per
1/6th of the side of the square field and sq. m. The breadth of the ground is 50 m.
the length is four times its breadth. What If the length of the ground is increased

n
is the cost (in Rs) of levelling the park at by 20 m. What will be the expenditure (in
rupees) at the same rate per sq. m?
Rs 30 per m2? ja ,d [sky dk eSnku vk;rkdkj gSA ml eSnk
,d oxkZdkj vkdkj okys eSnku dk {ks=kiQy 1764
R s
Hkwfe dks bLrseky yk;d cukus ds fy, 25 eS
ehVj2 gSA ,d vk;rkdkj ikdZ dh pkSM+kbZ] oxkZdkj
a th

oxZ eh dh nj ij 1000 #i;s [kpZ fd, x,A ml

eSnku dh Hkqtk dh 1@6 gS vkSj mldh yackbZ] bldh eSnku dh pkSM+kbZ 50 eh gSA ;fn eSnk
pkSM+kbZ dk pkj xquk gSA 30 :i;s izfr 2
dhehVj 20 eh c<+k nh tk, rks izfr oxZ eh dh nj
nj ls ikdZ dks lery djus dh ykxr (:i;s esa)
ty a

dqy fdrus #i;s [kpZ gksxsa\

Kkr djsaA (a) 1250 (b) 1000
di M

SSC CGL 23/08/2021 (Shift 01) (c) 1500 (d) 2250

(a) 5880 (b) 4768 94. A trapezium plate having two parallel sides
of length 15 cm and 9 cm. and distance
(c) 2940 (d) 6342 berween them is 6 cm, copper plating is to
91. The width of the path around a square field be done on the plate at a rate of ` 3 per
is 4.5 m and its area is 105.75 m². Find square cm. What will be the total cost of
copper plating on the upper side of the
the cost of fencing the path at the rate
plate?
of 100 per metre.
,d leyac IysV dh nks lekukarj Hkqtkvksa d ya
,d oXkkZdkj eSnku ds pkjksa vksj jkLrs dh pkSM+kbZ
lseh vkSj 9 lseh gS vkSj muds chp dh nwjh 6
4-5m gS vkSj bldk {ks=kiQy 105-75
m2 gSA
` 100 ml IysV ij ` 3 izfr oxZ ehVj ds nj ij dkWij Iys
A

izfr ehVj dh nj ls jkLrs ij ckM+ yXkkus dk O;; dh tkrh gSA IysV ds Åijh Hkkx ij dkWij Iys¯
Kkr dhft,A dh dqy ykxr fdruh gksxh\
SSC CGL 11/04/2022 (Shift- 02) SSC CHSL 21/03/2023 (Shift-01)

(a) Rs. 550 (b) Rs. 600 (a) ` 432 (b) ` 216
(c) ` 72 (d) ` 108
(c) Rs. 275 (d) Rs. 400
 Mensuration-2D

ANSWER KEY
1.(c) 2.(b) 3.(c) 4.(a) 5.(a) 6.(c) 7.(c) 8.(a) 9.(c) 10.(a)

11.(d) 12.(b) 13.(b) 14.(d) 15.(c) 16.(c) 17.(d) 18.(d) 19.(a) 20.(b)

21.(b) 22.(b) 23.(a) 24.(c) 25.(d) 26.(b) 27.(b) 28.(d) 29.(a) 30.(c)

31.(a) 32.(a) 33.(d) 34.(c) 35.(a) 36.(a) 37.(a) 38.(d) 39.(d) 40.(c)

41.(d) 42.(d) 43.(d) 44.(a) 45.(a) 46.(a) 47.(b) 48.(c) 49.(d) 50.(d)

51.(a) 52.(c) 53.(c) 54.(c) 55.(b) 56.(c) 57.(b) 58.(b) 59.(d) 60.(d)

61.(c) 62.(a) 63.(a) 64.(a) 65.(b) 66.(a) 67.(a) 68.(b) 69.(c) 70.(c)

71.(a) 72.(d) 73.(a) 74.(a) 75.(b) 76.(c) 77.(d) 78.(a) 79.(c) 60.(d)

r
81.(d) 82.(b) 83.(b) 84.(a) 85.(a) 86.(b) 87.(a) 88.(d) 89.(c) 90.(a)

si
91.(a) 92.(b) 93.(a) 94.(b)
an by
n
ja
R s
a th
ty a
di M
A
 4. The area of a field in the shape of a triangle
vkSj 97 m gSa] ,d vk;rkdkj ikdZ ds {ks=kiQy ds cjkcj gS with each side x metre is equal to the area of
ftldh Hkqtk,sa 5%13 ds vuqikr esa gSaA vk;rkdkj ikdZ dk another triangular field having sides 50m, 70m
ifjeki (ehVj esa) D;k gSa\ and 80m. The value of x is closest to:

r
(a) 108 (b) 180 ,d f=kHkqt ds vkdkj ds ,d [ksr dk {ks=kiQy ftldh izR;sd
Hkqtkx ehVj gS] ,d vU; f=kHkqtkdkj [ksr ds {ks=kiQy ds

si
(c) 216 (d) 144
Sol: (c) cjkcj gS] ftldh Hkqtk,¡ 50
m] 70m vkSj 80m gSaA
x dk eku
fudVre gS%

a n by
As the given triangle is right angle
(a) 65.5 (b) 63.2
1 (c) 62.4 (d) 61.8

n
Area of triangular field = × 65 × 72 Sol: (b)
2
ATQ,
= 2340 Sides of a triangle = 50m, 70m, 80m

ja
R s
Area of triangular field = Area of rectangular (50  70  80)
park Semi perimeter(s) = = 100
2

x2 =
2340
65
a th
2340 = 5x × 13x Area of  by heron's formula
Area = s(s – a)(s – b)(s – c)
= 100(100 – 50) (100 – 70) (100 – 80)
ty a
x=6 = 1000 3
 length = 13 × 6 = 78 and breadth = 5 × 6 = 30 Now,
di M

 Perimeter of the rectangular park 3 2

x = 1000 3
= 2(78 + 30) = 216 cm 4
2
2. If each side of a rectangle is inreased by 22%, x = 4000
then its area will increase by: x = 63.2
5. The perimeter of a square is equal to the pe-
;fn ,d vk;r dh izR;sd Hkqtk eas 22» dh o`f¼ dh tkrh gS] rimeter of a rectangle of length 16cm and
rks mlds {ks=kiQy esa fdruh o`f¼ gksxh\ breadth 14cm. Find the circumference of a
(a) 40% (b) 50% semicircle whose diameter is equal to the side
of the square.
(c) 46.65% (d) 48.84%
,d oxZ dk ifjeki 16 lseh yackbZ vkSj 14 lseh pkSM+kbZ oky
Sol: (d) vk;r ds ifjeki ds cjkcj gSaA ,d v/Zo`r dh ifjf/ Kkr
Each side is increased by 22% dhft, ftldk O;kl oxZ dh Hkqtk ds cjkcj gSaA
A

(a) 38.57 cm (b) 21.57 cm

22  22
Overall increase in area = 22 + 22 + (c) 23.57 cm (d) 25.57 cm
100 Sol: (a)
= 48.84% 4 × side = 2(16 + 14)
3. If each side of a rectangle is decreased by 11%, side = 15
then its area will decrease by: circumference of semicircle = r + d
;fn ,d vk;r dh izR;sd Hkqtk esa 11» dh deh dh tkrh gS] 22 15
rks mlds {ks=kiQy esa fdruh deh gksxh\ =
7

2
+ 15
(a) 21.69% (b) 20.79%
(c) 13.13% (d) 26.78% = 38.57 cm

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 [a = side] 1 1 1
Ratio of sides of triangle = : :
3 5 4
3 = 20 : 12 : 15
 (8.5)2  18.1 3
Perimeter = 141

r
4
7. Two sides of triangle are 12.8 m and 9.6 m. If 47 uinit  141

si
the height of the triangle is 12m, correspond- Difference b/w greatest and smallest side
ing to 9.6 m. Then what is its height (in m) cor- = 20 – 12 = 8

a n by
responding to 12.8 m? 141
,d f=kHkqt dh nks Hkqtk,sa m vkSj
12-8 9-6
m gSA ;fn f=kHkqt 1 unit =
47
dh ÅapkbZm 12gS] tks fd 9-6
m okys Hkqtk ds laxr gS] rks 12-

n
141
8m okys Hkqtk ds laxr ÅapkbZ (esa) D;k gksxh\ 8 unit = × 8 = 24 cm
47

ja
(a) 12 (b) 9

R s
10. If the angles of a triangle are 30° amd 45° and
(c) 10 (d) 8
Sol: (b) the included side is  
108 + 6 cm, then what
We know,

Area =
1
2
a th
× base × height
is the area of the triangle?
;fn fdlh f=kHkqt ds dks.k
30° vkSj 45° gSa vkSj mldh
lfEefyr Hkqtk 108 + 6  lseh gS] rks f=kHkqt dk {ks=kiQy
ty a
ATQ,
12 × 9.6 = 12.8 × H D;k gS\
di M

H=9
Height corresponding to 12.8 = 9
(a) 18  
3 +1 cm2 (b) 15 3 +18 cm 2

8. The base of a triangle is increased by 40%. By (c) 12  3 +1 cm 2

(d) 24  3 +1 cm 2

what percentage (correct to two decimal places)

should its height be increased so that the area Sol: (a)
increases by 60% A
,d f=kHkqt ds vk/kj esa 40» dh o`f¼ dh xbZA bldh ÅapkbZ
fdrus izfr'kr (n'keyo ds nks LFkkuksa rd lgh) c<+kbZ tkuh
pkkfg, rkfd {ks=kiQy esa 60» dh o`f¼ gks tk,\ 1
(a) 14.29% (b) 20.01%
(c) 15.54% (d) 18.62% 30 45
Sol: (a) B D 1 C
3
A

Area = 5 : 8 108 + 6
Base = 5 : 7
Now,  
3  1 unit  108  6
Height = 7 : 8 1 unit = 6

1
1  Area of triangle =
2
 
3  1  1  (6)2

1
 increase in height =
7
= 14.29% = 18  
3  1 cm2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 2
Sol: (d) three interior perpendiculars.
By value putting 2S
Let, a = b = 1 a=

r
3
Then, sides of triangle = 2, 5, 5

si
2
3 3  2S  S2
Area =  a2    
4 4  3 3

a n by
5 5
14. In the fig. given below ABC is a right-angled
triangle where A=90, AB=p cm and AC=q cm.

n
on the three sides as diameters semicircles are
drawn as shown in the fig. The area of the
1 1

ja
shaded portion, in sq.cm is

R s
2 2
fn, x, fp=k esaABC ,d ledks.k f=kHkqt gS tgka
A = 90, AB
= p cm vkSjAC = q cm rhuks Hkqtkvksa dks fod.kZ ij v/Zo`r

Then,
a th
Height of triangle = 5–
1
2

3
2 f•apk tkrk gS Nk;kafdr Hkkx dk {ks=kiQy Kkr djsa\
A
ty a
1 3
Area of triangle = 2  2 
2
di M

3 3
= or ab
2 2
12. The area of an equilateral triangle is 10.24 3 B C
m2. Its perimeter (in m) is:
1
fdlh leckgq f=kHkqt dk {ks=kiQy
10.24 3 m2 gSA bldk (a) 1 (b) pq
2
ifjeki (ehVj esa) Kkr djsaA
(a) 3.2 (b) 9.6 2 5
(c) pq (d)
(c) 6.4 (d) 19.2 3 3
Sol: (d) Sol: (b)
Area of the equilateral triangle = 10.24 3 m3 In this case area of triangle is equal to area of
shaded region.
A

3 2 1
a  10.24 3 Area of shaded portion = × AB × AC
4 2
a = 6.4 1
Perimeter = 3 × 6.4 = 19.2 = pq cm2
2
13. From an interior point of an equilateral tri-
15. In a triangle ABC, AB = AC and the perimeter of
angle, perpendiculars are drawn on all three
sides. The sum of the lengths of the three per- 5
pendicular is s. Then the area of the triangle triangle 544 cm, If equal sides are of the
6
is non- equal side, then find the area of triangle?

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 = 6 × 73 × 73 = 31974 cm2
B C 18. The area of a square shaped field is 1764 m².
3 D 3
AD divides BC in two equal half i.e 3 cm each 1

r
The breadth of a rectangular park is th of
(3, 4, 5 triplet) 6
the side of the square field and the length is

si
AD = 4
four times its breadth. What is the cost (in ') of
BD = 3 levelling the park at 30 per m²?

a n by
1 ,d oxkZdkj vkdkj okys eSnku dk {ks=kiQym²
1764
gSA ,d
So, area of ABC = 2 × × 3 × 4 = 12 unit 1
2 vk;rkdkj ikdZ dh pkSM+kbZ] oxkZdkj eSnku dh gS
Hkqtk dk

n
6
16 uinit  544
vkSj mldh yackbZ] bldh pkSM+kbZ dk pkj xqukm²
gSA 30 çfr

ja
1 unit  34 dh nj ls ikdZ dks lery djus dh ykxr ( esa) Kkr djsaA

R s
 Area of triangle = 12 × 342 (a) 5880 (b) 4768

16. a th
= 13872 cm2
The sum of three sides of an isosceles triangle Sol:
is 20 cm, and the ratio of equal sides to the
base is 3 : 4. The altitude of the triangle is :
(c) 2940
(a)
(d) 6342

Area of square field = 1764

side = 42
ty a
lef}ckgq f=kHkqt dh rhuksa Hkqtkvksa dkcm;ksxiQy
gS vkSj
20 Breadth of rectangle = 7, length of rectangle = 28
cjkcj yackbZ okyh Hkqtkvksa esa ls ,d Hkqtk vkSj vk/kj Area
dk of rectangle = 7 × 28
di M

Cost of levelling the park = 7 × 28 × 30 = 5880

vuqikr 3 % 4 gSA f=kHkqt dh ÅapkbZ Kkr djsa A
19. The cost of tilling the floor of a rectangular
(a) 3 3 cm (b) 3 5 cm room is 9100 at 65 per m2. The ratio of the
length and breadth of the floor is 7 : 5. The
(c) 4 5 cm (d) 2 5 cm perimeter (in m) of the floor of the room is :
Sol: (d) 65 çfr m2 dh nj ls] fdlh vk;rkdkj dejs dh iQ'kZ ij
Three sides of triangle are in ratio = 3 : 3 : 4 VkbYl yxkus dh ykxr 9]100 gSA dejs dh yackbZ vkSj pkSM+
dk vuqikr 7 % 5 gSA dejs ds iQ'kZ dk ifjeki (ehVj esa) Kkr
A djsaA
(a) 48 (b) 24
6 6 (c) 36 (d) 28.8
Sol: (a)
B C 9100
A

D 8
Area of floor = = 140
65
As Perimeter = 20 cm Ratio of length and breadth = 7 : 5
So, sides are 6, 6, 8 ATQ
10  20 7x × 5x = 140
12
140
x2 =
AD = (6)2 – (4)2 35
x=2
= 20  2 5 cm
Perimeter of floor of room = 2 (14 + 10) = 48

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs4

  L = 12 & B = 9 L – B = 40
Diagonal = 2 2 = 15  L = 90 and B = 50
L B
ATQ,
21. If the perimeter of a rectangle is 25 cm and

r
Area of rectangle = 90 × 50
the area is 25 cm², then its length is: Now, area of square
;fn ,d vk;r dk ifjeki 25 lseh gS vkSj {ks=kiQy 25 lseh

si
side = 90  50
gS] rks bldh yackbZ gS%

a n by
(a) 6.25 cm (b) 12.5 cm side of square = 30 5 m
(c) 7.5 cm (d) 10 cm 24. If length of a rectangle is decreased by 6 cm,

n
Sol: (d) we get a square and the area of the square so
formed is 252 square cm less than the area of
ATQ the square formed, when breadth of the origi-

ja
Perimeter 2(L + B) = 25 nal rectangle is increased by 6 cm. find the

R s
perimeter of the rectangle?
25
;fn ,d vk;r dh yackbZ esa 6 lseh dh deh dh tkrh gS] rks
(L + B) =

L × B = 25
Now,
a th
2
gesa ,d oxZ feyrk gS vkSj bl çdkj cus oxZ dk {ks=kiQy ml
oxZ ds {ks=kiQy ls 252 oxZ lseh de gksrk gS] tc ewy vk;r
dh pkSM+kbZ esa 6 lseh dh o`f¼ dh tkrh gSA vk;r dk ifjeki
ty a
(L – B)2 = (L + B)2 – 4LB Kkr dhft,\
(a) 42 cm (b) 88 cm
di M

15 (c) 80 cm (d) 84 cm
L–B=
2 Sol: (d)
 Length = 10 Let area of triangle = l b
22. The length of a rectangle is 4m more than side ATQ,
of a square and the breadth of the rectangle is l × (b + 6) – b(l – 6) = 252
4 m less than the side of the same square. If l b × 6 l – l b + 6b = 252
the area of this square is 576 sq.m, what is the 2 (l + b) = 84 cm
area of the rectangle? 25. A rectangular plot, 55m long and 45m broad,
fdlh vk;r dh yackbZ ,d oxZ ds Hkqtk lsm4vf/d gS vkSj has two concrete crossroads (of equal width)
pkSM+kbZ oxZ dh Hkqtkm de gSA
ls 4 ;fn oxZ dk {ks=kiQy
576 running in the middle of it one parallel to the
sq.m gSA rc vk;r dk {ks=kiQy D;k gksxk\ length and the other parallel to the breadth.
The rest of the plot is used as a lawn. If the
(a) 560 (b) 545
A

area of the lawn is 1911 m2, what is the width

(c) 557 (d) 551 of each of the crossroads (in m ) ?
Sol: (a) ,d vk;krdkj ikdZ dh eki 55 ehVj yack vkSj 45 ehVj
Side of square = 576 PkkSM+k gSA ikdZ ds chp ls gksdj nks jkLrs tkrs gS tks ikd
Side = 24 yackbZ vkSj pkSM+kbZ ds lekUrj gSA jkLrs ds vykok 'ks"k
Length of rectangle = 28, breadth of rectangle ykWu gSA ;fn ykWu dk {ks=kiQy21911 gSA ehVj
rc jkLrs dh
= 20 pkSM+kbZ D;k gksxh\
(a) 5 (b) 5.5
 Area of reactangle = 28 × 20 = 560
(c) 6 (d) 4

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs5

 2 cm, then what is the area of the trapezium? side = 2 13 cm
ABCD ,d Vªsisft;e gS] tgkaAB] CD ds lekukarj gSA ;fn28. The circumcentre of an equilateral triangle is
AB = 4 lseh]BC = 3 , CD = 7 lseh vkSjDA = 2 lseh] at a distance of 3.2 cm from the base of the
rks Vªsisft;e dk {ks=kiQy D;k gS\

r
triangle. What is the length (in cm) of each of
its altitudes?

si
2 3 ,d leckgq f=kHkqt dk ifjdsUæ f=kHkqt ds vk/kj ls 3-2 lseh
(a) 22 cm2 (b) 22 cm2
3 2 dh nwjh ij gSaA bldh Å¡pkbZ (lseh esa) D;k gS\

a n by
22 2 (a) 9.6 (b) 7.2
(c) 22 3 cm2 (d) cm2 (c) 6.4 (d) 12.8
3

n
Sol: (a)
Sol: (d)
ln equilateral triangle
A 4 B

ja
Height = circumradius + inradius

R s
Circumradius : Inradius
2 : 1
2 a th 3 Circumradius = 6.4 cm
Height of triangle = 6.4 + 3.2 = 9.6 cm
29. 'O' is a point in the interior of an equilatral
triangle. The perpendicular distance from 'O'
ty a
4 3
D F E C to the sides are 3 cm, 2 3 cm, 5 3 cm. The
di M

perimeter of the triangle is:

7 'O' ,d leckgq f=kHkqt ds vH;arj esa ,d fcUnq
'O'gSA
ls
From BFC,
Hkqtkvksa dh yEcor nwjh
3 lseh] 2 3 lseh] 5 3 lseh gSA
4 2 f=kHkqt dk ifjeki gSA
Height BE = (a) 48 cm (b) 32 cm
3
(c) 24 cm (d) 64 cm
Sol: (a)
1 4 2
Area of trapezium = (7  4)  2
2 3 Side = P1 + P2 + P3 
3
22 2 2
= cm2 = × 8 3 = 16
3
A

3
27. If the diagonal of a rhombus is 8 cm and its Perimeter = 3 × side = 16 × 3 = 48
area is 48 cm², then the length of each side of 30. ABC is an equilateral triangle. P, Q and R are
the rhombus is: the midpoints of sides AB, BC and CA, respec-
,d leprqHkZqt dk fod.kZ 8 lseh gS vkSj bldk {ks=kiQy 48tively. If the length of the side of the triangle
lseh gS] rks leprqHkZqt dh çR;sd Hkqtk dh yackbZ gS% ABC is 8cm, then the area of PQR is:
ABC ,d leckgq f=kHkqtP,gSQ vkSjR Øe'k% Hkqtkvksa
AB,
(a) 13 cm (b) 2 13 cm
BC vkSjCA ds eè; fcUnq gSA ;fn f=kHkqt
ABC dh Hkqtk dh
(c) 6 13 cm (d) 5 13 cm yEckbZ8 lseh gS] rks
PQR dk {ks=kiQy Kkr dhft, A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs6

 4 Q 4 (c) 22 cm (d) 21 cm
PQR also is an equilateral 
Sol: (c)
1 1 AB = BC = x cm
PR= BC= ×8 = 4 cm (by mid – point theorem)

r
2 2
 AC = AB2 + BC2 = x 2 + x 2 = 2x 2

si
3
Area of PQR = ×16 = 4 3 cm2 AC = 2x cm
4

a n by
31. Of The three angles of a triangle, one is five
times the smallest and another is three times 1
Area of ABC = × AB × BC
the smallest. Find the smallest angles, and the 2

n
type of the triangle
fdlh f=kHkqt ds rhuksa dks.kksa esa ls ,d lcls NksVs dks.k dkikap
2
x
= 121
xquk vkSj nwljk lcls NksVs dks.k dk rhu xquk gksrk gSA lcls2

ja
R s
NksVk dks.k vkSj f=kHkqt dk çdkj Kkr djsa x2 = 121 × 2
(a) 100, obtuse angled triangle

a th
(b) 50, obtuse angled triangle
(c) 60, acute angled triangle
(d) 20, obtuse angled triangle
Sol: (a)
x = 121 × 2 = 11 2 cm
Hypotenuse AC = 2x

= 2 ×11 2 = 22 cm
ty a
Let the smallest angle = x
Another angle = 5x 34. Taking any three of the line segments out of
segments of length 2 cm, 3 cm, 5 cm, and 6 cm,
di M

Third angle = 3x
Sum of the angles = 5x + 3x + x = 180 the number of triangles that can be formed is
x = 20 2 lseh] 3 lseh] 5 lseh] vkSj 6 lseh yackbZ okys •aMksa esa
Smallest angle = 20° fdUgha rhu js•k•aMksa dks ysdj cuus okys f=kHkqtksa dh la[
Greatest angle = 5 × 20 = 100 (a) 3 (b) 4
So triangle will be obtuse angle triangle. (c) 5 (d) 2
32. If the perimeter of an isosceles right triangle
Sol: (d)
 
is 16 2 +16 cm, then the area of the triangle 3+5>6
is: 2+5>6
;fn ,d lef}ckgq ledks.kh; f=kHkqt dk ifjeki 2 triangles can be formed by given line
  segments
16 2 +16 lseh gS] rks f=kHkqt dk {ks=kiQy fdruk gksxk\
35. 6, 8 and 'a' are sides of a triangle, How many
A

(a) 76 sq.cm (b) 64 sq.cm integer triangles are possible.

(c) 57 sq.cm (d) 66 sq.cm
Sol: (b) 6-8 vkSja ,d f=kHkqt dh Hkqtk,¡ gSa] fdrus iw.kkZd f=kH
Let triangle is a right angled isosceles laHko gSaA
Perimeter = x + x + 2 x (a) 11 (b) 12
(c) 14 (d) 15
 16 2 +16 = 2x + 2 x Sol: (b)
8–6<a<6+8
16 ( 2  1 ) = 2x 2  1   2 < a < 14
2x = 16 Possible integral values of a = 12

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs7

 S
10 Length of base = 13 cm
39. Find the area of triangle whose sides are 10
cm, 12 cm, and 18 cm.
10 cm] 12 cm vkSj 18 cm Hkqtkvksa okys f=kHkqt dk

r
Q 10 R {ks=kiQy Kkr dhft, A

si
PQ × QR (a) 22 2cm2 (b) 30 2cm2
QS =
PR

a n by
(c) 28 2cm2 (d) 40 2cm2
10 ×10 10
QS = = = 5 2 cm Sol: (d)
10 2 2

n
a = 10cm, b = 12cm, c = 18 cm, s = 20
37. ABC is a right angled triangle, right angled at
A. A circle is inscribed in it. The lengths of two Area = s s – as – b s – c

ja
sides containing the right angle are 48 cm and

R s
14 cm. The radius of the inscribed circle is: Area = 20 ×10 × 8 × 2
ABC ,d ledks.k f=kHkqt gS] ftlesa dks.kA ledks.k gSA

a th
blesa ,d var% o`Ùk •hapk x;k gSA ledks.k cukus okyh nks
Hkqtkvksa dh yackbZ;ka
f=kT;k gS%
(a) 4 cm
48 cm vkSj14 cm gSaA var% o`Ùk dh

(b) 8 cm
= 4 × 10 ×
2 40 2 = cm2
40. The difference between the semi- perimeter
and the sides of PQR are 18 cm, 17 cm and 25
cm, respectively. Find the area of the triangle.
PQR ds v/Z&ifjeki vkSj Hkqtkvksa ds chp dk varj Øe'k%
ty a
(c) 6 cm (d) 5 cm
Sol: (c) 18 lseh, 17 lseh vkSj 25 lseh gSA f=kHkqt dk {ks=kiQy K
dhft, A
di M

B
(a) 330 510 cm2 (b) 230 510 cm2
50 (c) 30 510 cm2 (d) 130 510 cm2
14 Sol: (c)

P
A 48 C
By triplet (14, 48, 50) r
Radius of incircle q
Base + Perpendicular – hypotaneous
=
2
Q R
A

14 + 48 – 50 62 – 50 12 p
= = = = 6 cm S – p = 18
2 2 2
38. The length of the base of a triangle is 3 cm S – q = 17
smaller than the length of its altitude. Its area S – r = 25
is 104 cm2. What is the length of the base? 3s – 60
,d f=kHkqt ds vk/kj dh yackbZ mldh ÅapkbZ dh yackbZ ls S3 = 2
lseh de gSA bldk {ks=kiQy 1042 gSA
lseh vk/kj dh yackbZ S = 60
fdruh gS\ Area = 60 ×18 ×17 × 25
(a) 14 cm (b) 13 cm
(c) 11 cm (d) 12 cm = 5 × 2 × 3 510 = 30 510 cm2

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs8

 Inradius = r, circumradius = R, Median = D Alternatively:
A 3a = 1 5 3 = 5 3
Inradius =
S
Where S = half perimeter 3 3 15

r
Height = a= ×5 3 = = 7.5 cm
A A 1 A r 2 2 2
×r

si
= = × 45. Twelve sticks, each of length one unit, are used
P 2S 2 S 2
to form an equilateral triangle. The area of the
42. Sides of a triangle are 7cm, 8cm and 9cm find

a n by
circumradius of triangle. triangle is:
bdkbZ yEckbZ dh çR;sd 12 NM+ksa dk mi;ksx ,d leckgq f=kH
,d f=kHkqt dh Hkqtk,¡ 7 lseh] 8 lseh] rFkk 9 lseh gS rks f=kHkqt
dh ifjf=kT;k Kkr djksA cukus ds fy, fd;k tkrk gSA f=kHkqt dk {ks=kiQy gS%

n
21 5 (a) 3 3 sq units (b) 2 3 sq units
cm

ja
(a) (b) cm
2 5 42 (c) 4 3 sq units (d) 8 3 sq units

R s
42 Sol: (c)
cm
(c) 42 5 cm
Sol: (a) a th
Area of triangle
(d)
5 Each side of the triangle = 4 units
Area of equilateral triangle

=
3 2
a =
3
× 4 = 4 3 sq. units
2

= s s – as – b s – c
ty a
4 4
46. A square card board of side 3m is folded on one
a+b+c
di M

S= = 12 of its diagonals to form a triangle. The height

2 of the triangle is:
= 12 × 5 × 4 × 3 = 12 5 3m Hkqtk okys oxkZdkj dkMZ cksMZ dks mlds ,d fod.kZ i
7×8×9 21 eksM+dj f=kHkqt cuk;k x;k gSA f=kHkqt dh Å¡pkbZ gS%
Circumradius = =
4 ×12 5 2 5 2 3
43. In a triangle ABC, the length of medians are (a) m (b) m
3 2
10 cm, 12 cm and 16cm. Find the Area of the
triangle. (c) 2 3 m (d) 3 2 m
,d f=kHkqt
ABC esa] ekfè;dkvksa dh yackbZ 10 lseh] 12 lseh
Sol: (b)
vkSj 16 lseh gSA f=kHkqt dk {ks=kiQy Kkr dhft,
(a) 3 399 (b) 4 399
(c) 5 399 (d) 5 399
A

Sol: (b)
3 3 2
Area of triangle
4
 m s m s – a m s – b m s – c
3
a+b+c 3
ms = , a, b, c are medians
2 We know,
ms = 19
Hypotenous 3 2 3
4 Height of traingle = = =
 19 × 9 × 7 × 3 = 4 399 2 2 2
3

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs9

 Area of triangle = s s – as – bs – c

= 8 8 – 68 – 58 – 5 = 8 × 2 × 3 × 3
A C
= 4 × 3 = 12 cm2

r
1
48. The difference between the two perpendicular × AC × (16 + 12) = 336
2

si
sides of a right-angled triangle is 17 cm and
its area is 84 cm2. What is the perimeter (in 336
cm) of the triangle? AC = = 24 m

a n by
14
,d ledks.k f=kHkqt dh nks yacor Hkqtkvksa ds chp dk51. varj In a circular grassy plot, a quadrilateral shape
17 lseh gS vkSj bldk {ks=kiQy 842
gSA
lsehf=kHkqt dk ifjeki with its corners touching the boundary of the

n
plot is to be paved with bricks. Find the area of
(lseh esa) D;k gS\ the Quadrilateral when the sides of the quad-
(a) 49 (b) 72

ja
rilateral are 36m, 77m, 75m, and 40m.

R s
(c) 56 (d) 65 ,d o`Ùkkdkj ?kkl okys Hkw[kaM esa] ,d prqHkqZt vkdkj ftld
Sol: (c) dksus Hkw[kaM dh lhek dks Nwrs gksa] dks bZaVksa ls iDdk

By triplate
a th
The difference between two perpendicular
sides of a triangle = 17

The sides of triangle respectively 24, 7, 25

gSA prqHkqZt dk {ks=kiQy Kkr dhft, tc prqHkqZt dh Hkq
36m, 77m, 75m, vkSj40m gksaA
(a) 2886
(c) 1443
(b) 114
(d) 1456
ty a
The perimeter of the triangle = (24 + 7 + 25) = 56 Sol: (a)
49. A circle is inscribed in a right-angled triangle.
di M

The lengths of the two sides containing the 36 + 77 + 40 + 75 228

s= = = 114
right angle are 15 cm and 8 cm. What is the 2 2
radius of the incircle?
Area = s – a  s – b s – cs – d
,d ledks.k f=kHkqt esa ,d o`r vafdr gSA ledks.k okyh nks
Hkqtkvksa dh yackbZ 15 lseh vkSj 8 lseh gSA o`r dh f=kT;k= D;k
114 – 36114 – 77114 – 75114 – 40
gS\
(a) 4.5 cm (b) 4 cm = 78 × 37 × 39 × 74 = 2886
(c) 3.75 cm (d) 3 cm 52. The area of a parallelogram is 338 m2. If its
height is two times the corresponding base
Sol: (d) then its base is:
,d lekUrj prqHkqZt dk {ks=kiQy 2338
gSAeh;fn mldh Å¡pkbZ
laxr vk/kj ls nqxquh gS rks bldk vk/kj gSA
(a) 14 (b) 28
A

17
(c) 13 (d) 26
8 r Sol: (c)

Let base = x
Height = 2x
15 Area of parallelogram = 338
P+B– H x × 2x = 338
Inradius of right triangle =
2
338
8 + 15 – 17  x2 = = 169
r= =3 2
2  x = 13 m

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs10

 (a) (b)
a = 40 4 4
So, perimeter of rectangle = 2(50 + 32) = 164 1 2 1 2
(c) (L – 4P 2 ) (d) (L – 3P 2 )
54. One side of a rhombus is 26 cm. and its one 2 4
diagonal is 48 cm. What will be area of rhom- Sol: (b)

r
bus? 1 2
(L – 4P2)

si
,d leprqHkZqt dh ,d Hkqtk 26cm gS vkSj bldk ,d 4
d1 + d2 = L
fod.kZ 48cm gSA leprqHkZqt dk {ks=kiQy fdruk gksxk\

a n by
 d12  d22 = 4P2
(a) 580 cm2 (b) 520 cm2
(c) 624 cm2 (d) 480 cm2 d12  d22 + 2d1 d2 = L2

n
Sol: (d) 1 2
In AOB, d1d2 = (L –4P2)
2

ja
1 1

R s
D Area = d 1 d 2  (L2 – 4P2)
2 4

a th 24 O 24
57. One diagonal of a rhombus is 8 3 cm. If the
other diagonal is equal to its side, then the
area (in cm²) of the rhombus is:
,d leprqHkZqt dk ,d fod.kZ 8 3 lseh gSA ;fn nwljk
fod.kZ bldh Hkqtk ds cjkcj gS] rks leprqHkZqt dk {ks=kiQ
ty a
A C
(lseh esa) gS
di M

(a) 16 3 (b) 12 3
26
(c) 32 3 (d) 24 3
Sol: (c)
B
4
OB = 26 – 24  676 – 576  100 = 10
2 2

 BD = 10 × 2 = 20 cm
4 3
1 4
Area of rhombus = × product of diagonals
2
1 8
= × 20 × 48 = 480 cm2
2 1
So, area =  8 3 × 8 = 32 3
55. The area of a square and rectangle are equal.
A

2
The length of the rectangle is greater than the 58. A field is in the shape of a trapezium whose
length of a side of the square by 10 cm and the parallel sides are 200 m and 400 m long,
breadth is less than 5 cm. The perimeter (in whereas each of other two sides is 260 m long.
cm) of the rectangle is: What is the area (in m²) of the field?
,d oxZ vkSj vk;r dk {ks=kiQy cjkcj gSA vk;r dh yackbZ oxZ ,d eSnku leyEc prqHkZqt dh vkÑfr tSlk gS] ftldh lekukarj
dh ,d Hkqtk dh yackbZ ls 10 lseh vf/d vkSj pkSM+kbZ 5 lsehHkqtkvksa dh yackbZ 200 eh vkSj 400 eh gSa] tcfd vU; nk
ls de gSA vk;r dk ifjeki (lseh esa) gS% Hkqtkvksa esa ls çR;sd dh yackbZ 260 eh gSA eSnku dk {k
(eh2 esa) fdruk gS\
(a) 50 (b) 40
(a) 48000 (b) 52000
(c) 80 (d) 100
(c) 72000 (d) 60000

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs11

 height  Side of triangle = y units
1  2x+ 2y = 3y ...(i)
= × (200 + 400) × 240
2  2x = y

r
= 600 × 120 = 72000 Area of rectangle
59. The two parallel sides of a trapezium are 17 

si
cm and 15 cm, respectively. if the height of Area of triangle
the trapezium is 6 cm, then its area (in m²) is: xy x  2x
fdlh leyac prqHkZqt dh nks lekarj Hkqtk,¡ Øe'k%
cm 17

a n by
 2: 3
= 3 2 3
vkSj 15cm gSA ;fn ml leyac prqHkZqt dh Å¡pkbZ
cm gS]
6 4
y
4
 (2x)2
rks bldk {ks=kiQy
m2(esa) Kkr djsaA

n
62. A bucket is drawn from a well by a wheel of
(a) 9.6 (b) 960 radius 48 cm rounded rope. If bucket is drown
(c) 0.96 (d) 0.0096

ja
by speed of 1.2 m/sec in 1 minute 12 second

R s
Sol: (d) then what is length of rope?
15 ,d ckYVh 48 lseh f=kT;k ds ifg;s ij fyiVh jLlh ls dq,¡
a th 2
6
ls [khaph tkrh gSA ;fn ckYVh dks 1-2 eh@lsd.M dh xfr ls
1 feuV 12 lsd.M esa Åij [khapk tkrk gS rks jLlh dh yackbZ
D;k gksxh\
ty a
17 (a) 8640 cm (b) 864 cm
1 (c) 86.4 cm (d) 8.64 cm
Area of trapezium = × 32 × 6 = 96 cm2 Sol: (a)
di M

2
= 0.0096 m2 Length of rope = Speed × Time = 1.2 × 100 ×
60. In a trapezium PQRS, PQ is parallel to RS and 72 cm = 8640 cm
diagonals PR and QS intersect at O. If PQ = 63. The sum of the lengths of the radius and the
4cm SR = 10 cm, then what is area (POQ): area diameter of circle is 84 cm. What is the differece
(SOR)? between the lengths of the circumference and
,d leyEc PQRS esa]PQ, RS ds lekukarj gS vkSj fod.kZ 22
PR vkSjQS, O ij çfrPNsn djrs gSaA ;fnPQ = 4 lseh] the radius of this circle? [Use  = ]
SR = 10 lseh] rks {ks=kiQy
(POQ) : {ks=kiQy
SOR) D;k 7
gS\ ,d o`r dh f=kT;k vkSj O;kl dh yackbZ dk ;ksx 84 lseh gSA
(a) 4 : 25 (b) 2 : 3 bl o`r dh ifjf/ dh yackbZ vkSj f=kT;k ds chp fdruk varj
(c) 4 : 9 (d) 2 : 5 22
Sol: (a) gS\ (fn;k gS] = )
7
A

P 4 Q (a) 156 cm (b) 172 cm

(c) 148 cm (d) 128 cm
O Sol: (c)
2r + r = 84
 r = 28
S 10 R Circumference = 2r
Area of POQ  4 
2
4 22
   = 2× × 28 = 176
Area of SOR  10  25 7
Difference between circumference and radius
= 176 – 28 = 148 cm

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs12

 side = 15 cm
22 Perimeter of semicircle = r + 2r
radius of the circle ?[Use  = ]
7
15  22 
lseh o`Ùk dh f=kT;k D;k gksxh\= 2  7 + 2 = 38.57 cm
,d o`Ùk dk {ks=kiQy 1]3862 gSA

r
22 69. The inner and outer radius of two concentric

si
[Use  = ]
7 circle are 6.7 cm and 9.5 cm, respectively. What
(a) 7 cm (b) 14 cm is the difference between their circumferences

a n by
(c) 18 cm (d) 21 cm 22
(in cm) ? (Take  = )
Sol: (d) 7

n
R2 = 1386 nks ladsafnzr o`Ùkksa ds vkarfjd vkSj ckgjh f=kT;k Øe'k%
lseh vkSj 9-5 lseh gSaA mudh ifjf/ (lseh esa) ds chp D;k

ja
22
 × R2 = 1386

R s
22
7 varj gS\( = yhft,)
 R = 21 cm 7
66.
a th
If the area of a circle is 154 sq. cm. the ratio
between the circumferece of this circle and
that of another circle of radius 21 cm is:
,d o`Ùk dk {ks=kiQy 1542 lseh
(a) 6.5
(c) 10.4
Sol: (b)
gSA bl o`Ùk dh ifjf/ dk vkSj
(b) 17.6
(d) 20.5

Difference between circumference = 2 (R – r)

ty a
nwljs o`Ùk ftldh f=kT;k 21 lseh gS] dk vuqikr D;k gksxkA 22
=2× × 2.8 = 17.6 cm
(a) 1 : 3 (b) 2 : 3 7
di M

(c) 2 : 1 (d) 1 : 2 70. A wire in the shape of a circle of radius 28 cm

Sol: (a) is bent in the form of a square, what is the dif-
ference of their areas?
Area of first circle = 154 sq. cm.
f=kT;k 28 lseh ds ,d o`Ùk ds vkdkj dk ,d rkj oxZ ds :i
2
 r =
22
× r2 = 154 esa eqM+k gqvk gS] muds {ks=kksa dk varj D;k gS\
7 (a) 530 sq. cm (b) 532 sq. cm
(c) 538 sq. cm (d) 528 sq. cm
r = 7 cm Sol: (d)
C1 = 2r = 2× 7 = 14 cm A.T.Q,
C2 = 2R = 2× 21 = 42 cm 2r = 4a
C1 : C2 = 14 : 42= 1 : 3 4a = 44 × 4
67. Find the perimeter (in cm) of a semicircle of a = 44
A

radius 7 cm. Required difference = (r2 – a2)

7 ls-eh- f=kT;k okys ,d v/Zo`Ùk dk ifjeki (ls-eh-) Kkr  22 2
 28 – 44 
2

djsaA 7 
(a) 36 (b) 72 = 528 sq. cm
(c) 44 (d) 88 71. ABCD is a square of side 21 cm. A circle is in-
Sol: (a) scribed in the square, which touches the sides
Perimeter of semicircle = r + 2r of the square at P, Q, R and S as shown below
in the figure. What is the area (in cm2) of the
 22  36 non-shaded region? (Figure is not drawn to
= r  + 2 = 7 × = 36 cm scale.)
7  7

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs13

 22
= × 7 11.5 +1011.5 – 10 × 7
7
= 154 (1.5) (21.5)
S Q

r
3 43
= 154 × ×

si
2 2
The cost of levelling the track = 77 × 129 × 3

a n by
C = Rs 29,799
D R
74. The sum of the radii of two circles is 286 cm
21

n
Side of square =21 cm, radius of circle = and the area between the concentric circles is
2 50336 cm2. What are the radii ( in cm) of the
Area of shaded region = (a2 – r2)

ja
22

R s
 22 21 × 21 two circles ? ( = )
= 441 –
7
× 
 7 4  nks o`Ùk dh f=kT;kvksa dk ;ksx 286 lseh gS vkSj ladsafnzr

72.

= 4411 –

a th
22  378
=
28  4
= 94.5 sq. cm
Find area of the shaded region? If side of the
ds eè; dk {ks=kiQy 50336 lseh
2
gSA nksuksa o`Ùkksa dh f=kT
(lseh- esa) fdruh&fdruh gksaxh\
( =
22
7
) eku ysaA
ty a
square ABCD is 10. (a) 115 and 91 (b) 115 and 171
Nk;kafdr {ks=k dk {ks=kiQy Kkr dhft;s\ ABCD
;fn oxZdh (c) 91 and 84 (d) 171 and 84
di M

Hkqtk 10 gSA Sol: (b)

The area between the concentric circles
D C =  (R + r) (R – r)
22
= (286) (R – r) = 50336
7
= R – r = 56
R = 171 cm
A r = 115 cm
B
75. The area of circular park is 12474 m². There is
(a) 100 (b) 50 3.5 m wide path around the park. What is the
A

(c) 75 (d) 120 22

area (in m2) of the path? (Take  = )
Sol: (a) 7
Area of shaded region = Area of square ,d o`Ùkkdkj ikdZ dk {ks=kiQy 12474
2
gSA
ehikdZ ds pkjksa
Area of shaded region = a2 vksj 3-5 eh pkSM+k ekxZ gSA ekxZ dk {ks=kiQy
2
esa) Kkr(ehVj
= 100 sq. cm
22
73. A race track is in the shape of a ring whose djsaA(= ysa)
inner and outer circumference are 440 m and 7
506 m, respectively. What is the cost of level- (a) 1424.5 (b) 1435.5
22 (c) 1380.5 (d) 1440.5
ling the track at 6/sq.m? ( = )
7

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs14

 7 4
= 1424.5 m2 = 66 + 28 = 94 cm
76. If perimeter of a semicircle is 54cm. Then find 79. An arc on a circle that is 15 cm long subtends a
the area? 24° angle at the centre. What is the circumfer-

r
v/Zo`Ùk dk ifjeki 54 lseh gSaA rc v/Zo`Ùk dk {ks=kiQy gksxk&
ence of the circle?
(a) 192.5 (b) 115.5 ,d o`Ùk dh 15 lseh yach pki] dsUnz 24°ij dk dks.k

si
(c) 173.25 (d) 134.75 vaarfjd djrh gSA o`Ùk dh ifjf/ D;k gksxh\
Sol. (c) (a) 240 cm (b) 220 cm

a n by
We know, (c) 236 cm (d) 225 cm
Perimeter of semicircle = r + 2r
Sol: (d)
 22 

n
r  + 2 = 54 We know,
 7  360° = Circumference of the circle

ja
36 Given that,

R s
54 = r × 24° = 15 cm
7
15 × 360
r=
21
2

Area =
a th
r 2 22  21
= ×  
2

= 173.25 cm 2
360° =
24
= 225 cm
80. The sides of a triangle are 24 cm, 26 cm and
10 cm. A circle of radius 4.2 cm is drawn touch-
7  2  ing each of its vertices. Find the area (in cm2)
ty a
2
of the triangle, except for the part covered by
77. The area of the quadrant of a circle whose cir- the segments of the circle.
cumference is 22 cm, will be:
ml o`Ùk ds prqFkkZa'k dk {ks=kiQy Kkr djsa ftldh ifjf/ 22fdlh f=kHkqt dh Hkqtk,a 24 lseh] 26 lseh vkSj 10 lseh gSa
di M

lseh gSA blds çR;sd 'kh"kZ dks Nwrk gqvk 4-2 lseh f=kT;k okyk o`
(a) 3.5 cm2 (b) 10 cm2 •hapk tkrk gSA o`Ùk ds •aMks }kjk doj fd, x, Hkkx dks
(c) 38.52 cm 2
(d) 9.625 cm2 NksM+dj] f=kHkqt dk {ks=kiQy
2
esa) (lseh
Kkr djsaA
Sol: (d) (a) 105.86 (b) 92.28
2r = 22 (c) 27.72 (d) 120
7 Sol: (b)
r=
2
22  7 
2

×  26
r 2
Area of quadrant = = 7  2  24
4 4
= 9.625 cm2
A

78. One- quarter of a circular pizza of diameter 28

cm was removed from the whole pizza. What is
the perimeter (in cm) of the remaining pizza 10
Area of the remaining part
22
( Take = ) 1 180 22
7 = ×10 × 24 – × × 4.2 × 4.2
28 lseh O;kl okys o`Ùkkdkj fiTtk dk ,d& pkSFkkbZ Hkkx iwjs2 360 7
fiTtk ls fudkyk tkrk gSA 'ks"k fiTtk dk ifjeki Kkr djas = 120 – 11 × .6 × 4.2
= 120 – 27.72
22 = 92.28 cm2
(lseh esa
) ( = ysa
)
7

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs15

 O D Q
O
21 21 We know,

r
R 1 + Sin
 =

si
21 2 r 1 – Sin
4 1+ 2

a n by
270° 1  =
Area of major segment = r2 × + × 21 × 21
360° 2 r 1– 2

n
=
22 3
× 21 × 21 × + 220.5
⇒r =

4 1– 2 
7 4
1+ 2

ja
22 × 63 × 3

R s
= + 220.5  r = 4 (3 – 2 2 ) cm
4

= 1260 cm2
a th
= 1039.5 + 220.5

82. The area the sector of a circle with radius 4 cm

and of angle 30° is? (take  = 3.14)
4 lseh f=kT;k vkSj
84.

30° dks.k okys ,d o`Ùk ds f=kT;•aM dk

In the given figure the ratio of radii of the
sector and the incircle is 3:1. What is the ratio
of their areas?
fn, x, fp=k esa f=kT;•aM dh f=kT;k vkSj o`Ùk dh f=kT;k dk
vuqikr 3 % 1 gSA muds {ks=kiQy dk vuqikr D;k gS\
ty a
{ks=kiQy D;kgS\= 3.14 ysa)
(a) 4.19 sq. cm (b) 4.49 sq. cm
di M

(c) 4.39 sq. cm (d) 4.29 sq. cm

Sol: (a)
1
30 O
Area of sector = × r 2
360 2
30 22 30° 60° 1
= × ×4×4
360 7 30°
88 (a) 3 : 2 (b) 9 : 4
= = 4.19 cm2
21 (c) 2 : 1 (d) 6 : 5
83. A circle of diameter 8 cm is placed in such a Sol: (b)
manner that it touches two perpendicular
lines. Then another smaller circle is placed in Required ratio
the gap such that it touches the lines and the 90
A

circle. What is the diameter of the smaller = × 3 × 3 :1

circle ? 360
O;kl 8 lseh dk ,d o`Ùk bl rjg ls j•k x;k gS dh ;g nks =9:4
yacor js•kvksa dks Nwrk gSA fIkQj ,d vkSj NksVk o`Ùk The
85. varjkyinterior angle of a regular polygon exceeds
its exterior angle by 140°. The number of sides
esa j•k tkrk gS tSls dh og js•kvksa vkSj cM+s o`Ùk dks Li'kZ
of the polygon is:
djrk gSA NksVs o`Ùk dk O;kl D;k gS\
,d le cgqHkqt dk vkarfjd dks.k mlds ckádks.k140°
ls
(a) 4(3 – 2 ) cm (b) 4(3 – 2 2 ) cm vf/d gSA cgqHkqt esa Hkqtkvksa dh la[;k fdruh gS\
(c) 8(3 – 2 ) cm (d) 8(3 – 2 2 ) cm (a) 16 (b) 12
(c) 20 (d) 18

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs16

 vkSj ,d nwljs lkekU; cgqHkqt
B dk çR;sd vkarfjd dks.k more than the number of sides. Find the num-
ber of sides in polygon?
4
128 ° gSA cgqHkqt A vkSjB dh Hkqtkvksa dh la[;kvksa dk ,d cgqHkqt esa fod.kks± dh la[;k Hkqtkvksa dh la[;k
vf/ dls 12
7 gSA cgqHkqt esa Hkqtkvksa dh la[;k Kkr dhft,\
;ksx fdruk gksxk\

r
(a) 8 (b) 6
(a) 18 (b) 16 (c) 5 (d) 10

si
(c) 19 (d) 17
Sol: (a)
Sol: (b)
Polygon (A) n – 3

a n by
(n – 2) 180 = 1260 n = n +12
2
n=9 If n = 8

n
Polygon (B)
8×5
3 360 then, = 8 +12
Exterior angle = 51 =

ja
2
7 7

R s
LHS = RHS = 8
360 × 7 No. of sides in polygon
Sides of polygon (B) = =7

a th 360
Sum of the sides of polygons = (9 + 7) = 16
87. The ratio of the numbers of sides of two regu-
lar polygon is 5 : 3 if each interior angle of the
first polygon is 156°, then the measure of each
90. The area of a field in the shape of a hexagon is
1944 3 m². What will be the cost (in ) of fenc-
ing it at the rate of 11.50 per metre?
,d "kV~Hkqtkdkj •sr dk {ks=kiQy1944 3 m² gSA mlds
ty a
interior angle of the second polygon is? pkjksa vksj # 11-50 çfr ehVj dh nj ls ckM+ yxkus esa fdruh
nks lecgqHkqt dh Hkqtkvksa dh la[;k dk vuqikr 5%3 gS ;fn
ykxr (# esa) vk,xh\
di M

igys cgqHkqt dk çR;sd var%dks.k

156° gS] rks nwljs cgqHkqt (a) 2,785 (b) 2,484
ds çR;sd var% dks.k dk eki gS\ (c) 3,200 (d) 2,256
(a) 136° (b) 150° Sol: (b)
(c) 140° (d) 135° A.T.Q,
Sol: (c)
3
Exterior angle of 1st polygon = 180 – 156 = 24 × a2 × 6 = 1944 3
360 30 4
Sides of 1st polygon = = = 15 a2 = 324 × 4
24 2 a = 36
Ratio of sides 5 : 3 23
Required amount = 6 × 36 ×
3×  ×3 2
= 108 × 23 = 2484
No. of sides 15 91. A regular hexagon is circumscribed in a circle
A

9
No. of sides of 2nd Polygon = 9 of radius 4.5 cm. What is the area (in cm²) of
the hexagon?
360
Exterior angle of 2nd polygon = = 40° ,d le"kV~Hkqt] 4-5 lseh f=kT;k ds ,d o`Ùk }kjk ifjc¼ gSA
9 "kV~Hkqt dk {ks=kiQy2
esa)
(lseh
D;k gksxk\
Interior angle of 2nd polygon = (180 – 40) = 140°
88. A regular polygon is having 4p + 2 as the num- 243 135
ber of its sides where p is a positive integer. (a) 3 (b) 2
8 4
What will be the ratio of the measure of its
interior angle to that of its exterior angle ? 135 243
(c) 3 (d) 2
4 8

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs17

 1215 243 3 = × 24 × 24 × 3
= 3= 4
40 8 = 432 3 cm2
92. ABCDEF is a regular polygon. Two poles at C 94. The length of each side of a square is twice the
and D are standing vertically and subtend

r
radius of a cirlcle. If the radius of the circle is
angles of elevation 30° and 60° at point A re- 7 cm, then what is the difference between the
spectively. What is the ratio of the height of

si
the pole at C to that of the pole at D? 22
areas of the square and the circle? [Use  = ]
ABCDEF ,d lecgqHkqt gSA blds fcanq C vkSjD ij nks 7

a n by
LrEHk •M+s gS vkSj ;s LrEHk
A ij Øe'k%
fcanq 30° vkSj60° ,d oxZ dh çR;sd Hkqtk dh yackbZ ,d o`r dh f=kT;k dh
nksxquh
dk mUu;u dks.k cukrs gSA rc nksuks LrEHkksa ds Å¡pkbZ dk vuqikr gSA ;fn o`r dh f=kT;k 7 lseh gS] rks oxZ vkSj o`Ù

n
gksxk& 22
(a) (b) {ks=kiQyksa dk varj fdruk [gksxk\
= ] ekfu,º
1: 3 1: 2 3 7

ja
(c) 2 3 : 1 (d) 2 : 3 (a) 44 cm2 (b) 46 cm2

R s
Sol: (c) (c) 42 cm2 (d) 48 cm2
Sol: (c)

F
A
30°
B
a th
60°
r = 7 cm
sides of square (a) = 2 × 7 = 14 cm
area of square = (14)2 = 196
area of circle =  × 7 × 7 = 154
ty a
C difference = 42 cm2
95. ABCD is a square whose side's length is 30 m.
What will be the area of the least sized square
di M

that can be constructed inside ABCD with its

E D vertices on its sides?
A A ABCD ,d ,slk oxZ gS ftldh Hkqtk dh yEckbZm
30gSA ml
U;wure vkdkj ds oxZ dk {ks=kiQy fdruk gksxk ftls
ABCD
60
30° ds vanj bldh Hkqtkvksa ij blds 'kh"kks± ds lkFk fufeZr fd;k
3
2
tk ldrk gS\
30° (a) 550 m2 (b) 500 m2
1 D (c) 450 m2 (d) 400 m2
C 2 3 Sol: (c)
Required ratio = 1 : 2 3 D R C
93. A is the center of given regular hexagon. Find
the area of shaded region? (PQ = 24 cm)
A fn, x, fu;fer "kV~Hkqt dk dsanz gSA Nk;kafdr {ks=k dk
A

S Q
{ks=kiQy Kkr dhft,\
T S
A P B
Side of ABCD = a = 30
U R a
Side of PQRS = = 15 2
2
Area of square = 450 m2
P Q
24 cm

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs18

 Sol: (c) 2 2
Q 5 C R
2nd
30 3 24
Area (1) = = =

r
24cm 10 ×10 10 80
10 10

si
12cm
Area (2) = =
5th 1st 4th 10 × 8 80

a n by
12 12
Area (3) = =
8 × 10 80
Area (ABCD) = 80 – (24 + 10 + 12) = 34

n
3rd P

ja
Area of 1st = 12 × 24 = 288 cm2

R s
Area of 2nd & 3rd = r2 = 3.14 × 144
10 10

a th
Area of 4th of 5th = 3.14 × 36
[r =
24

[r =
2
= 12 ]

12
=6]
ty a
2 Q M 4 R
Total area = 288 + 3.14 × 180 8
2 2 2
Cost = 853.2 × 100 = 85320 PM = 10 – 4 = 84
di M

97. In the given figure, PQR is a triangle and quad- PM = 84

rilateral ABCD is inscribed in it. QD = 2 cm, QC 1
= 5 cm, CR = 3 cm, BR = 4 cm, PB = 6 cm, PA = 5 Area (PQR) = × 8 × 84 = 4 84
cm and AD = 3 cm. What is the area (in cm²) of 2
the quadrilateral ABCD? 4 84 × 34 17 21
nh xbZ vkÑfr esa]PQR ,d f=kHkqt gS rFkk prqHkZqtABCD Area (ABCD) = = cm2
80 5
mlesa vafdr fd;k x;k gSA
QD = 2 ls-eh-, QC = 5 ls-eh-
, CR 98. In the given figure, a circle touches the sides
= 3 ls-eh-
, BR = 4 ls-eh-
, PB = 6 ls-eh-
, PA = 5 ls-eh-rFkk of the quadrilateral PQRS. The radius of the
AD = 3 ls-eh- gSA prqHkZqt
ABCD dk {ks=kiQy (ls-eh- 2
esa) circle is 9 cm. RSP =SRQ = 60° and PQR =
D;k gS\ QPS = 120° What is the perimeter (in cm) of
the quadrilateral?
nh xbZ vkÑfr esa] ,d o`Ùk prqHkZqt
PQRS ds Hkqtkvksa dks Li'kZ
P
dj jgk gSaA o`Ùk dh f=kT;k
9 ls-eh- gSaA
RSP =SRQ =
A

60° rFkkPQR =QPS = 120° gSaA prqHkZqt dk ifjeki

(ls-eh-) D;k gS\
A Q
P
B

D
O
Q C R

S R

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs19

 30° 6
S D R
3

r
In  ODS
If 1 unit 9 cm  
Area = (1)2 –  +1

si
2 
3 units  9 3 cm
So, D is the mid point of SR  –2

a n by
= – –1=
So, DS = 9 3 2 2
100. The sides of a triangle are 10 cm, 24 cm and

n
SR = 18 3 and then 26 cm. At each of its vertices, circles of radius
3.5 cm are drawn. What is the area of the tri-
In  PCO angle excluding the portion covered by the sec-

ja
 22 

R s
3 units 9 unit
tors of the circles?   = 
9  2 
1 units =

PQ = 6 3
a th3
= 3 3 = PC

Sum of pair of 2 opposite side is equal to sum

,d f=kHkqt dh Hkqtk,¡ 10 lseh] 24 lseh vkSj 26 lseh gSA blds
çR;sd dksus ij 3-5 lseh f=kT;k ds o`Ùk •haps tkrs gSA o`Ùk
{ks=kksa }kjk ?ksjs x, fgLls dks NksM+dj f=kHkqt dk {ks
 
ty a
  = 22 
gksxk\
of pair of other two opposite side  2 
So, PQ + SR + SP + QR (a) 81.5 cm2 (b) 100.75 cm2
di M


Perimeter = 2 18 3 + 6 3 = 48 3  (c) 75.75 cm2
Sol: (b)
(d) 78.25 cm2

Alternatively
A

9
3 3 3 3
P Q
M
3 3 3 3
9
9 3
 9 3 B C
60 9 60° 180°
°
A

Area of three sectors = × r 2

R 360°
S 9 3 9 3 1 22
= × × 3.5 × 3.5 = 19.25 sq. cm
2 7
3 For ABC = 102 + 242 = 262
a = 27 a = 18 3
2 ABC is right angled
PM 9 1
=  PM = 3 3 Its area = × 10 × 24 = 120 sq. cm
9 3 27 2
Area of shaded region = (120 – 19.25)
12 3 + 36 3 = 48 3 cm sq. cm. = 100.75 sq. cm

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs20

 + area of ABC – area of quarter made on BC

28 c
1 22 35 35 1 1 22
× × × + × 28 × 21 – × × 21 × 21
2 7 2 2 2 4 7

r
11 × 5 × 35 11 × 3 × 21
= + 14 × 21 –
B 21 cm C 4 2

si
= 481.25 + 294 – 346.5 = 428.75 cm2
(a) 428.75 cm2 (b) 857.50 cm2

a n by
(c) 214.37 cm2 (d) 371.56 cm2

n
ja
R s
a th
ty a
di M
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs21

 Mensuration-3D

MENSURATION -3D/f=kfoeh;
{ks=kfefr
[CLASSROOM SHEET]

3D Figure/f=kfoeh; vkÑfr 1. A solid has 12 vertices and 30edges. How

many faces does it have?
A 3-D solid is a figure that is not flat, it is ,d Bksl esa 12 'kh"kZ vkSj 30 fdukjs gSa bld
three dimensional, namely length, breadth and iQyd gSa\
height or thickness. The flat surface that binds (a) 22 (b) 24
it is called faces. The intersection of surfaces (c) 18 (d) 20
gives us a line known as edge. The intersection
In general a solid has two types of surface
of edges gives point known as vertices.
areas:
f=kfoeh; Bksl og vkÑfr gS tks likV ugha gS] ;g
lekU;r fdlh Bksl ds lrg dk {ks=kiQy nks izdk
rhu vk;keh gS] vFkkZr~ yackbZ] pkSM+kbZ vkSj ÅapkbZ ;k eksVkbZA
gksrk gSA

r
og likV lrg tks bls cka/rh gS] iQyd dgykrh gSA iQydksa
Lateral Surface Area (LSA) ik'oZ i`"Bh; {ks=
dk çfrPNsnu gesa ,d js•k nsrk gS ftls fdukjk dgk tkrk

si
LSA of a solid is the sum of areas of all the
gSA fdukjksa ds çfrPNsnu ls ,d fcanq feyrk gS ftls 'kh"kZ
an by surface excluding top and bottom surface.
dgk tkrk gSA fdlh Bksl dk ik'oZ i`"Bh; {ks=kiQy Åijh vkS
Euler derives a law to establish relationship
lrg dks NksM+dj lHkh lrgksa ds {ks=kiQy dk ;

n
between number of vertices, faces and edges. It
states that number of vertices plus the number
ja Total Surface Area (TSA) dqy i`"Bh; {ks=kiQ
of faces in every 3-D solid will will always equal TSA of a solid is the sum of the lateral
R s
to number of edges plus two. If V, F and E surface area and the areas of the base amnd
denotes the number of vertices, faces and edges the top.
a th

respectively then, by Euler's law fdlh Bksl dk dqy i`"Bh; {ks=kiQy ik'oZ i`"Bh;
;wyj us 'kh"kks±] iQydksa vkSj fdukjksa dh la[;k ds vkSj
chp vk/kj vkSj 'kh"kZ ds {ks=kiQyksa dk ;k
laca/ LFkkfir djus ds fy, ,d fu;e fudkykA blesa dgk TSA = LSA + Area of top surface + Area of
ty a

x;k gS fd çR;sd f=kfoeh; Bksl esa 'kh"kks± dh la[;k vkSj bottom surface
iQydksa dh la[;k ges'kk fdukjksa dh la[;k esa nks tksM+sdqy ij i`"Bh; {ks=kiQy
= ik'oZ i`"Bh; {ks=kiQy
+ Åijh lrg
di M

] F vkSj E Øe'k% 'kh"kks±] dk {ks=kiQy $ fupyh lrg dk {ks=kiQy

izkIr ;ksxiQy cjkcj gksxhA V;fn
iQydksa vkSj fdukjksa dh la[;k n'kkZrs gSa] rks ;wyj ds fu;e Cuboid/?kukHk
ds vuqlkj
A cuboid is a rectangular solid object
VF  E2 having six rectangular surfaces. It is
Let us consider an example of cube: sometimes also called as rectangular
parallelopiped.
bls ?ku ds mnkgj.k ds ekè;e ls le>rs gS%
?kukHk ,d vk;rkdkj Bksl gS ftlesa Ng vk;r
Face lrgsa gksrh gSaA bls dHkh&dHkh vk;rkdk
prqHkZqt Hkh dgk tkrk gSA
Edge
A

Vertex
Height

Clearly,
Br
ea

Length
V= 8, F = 6 and E = 12
dt
h
 Mensuration-3D

yackbZl)]( pkSM+kbZ
b) vkSj
( ÅapkbZ h)(okys ?kukHk ds
6. Volume of a cuboid is 4800 cm3, If the
fy, height of this cuboid is 20 cm, then what
will be the area of the base of cuboid?
(i) Lateral Surface Area /ik'oZ i`"Bh; {ks=kiQy
= 2(l + b)h ,d ?kukHk dk vk;ru4800 cm³ gSA ;fn bl ?ku
(ii) Total Surface Area/dqy i`"Bh; {ks=kiQy dh Å¡pkbZ20 cm gks] rks ?kukHk ds vk/
= 2(lb + bh + hl) {ks=kiQy fdruk gksxk\
(iii)Volume of Cuboid/?kukHk dk vk;ru SSC CGL MAINS (08/08/2022)
=l×b×h (a) 480 cm² (b) 150 cm²
(iv) Diagonal of Cuboid/?kukHk dk fod.kZ
(c) 240 cm² (d) 120 cm²
= l 2  b2  h 2
7. The breadth of a cuboidal box half its
Note:- Length of longest rod that can be
length and one-fifth its height. If the
placed in the room/fdlh dejs esa j[kh tk ldus lateral surface area of the cuboid is 4320
okyh lcls cM+h NM+ dh yEckbZ= Diagonal fod.kZ cm2, then its volume (in cm2) is :
2. Find the total surface area (in cm2) of a
cuboid having dimensions 5 cm, 7 cm and ,d ?kukdkj ckWDl dh pkSM+kbZ mldh

r
11 cm. 1
vk/h gS vkSj bl
dh Å¡pkbZ dh 5 gSA ;fn ?kukH

si
5 cm, 7 cm vkSj11 cm foekvksa okys ,d ?kukHk
dk oqQy i`"Bh; {ks=kiQy
an by cm2 esa)
( Kkr djsaA ik'oZ i`"Bh; {ks=kiQy 4320lsehoxZgks] rks b
SSC CGL TIER I 20/07/2023 (Shift-02) vk;ru (lseh esa) fdruk gksxk\
3

(a) 385 (b) 334 ICAR Mains, 08/07/2023 (Shift-2)

n
(c) 343 (d) 167
3. The length of a cuboid is 4 cm. If the breadth
ja (a) 17280 (b) 18720
of the cuboid is four times of its length and (c) 16704 (d) 15840
R s
height of the cuboid is twice of its length,
then what is the lateral surface area of the 8. The ratio of the length, width and height
a th

cuboid? of a closed cuboid is given as 6 : 3 : 2. The

total surface area of this cuboid is given
,d ?kukHk dh yackbZ 4 lseh gSA ;fn ?kukHk dh pkSM+kbZ
as 1800 cm2. Find the volume (in cm3) of
mldh yackbZ dh pkj xquk gS vkSj ?kukHk dh ÅapkbZ mldh
this cuboid.
ty a

yackbZ dh nksxq
uh gS] rks ?kukHk dk ik'oZ lrg {ks=k D;k gS\
SSC MTS 15/06/2023 (SHIFT-02) ,d lao`r ?kukHkdh yackbZ] pkSM+kbZ vkSj
di M

(a) 380 cm 2
(b) 440 cm 2
vuqikr 6 % 3 % 2 fn;k x;k gSA bl ?kukHk d
(c) 260 cm2 (d) 320 cm2 i`"Bh; {ks=kiQy 18002 lseh
fn;k x;k gSA bl ?ku
4. Area of a cardboard (in cm2) needed to
dk vk;ru (lseh 3 esa) Kkr dhft,A
make a closed box of size 20 cm × 10 cm
× 8 cm will be: SSC CGL TIER II 26/10/2023
20 cm × 10 cm × 8 cm vkdkj ds ,d can ckWDl (a) 4650 (b) 4500
dks cukus ds fy, vko';d dkMZcksMZ dk {ks=kiQy
(cm2
(c) 4200 (d) 4800
esa
) fdruk gksxk\
SSC CGL 18/04/2022 (Shift-03) 9. The length and breadth of a cuboidal store
(a) 960 (b) 690 are in the ratio 2 : 1 and its height is 3.5
(c) 880 (d) 750 meters. If the area of its four walls
(including doors) is 210 m2 then its volume
A

5. What is the length (in cm) of the longest

rod that can be fitted in box of dimensions is ......
28cm × 4cm × 10cm? ?kukHk ds vkdkj okys LVksj dh yackbZ vkS
28cm × 4cm × 10cm ds vk;ke ds ,d ckWDl esa vuqikr 2 % 1 gS vkSj bldh Å¡pkbZ 3-5 ehVj
fiQV dh tk ldus okyh lcls yach NM+ dh yackbZ bldh pkj nhokjksa (njoktksa lfgr) dk {ks=kiQ
(cm esa
) fdruh gksxh\ ehVj2 gS] rks bld vk;ru ------- gksxkA
 Mensuration-3D

Cube/?ku 13. The length of a cuboid is double of its

breadth and its height is half of its breadth.
A solid object having all the six surfaces If the height of the cuboid is 2 cm, then
are square is known as cube. Thus, length, what will be the edge of a cube whose
breadth and height of a cube are equal. volume is the same as that of the cuboid
,d Bksl ftldh lHkh Ng lrgsa oxkZdkj gksa] ?ku dgykrh mentioned above ?
gSA bl çdkj] ,d ?ku dh yackbZ] pkSM+kbZ vkSj ÅapkbZ fdlh ?kukHk dh yackbZ viuh pkSM+kbZ dh n
cjkcj gSA mldh ÅapkbZ] mldh pkSM+kbZ dh vk/h gSA
ÅapkbZ 2 lseh gS] rc fdlh ?ku dk dksj (,t) fdr
gksxk ftldk vk;ru mDr ?kukHk ds leku gS\
SSC MTS 5/08/2019 (Shift-02)
(a) 4 cm (b) 6.4 cm
(c) 8 cm (d) 7.2 cm

Consider 'a' is the side of the cube. 14. The side of a cube is 15 cm. What is the

r
base area of a cuboid whose volume is 175
eku ysa fd'a' ?ku dh Hkqtk gS
cm3 less than that of the cube and whose

si
(i) Lateral Surface Area/ik'oZ i`"Bh; {ks=kiQy
height is 32 cm?
= 4a² an by
(ii) Total Surface Area/dqy i`"Bh; {ks=kiQy ,d ?ku dh Hkqtk 15 lseh gSA ml ?kukHk ds v
= 6a² {ks=kiQy Kkr djsa ftldk vk;ru ?ku ds vk;ru ls 1

n
(iii)Volume/vk;ru = a³ ?ku lseh de gS rFkk ftldh ÅapkbZ 32 lseh gS
(iv) Diagonal/ fod.kZ= 3a
ja SSC MTS 2/08/2019 (Shift-03)
10. If the side of a cubical box is 12 cm, then
R s
find its total surface area. (a) 200 cm² (b) 100 cm²
a th

;fn ,d ?kukdkj fMCcs dh Hkqtk

12 lseh- gS] rks bldk (c) 160 cm² (d) 325 cm²
laiw.kZ i`"Bh; {ks=kiQy Kkr dhft,A 15. 2 cubes each of volumes 125 cm 3 are
SSC CGL (PRE) 26/07/2023 (Shift-3) joined end to end. The surface area of the
ty a

(a) 952 cm2 (b) 864 cm2 resulting cuboid is:

(c) 664 cm2 (d) 792 cm2 125 lseh3 vk;ru okys 2 ?ku ,d fljs ls nwljs fljs
di M

11. If the total surface area of a cube is 24 rd tqM+s gq, gSaA ifj.kkeh ?kukHk dk lrg {k
sq.units, then what is the volume of the
cube? SSC MTS 18/05/2023 (Shift-01)
;fn ,d ?ku dk laiw.kZ i`"Bh; {ks=kiQy 24 oxZ bdkbZ gS]
(a) 325 cm2 (b) 350 cm2
rks ?ku dk vk;ru D;k gksxk\ (c) 125 cm2 (d) 250 cm2
CGL PRE, 14/07/2023 (Shift-2)
16. 8 cubes, each of edge 5 cm, are joined end
(a) 8 cu.units/?ku bdkbZ to end. What is the total surface area of the
(b) 16 cu.units/?ku bdkbZ resulting cuboid?
(c) 10 cu.units/?ku bdkbZ 8 ?ku] ftuesa ls izR;sd dk fdukjk 5 lseh gS
(d) 4 cu.units/?ku bdkbZ fljs ls nwljs fljs rd tqM+s gq, gSaA ifj.kkeh
A

12. The surface area of a cube is 13.5 m². What dqy i`"Bh; {ks=kiQy fdruk gS\
is the length (in m) of its diagonal? (a) 850 sq.cm (b) 825 sq.cm
fdlh ?ku dk i`"Bh; {ks=kiQy 13-5 ehVj
2
gSA blds (c) 1200 sq.cm (d) 800 sq.cm
fod.kZ dh yackbZ (ehVj esa) Kkr djsaA 17. Four solid cubes, each of volume 1728 cm³,
SSC CGL 24/08/2021 (Shift 03) are kept in two rows having two cubes in
 Mensuration-3D

izR;sd 1728 lseh

3
vk;ru okys pkj Bksl ?kuksa dks 21.
nks The sum of the length, breadth and depth
iafDr;ksa esa j[kk tkrk gS vkSj izR;sd iafDr esa nks ?kuofj[ks
a cuboid is 23 cm, and its diagonal is
tkrs gSaA muls oxkZdkj vk/kj okyk vk;rkdkj Bksl 5 7 cm . Its surface area is:
curk gSA ifj.kkeh Bksl dk dqy i`"Bh; {ks=kiQy 2
(lseh
,d ?kukHk dh yackbZ] pkSM+kbZ vkSj x
esa) Kkr djsaA
SSC MTS 18/10/2021 23 cm gS] vkSj bldk fod.kZ5 7 cm gSA bld

(a) 576 (b) 1152 i`"Bh; {ks=kiQy D;k gS\

(c) 2304 (d) 1440 SSC CHSL 31/05/2022 (Shift- 1)
 Relation between area of faces and volume (a) 288 cm2 (b) 354 cm2
iQydksa ds {ks=kiQy vkSj vk;ru esa lEca/ (c) 372 cm2 (d) 222 cm2
Volume  A1  A 2  A 3 22. The sum of length breadth and height of a
18. If the areas of three adjacent faces of a cuboid is 14 cm and its total surface area
cuboidal box are 729 cm2, 529 cm2 and 289 is 96 cm2 then find maximum length of a
cm2, respectively, then find the volume of stick that can placed inside the box?

r
the box.
,d pkSM+kbZ dh yackbZ vkSj ,d ?kukHk d
;fn ?kukHk fMCcs ds rhu vklUu iQydksa dk {ks=kiQy

si
Øe'k%729 cm2 ,529 cm2 vkSj289 cm2 gS] rks ;ksx 14 lseh gS vkSj bldh lrg dk dqy {ks=kiQ
fMCcs dk vk;ru Kkr dhft,A lseh2 gS fiQj ,d NM+h dh vf/dre yackbZ Kkr
an by
SSC CHSL 03/06/2022 (Shift- 3) tks ckWDl ds vanj j[kh tk lds\

n
(a) 10557 cm3 (b) 10560 cm3 (a) 15 cm (b) 12 cm
(c) 10555 cm3 ja (d) 10551 cm3 (c) 10 cm (d) 20 cm
19. If the area of three adjacent faces of a
R s
 Volume of hollow cuboid/[kks[kys ?kukH
rectangular box which meet in corner are
vk;ru = lbh – (l – 2x) (b–2x) (h – 2x)
a th

32 cm2, 24 cm2 and 48 cm2 respectively.

Then the volume of the box is? Where, x is the thickness of walls of the
;fn ,d ?kukHk ds rhu iQydks tks ,d fdukjs ij cuboid
feyrs gS] muds {ks=kiQy Øe'k% 232] 24lseh
lseh2 tgk¡]x ?kukHk dh nhokj dh eksVkbZ gS
ty a

vkSj 48 lsehgSA rc ckWDl dk vk;ru Kkr djsaA 23. A wooden box measures 20 cm by 12 cm by
2
di M

(a) 192 (b) 216 10 cm. Thickness of wood is 1 cm. Volume

(c) 144 (d) 256 of wood to make the box (in cubic cm) is
 Relation between diagonal and total surface ,d vk;rkdkj cDls dh foek,¡ Øe'k% 20 lseh × 12
area of a cuboid lseh× 10 lseh gSA ydM+h dh eksVkbZ 1 lseh g
?kukHk ds fod.kZ vkSj dqy i`"Vh; {ks=kiQy esa lEca/cukus esa yxh ydM+h dk vk;ru Kkr djsaA
(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)
(Sum of dimensions)² = (Diagonal)2 + Total (a) 960 (b) 519
Surface Area (c) 2400 (d) 1120
(foekvksa dk ;ksxiQy)2
= (fod.kZ)2 + (dqy i`"Bh; 24. The length, breadth and height of a wooden
{ks=kiQy) box with a lid are 10 cm, 9 cm and 7 cm,
20. The sum of length, breadth and height of respectively. The total inner surface of the
A

a cuboid is 20 cm. If the length of the closed box is 262 cm2. The thickness of the
diagonal is 12 cm, then find the total wood (in cm.) is
surface area of the cuboid.
,d yM+dh ds cDls dh yackbZ] pkSM+kbZ vkSj
,d ?kukHk dh yackbZ] pkSM+kbZ vkSj ÅapkbZ dk ;ksx 20
10 lseh] 9 lseh vkSj 7 lseh gSA can ckWDl dh dq
lseh gSA ;fn fod.kZ dh yackbZ 12 lseh gS] rks ?kukHk
lrg 262 lseh2 gSA ydM+h dh eksVkbZ (lseh esa
dk dqy lrg {ks=kiQy Kkr djsaA
(a) 2 (b) 3
 Mensuration-3D

25. A water tank has 8360 litres of capacity. 28. A square of side 3 cm is cut off from each
It is made up of a material and the corner of a rectangular sheet of length 24
thickness of material for all four walls is 5 cm and breadth 18 cm and the remaining
cm. Find the thickness of material in the sheet is folded to form an open rectangular
bottom of that tank. The dimension of box. The surface area of the box is
water tank is 3.3m × 2.6m × 1.1m. pkjksa dksuksa ls 3 lseh Hkqtk okyk oxZ ,d 2
,d ikuh ds VSad dh {kerk 8360 yhVj gS] ;g ,sls rFkk 18 lseh pkSM+s vk;rkdkj 'khV ls dkV f
/krq ls cuk gS ftldh pkjksa nhokjksa dh eksVkbZ 5 lseh
rFkk 'ks"k Hkkx dks eksM+dj ,d •qyk cDlk cu
gS] rc bl VSad ds iQ'kZ dh eksVkbZ D;k gksxh] ;fn cDls dk i`"Bh; {ks=kiQy Kkr djsaA
VSad dh eki3.3 ehVj× 2.6 ehVj× 1.1 ehVjgSA (a) 468 cm² (b) 396 cm²
(a) 4.5 cm (b) 5.5 cm
(c) 612 cm² (d) 423 cm²
(c) 6.5 cm (d) 7.5 cm
29. A room is in the shape of a cuboid, with
 Making a box from rectangular sheet
dimensions 12m × 10m × 3m. What is the
vk;rkdkj 'khV ls [kqyk fMCCkk cukuk cost of painting the four walls of the room
We can make an open rectangular box by at the rate of Rs. 50 per sq.m?

r
cutting off equal squares of side x unit at
four corners and the remainder is folded ,d dejk ?kukHk ds vkdkj esa gS ftldh yackbZ]

si
up vkSj ÅapkbZ 12× eh
10 eh × 3 eh gSA 50 :i;s izfr o
pkjksa dksuksa ls cjkcjxHkqtk
bdkbZ dk oxZ dkVus ds ehVj dh nj ls bl dejs dh pkj nhokjksa dks jaxu
an by
ckn 'ks"k dks eksM+dj ge ,d [kqyk vk;rkdkj fMCck ykxr Kkr djsaA
cuk ldrs gSa

n
SSC MTS 7/08/2019 (Shift-03)
x x Folded up
x x ja x (a) Rs. 15000 (b) Rs. 15600
R s
b b – 2x (c) Rs. 6600 (d) Rs. 7500
l – 2x
x x 30. The length, breadth, and height of a room
a th

x x are 10 m, 8 m and 6 m respectively. Find

l
the cost of white washing the walls of the
Volume of rectangular box/vk;rkdkj fMCcs dk
room and the ceiling at the rate of Rs. 7.50
vk;ru
ty a

per m².
= (l – 2x) (b – 2x)x
26. From the four corners of a rectangular sheet
,d dejs dh yackbZ] pkSM+kbZ vkSj ÅapkbZ 10m
di M

of dimensions 25 cm × 20 cm, square of 8m vkSj6m gSA Rs.7.50 çfr oxZ ehVj dh nj ls

side 2 cm is cut off from four corners and a dejs dh nhokjksa vkSj Nr ij liQsnh djus dh ykx
box is made. The volume of the box is. djsaA
25 lseh× 20 lseh vk;ke okyh ,d vk;rkdkj 'khV ds (a) Rs. 2,220 (b) Rs. 1,850
pkjksa dksuksa ls 2 lseh Hkqtk okyk oxZ dkV fn;k tkrk gS
(c) Rs. 2,150 (d) Rs. 2,000
vkSj ,d ckWDl cuk;k tkrk gSA ckWDl dk vk;ru gSA
31. The internal measures of a cuboidal room
(a) 672 cm³ (b) 372 cm³
(c) 560 cm³ (d) None of these are with length as 12 m, breadth as 8 m
27. A rectangular sheet of metal is 40cm by and height as 10 m. The total cost (in Rs.)
15cm. Equal squares of side 4 cm are cut of whitewashing all four walls of the room
off at the corners and the remainder is and also the ceiling of the room, if the cost
A

folded up to form an open rectangular box. of whitewashing is Rs. 25 per m2 is:

The volume of the box is: fdlh ?kukHk ds vkdkj okys dejs dh vkrafjd
,d vk;rkdkj /krq dh 'khV dk vk;ke 40 lseh × dh yackbZ 12 ehVj] pkSM+kbZ 8 ehVj vkS
15 lseh gS] bl 'khV ds izR;sd fdukjs ls 4 lseh Hkqtk ehVj gSA ;fn liQsnh dh ykxr 25 :i;s çfr ehV
2

ds oxZ dkVs x, gS] cps Hkkx dks eksM+dj ,d vk;rkdkj gS] rks dejs dh lHkh pkjksa nhokjksa ds lkFk
ckWDl cuk;k x;k gS] bl ckWDl dk vk;ru D;k gksxk\ liQsnh djkus dh dqy ykxr (:- esa) Kkr djsaA
 Mensuration-3D

 A rectangular tank is 'l' metres long and 'h' (a) 12 (b) 16

metres deep. If 'x' cubic metres of water be (c) 20 (d) 24
draw in off the tank, the level of the water
in the tank goes down by 'd' metres, then 35. If the rectangular faces of a brick have their
the amount of water (in cubic metres) the diagonals in the ratio 3 : 23 : 15, then
the ratio of the length of the shortest edge
x  h
tank can hold is given by  cubic of the brick to that of its longest edge is
 d 
;fn ,d bZaV ds vk;rkdkj iQydksa esa muds fod3:
x 23 : 15 ds vuqikr esa gSa] rks bZaV ds lcls N
metres and the breadth of the tank is  
 ld  dh yackbZ dk vuqikr mlds lcls yacs fdukjs ls gS
metres. (a) 2 : 5 (b) 2 : 3
,d vk;rkdkj VSad'l' ehVj yack vkSj'h' ehVj xgjk (c) 1 : 3 (d) 3 : 2
gSA ;fn VSad'x'ls?ku ehVj ikuh fudyk tkrk gS] rks
VSad esa ikuh dk 'd' Right Circular Cylinder/leo`Ùkh;
LrjehVj de gks tkrk gS] rks VSad csy
esa j•s tk ldus okys ikuh dh ek=kk (?ku ehVj esa)
A solid which has uniform circular cross-
x  h x
?ku ehVj gksrh gS rFkk Vadh dh pkSM+kbZ section is called a cylinder (or, a right

r
 
d   ld  circular cylinder)
ehVj gksrh gSA

si
,d Bksl ftldk vuqçLFk dkV le o`Ùkkdkj gksrk g
32. A rectangular tank is 50 metres long and
an by csyu (;k yac yEco`Ùkh; csyu) dgk tkrk gSA
29 metres deep. If 1000 cubic metres of
water be drawn off the tank, the level of

n
the water in the tank goes down by 2
metres. How many cubic metres of water
can the tank hold? And also find the
ja
R s
breadth of the tank.
,d vk;rkdkj VSad 50 ehVj yack vkSj 29 ehVj xgjk h
a th

gSA ;fn VSad ls 1000 ?ku ehVj ikuh fudkyk tkrk gS] rks
VSad esa ikuh dk Lrj 2 ehVj de gks tkrk gSA VSad fdrus
?ku ehVj ikuh j• ldrk gS\ vkSj VSad dh pkSM+kbZ Hkh
ty a

Kkr djsaA
(a) 14500 m³, 10m (b) 15500 m³, 12m
r
di M

(c) 15400 m³, 15m (d) 10500 m³, 10m

33. There is a cuboid of dimension 6 cm by 4 Let r be the radius of circular cross-section
cm by 3 cm. the minimum such cuboids and h be the height of cylinder, then
are arranged to make a cube. Find the eku yhft, r o`Ùkkdkj vuqçLFk dkV dh f=kT;
volume of the cube. h csyu dh ÅapkbZ gS] rks
6 lseh 4 lseh vkSj 3 lseh vk;ke dk ,d ?kukHk gSA (i) Area of cross-section/ vuqizLFk dkV
U;wure ,sls ?kukHkksa dks ,d ?ku cukus ds fy, O;ofLFkr {ks=kiQy = r²
fd;k tkrk gSA ?ku dk vk;ru Kkr dhft,A (ii) Perimeter (circumference) of cross-
CRPF HCM 11/03/2023 (Shift - 02) section/vuqizLFk dkV dh ifjf/= 2r
(a) 1728 cm³ (b) 1000 cm³ (iii) Curved Surface area/oØ (ik'oZ) i`"Bh
(c) 512 cm³ (d) 216 cm³ {ks=kiQy= Perimeter of cross-section/
A

34. A rectangular block of length 20 cm, vuqçLFk dkV dk ifjeki

× height/ÅapkbZ
= 2rh
breadth 15 cm and height 10 cm is cut up (iv) Total Surface area lEiw.kZ i`"Bh; {ks= =
into exact number of equal cubes. The least Curved surface area oØ (ik'oZ) i`"Bh
possible number of cubes will be
{ks=kiQy
+ 2 × Area of cross-section/vuqçL
yackbZ 20 lseh] pkSM+kbZ 15 lseh vkSj Å¡pkbZ 10 lseh dkV dk {ks=kiQy
ds vk;rkdkj CykWd dks cjkcj ?kuksa dh lVhd la[;k = 2rh + 2(r2) = 2r(r + h)
 Mensuration-3D

36. What is the volume of a cylinder if the ra- 40. The curved surface area of a cylinder is five
dius of the cylinder is 10 cm and height is times the area of a base. Find the ratio of
20 cm? (Take  = 3.14) radius and height of the cylinder.
csyu dk vk;ru D;k gksxk ;fn csyu dh f=kT;k
10 ,d flysaMj dk ?kqekonkj lrg dk {ks=kiQy mlds
cm vkSj Å¡pkbZ
20 cm gks\( = 3.14 ysa
)
ds {ks=kiQy dk ik¡p xquk gSA flysaMj ds
Å¡pkbZ ds vuqikr Kkr djsaA
SSC CGL TIER- II 07/03/2023
(CGL MAINS 18/10/2020)
(a) 6280 cm³ (b) 5306 cm³ (a) 2 : 5 (b) 2 : 3
(c) 6260 cm³ (d) 5280 cm³ (c) 3 : 4 (d) 3 : 5
37. The curved surface area of a solid cylinder 41. The sum of the radius of the base and the
of height 15 cm is 660 cm2. What is the height of a closed solid cylinder is 12.5 cm.
volume (in cm3) of the cylinder? If the total surface area of the cylinder is
22 275 m2, then its radius is:
(Take = )
7
 22 
 Take  =
15 lseh Å¡pkbZ okys ,d Bksl csyu dk oØ i`"Bh; 

7 

r
{ks=kiQy 660 2lseh
gSA flysaMj dk vk;ru (lseh
3
esa) fdlh lao`r Bksl csyu dh vk/kj f=kT;k vkSj m
dk ;ksxiQy12.5cm gSA ;fn csyu dk lEiw.kZ

si
22
D;k gS\ (= 7 yhft,)
an by {ks=kiQy
275cm2 gS] rks bldh f=kT;k Kkr djsaA
SSC CPO 05.10.2023 (Shift-2)
22
(a) 2060 (b) 3210 ( = ysa)

n
7
(c) 2540 (d) 2310
ja SSC CHSL 12/04/2021 (Shift- 3)
38. The height of a cylinder is 45 cm. If
R s
circumference of its base is 132 cm, then (a) 3.5 cm (b) 3 cm
what is the curved surface of this cylinder? (c) 7 cm (d) 5 cm
a th

 22  42. The sum of the curved surface area and

 Use   66 total surface area of a solid cylinder is 2068
 7 
,d csyu dh Å¡pkbZ 45 lseh gSA ;fn blds vk/kj dh cm2. If radius of its base is 7 cm, then what
ty a

ifjf/ 132 lseh gks] rks bl csyu dk oØ o`"Bh; {ks=kiQy is the volume of this cylinder?  Use  22 
 7 
fdruk gS\
di M

,d Bksl csyu ds oØ i`"Bh; {ks=kiQy rFkk

 22 
   yhft,  i`"Bh; {ks=kiQy dk ;ksx 2068cm² gSA ;fn bld
 7 
vk/kj dh f=kT;k 7cm gks] rks bl csyu dk vk;r
SSC CGL MAINS (08/08/2022)
(a) 5720 cm² (b) 5940 cm²
D;k gksxk\
 22
(c) 6270 cm² (d) 6360 cm²
 = yhft, 
39. Find the ratio of the curved surface area  7 
to the total surface are of a cylinder with SSC CGL MAINS (08/08/2022)
diameter of base 14 cm and height 10 cm. (a) 2480 cm³ (b) 2760 cm³
14 cm vkèkkj ds O;kl vkSj
10 cm Å¡pkbZ okys (c) 3080 cm³ (d) 2060 cm³
A

osyu ds oØ i`"Bh; {ks=kiQYk dk] laiw.kZ i`"Bh; {ks=kiQYk

43. The ratio of TSA and CSA of cylinder is 7:4
ls vuqikr Kkr dhft,A
3
and its volume is 4851cm then what is the
sum of area of two bases of this cylinder?
SSC Phase X 05/08/2022 (Shift- 03)
yEco`Ùkh; csyu ds
TSA vkSjCSA dk vuqikr 7%
(a)
10
(b)
5 gS vkSj bldk vk;ru 4851lseh
3
gS rks bl csyu d
17 17 nksuksa vk/kjksa ds {ks=kiQyksa dk ;ksx D;k
 Mensuration-3D

44. The ratio of the volume of two cylinders is  Folding and revolving a rectangular sheet
27: 25 and the ratio of their heights is 3 : 4
If the area of the base of the second cylinder vk;rkdkj 'khV dks eksM+uk vkSj ?kqekuk
is 3850 cm2, then what will be the radius of Rectangular sheet to be fold
the first cylinder?
nks flysaMjksa ds vk;ru dk vuqikr 27 % 25 gS vkSj
mudh ÅapkbZ dk vuqikr 3 % 4 gSA ;fn nwljs flysaMj ds b
vk/kj dk {ks=kiQy 3850 2lseh
gS] rks igys flysaMj dh
f=kT;k D;k gksxh\
SSC MTS 16/06/2023 (SHIFT-01) l
(a) 42 cm
r
(b) 56 cm h
h =b
h 2 r = l r
(c) 63 cm
r =
l
h =l, 2  r = b r = b
(d) 34 cm 2 2

r
45. What will be the total cost (in Rs.) of Folding along length Folding along Breadth

polishing the curved surface of a wooden

si
cylinder at rate of 50 per m 2 , if its
Rectangular sheet to be revolve
diameter is 70cm and height is 6 m?
22 
an by

 Take π = 7 

n
  b
,d ydM+h ds csyu dh oØ i`"B dks #- 50/m2 dh
ja
R s
nj ls ikWfy'k djus dh dqy ykxr (#- esa) D;k gksxh]
l
;fn bldk O;kl 70cm vkSj mQapkbZ 6m gS\
a th

l
SSC CGL 13/04/2022 (Shift- 02) l
h =b
(a) 612 b r =l b
ty a

(b) 675 h =l , r = b
di M

(c) 660 Along length Along Breadth

(d) 624
47. A rectangular piece of paper is 52 cm long
46. The cost of painting the total surface area and 22 cm wide. A cylinder is formed by
of a 30 m high solid right circular cylinder rolling the paper along its breadth. Find
at the rate of 25 per m² is Rs. 18,425. What the volume of the cylinder.
is the volume (in m³) of this cylinder (Take  = 22/7)
 22  dkxt dk ,d vk;rkdkj VqdM+k52 cm yEck vk
 Use   7  ?
22 cm pkSM+k gSA dkxt dks mldh pkSM+
30 ehVj Åaps Bksl yac o`Ùkkdkj flysaMj ds dqy,d csyu cuk;k tkrk gSA csyu dk vk;ru Kkr dhft,
lrg {ks=k dks 25 çfr oxZ ehVj dh nj ls isaV djus SSC CHSL 09/06/2022 (Shift- 3)
dh ykxr 18]425 #i;s gSA bl flysaMj dk vk;ru (a) 1870 cm3
A

3
(b) 2002 cm
 22 
(ehVj esa) D;k
3
gS\ Use   7  (c) 1030 cm3
(d) 1290 cm3
SSC CPO 03/10/2023 (Shift-02) 48. Two rectangle sheets of paper each 30cm
(a) 1210 X 18cm are made into two right circular
(b) 1155 cylinders, one by rolling the paper along its
 Mensuration-3D

çR;sd 30 lsehX 18 lseh ds dkxt ds nks vk;rkdkj Hollow Cylinder/[kks[kyk csyu

'khV dks nks yEco`Ùkh; csyuksa esa cuk;k x;k gSa] ,d
viuh yackbZ ds ifjr% dkxt dks jksy djds vkSj nwljk r
pkSM+kbZ ds ifjr% jksy djds bl çdkj cuh nks csyuksa
ds vk;ru dk vuqikr gS&
(a) 2 : 1 (b) 3 : 5
(c) 4 : 3 (d) 5 : 3 h
(i) If curved surface area (c) and volume (v) are
given then ratio of radius to height
;fn ik'oZ i`"Bh; {ks=kiQy
(c) rFkk vk;ru (v) fn, gks
rks f=kT;k dk ÅapkbZ ls vuqikr
R
r 8V²
=
h c3 (i) Curved Surface Area/ik'oZ i`"Bh; {ks=

r
49. The curved area of a cylindrical pillar is 264 = 2rh + 2Rh = 2h(R + r)
m2 and its volume is 924 m3. Find the ratio
(ii) Total Surface Area/dqy i`"Bh; {ks=kiQ

si
 2 
of its diameter to its height.  Take   = 2h(R + r) + 2(R² – r²)
an by 7
(iii) Volume of material of hollow Cylinder
,d csyukdkj LraHk dk ik'oZ i`"Bh; {ks=kiQy
264 m2
gS vkSj bldk vk;ru924 m3 gSA blds O;kl vkSj [kks[kys csyu esa mi;ksx fd, x, inkFkZ dk

n
= (R² – r²)h
bldh Å¡pkbZ dk vuqikr Kkr dhft,A
(iv) Mass (weight) of hollow cylinder/[kks[
ja
R s
4 csyu dk æO;eku (Hkkj)
(a)
7
= Density × Volume of material/?kuRo
a th

×
7 inkFkZ dk vk;aru
(b)
4 51. A hollow cylindrical tube made of plastic
ty a

is 3 cm thick. If the external diameter is

3
(c) 20 cm and length of the tube is 49 cm, find
7
the volume of the plastic.
di M

(d)
7 IykfLVd ls cuh ,d •ks•yh csyukdkj VÔwc 3
3 eksVh gSA ;fn ckgjh O;kl 20 lseh gS vkSj
(ii) If curved surface area (c) and height (h) are yackbZ 49 lseh gS] rks IykfLVd dk vk;ru Kk
given the volume of cylinder/ ;fn ik'oZ i`"Bh; SSC CPO 05.10.2023 (Shift-3)
{ks=kiQYk
(c) rFkk ÅapkbZ
(h) fn, gks rks csyu dk vk;ru (a) 7644 cm2 (b) 7238 cm3
(c) 7854 cm3 (d) 7546 cm3
c2
V 52. The volume of a metallic cylindrical pipe
4 h
is 3564 cm3. If its external radius is 12 cm
50. A right circular cylinder of height 16 cm is and thickness is 3 cm, then the length of
A

covered by a rectangular tin foil of size

16cm × 22 cm. The volume of the cylinder  22 
the pipe will be:  Take  = 7 
is:  

16 lseh ÅapkbZ dk ,d yac o`Ùkkdkj csyu 16×lseh

22
,d /krq ds csyukdkj ikbi dk vk;ru 3564 lseh 3
gSA ;fn bldh ckgjh f=kT;k 12 lseh vkSj ek
lseh vkdkj ds ,d vk;rkdkj fVu dh iryh iUuh ls
lseh gS] rks ikbi dh yackbZ gksxh%
<dk gqvk gSA csyu dk vk;ru gS%
SSC CHSL 04/08/2021 (Shift- 2)
 Mensuration-3D

53. A hollow iron pipe is 35 cm long and its 56. Perimeter of a base of circular cylinder is
external diameter is 14 cm and the 35cm and CSA is 9660 cm2, a thread is
thickness of the pipe is 1 cm and the iron wound on a cylinder such that it makes
used to make the pipe weight 10 g/cm³, exactly 23 turns around the cylinder then
then the weight of the pipe in kg is: find length of string?

 use  
22 
 yEco`Ùkh; csyu ds vk/kj dh ifjf/ 35 lseh gS v
7 
oØ i`"B {ks=kiQy 96602 lseh
gS] ,d /kxk bl çdkj
,d •ks•ys yksgs dk ikbi 35 lseh yack gSA bldh yisVk gqvk gS fd ;g yEco`Ùkh; csyu ds pkj
ckgjh O;kl 14 lseh gS vkSj ikbi dh eksVkbZ 1 lseh gS23 pDdj yxkrk gS fiQj /kxs dh yackbZ gksxh
ikbi dks cukus esa yxs yksgs dk otu 10 xzke@lseh
3
(a) 851 cm
 22  (b) 828 cm
gS] rks ikbi dk otu fdyks esa gS%
 use  
7 

(c) 1380 cm
CRPF HCM 26/02/2023 (Shift - 03) (d) 925 cm

Right Circular Cone/le o`Ùkh; 'kad

(a) 13.4 (b) 15.2

r
(c) 12.6 (d) 14.3
The solid obtained by revolving a right-

si
54. A hollow cylinder is made up of metal. The
angled triangle about oneb of its sides
difference between outer and inner curved
an by (other than hypotenuse) is called a cone or
surface area of this cylinder is 352 cm2.
right circular cone.
Height of the cylinder is 28cm. If the total
fdlh ledks.k f=kHkqt dks mldh fdlh ,d Hkqtk (

n
surface area of this hollow cylinder is 2640
cm2, then what are the inner and outer ds vfrfjÙkQ) ds ifjr% ifjØe.k djus ij çkIr Bk
radius (in cm)? dks 'kadq ;k yac o`Ùkh; 'kadq dgrs gSaA
ja
R s
,d •ks•yk csyu /krq ls cuk gSA bl csyu ds ckgjh Let the right angled triangle ABC be
vkSj vkarfjd oØ i`"B ds {ks=kiQy dk varj 352lseh
2 revolved about its side AB to form a cone;
a th

then AB is the height (h) of the cone

gSA csyu dh ÅapkbZ 28 lseh gSA ;fn bl •ks•ys csyu
formed, BC is the radius (r) of its base and
dk dqy i`"B {ks=kiQy 26402 gS]
lseh rks vkarfjd vkSj AC is the slant height(l).
ckgjh f=kT;k (lseh esa) D;k gSa\
ty a

eku yhft, fd ,d 'kadq cukus ds fy, ledks.k f=kH

(a) 4, 6 (b) 10, 12 ABC dks mldh Hkqtk AB ds pkjksa vksj ?kqek;
di M

(c) 8, 10 (d) 6, 8 gS_ rcAB cus 'kadq dh ÅapkbZ(h) gS]BC blds vk/
55. A field roller, in the shape of a cylinder, kj dh f=kT;k(r) gS vkSj
AC frjNh ÅapkbZ (l) gSA
1 A
has a diameter of 1 m and length of 1
4
m. If the speed at which the roller rolls is
14 revolutions per minute, then the l
maximum are (in ) that it m2 can roll in 1 h
hour is :
(Take  = 22/7)
,d eSnkuh jksyj] tks cysu ds vkdkj dk gS] mldk r
A

B C
1
O;kl 1 eh vkSj yackbZ
1 eh gSA ;fn bl jksyj ds (i) Slant height/fr;Zd Å¡pkbZ
l = r²  h²
4
?kweus dh pky 14 pDdj izfr feuV gS] rks 1 ?kaVs esa(ii) Curved Surface Area/ ik'oZ i`"Bh; {ks=
;g vf/dre fdrus {ks=kiQy (oxZ eh esa) ij ?kwe ldrk = rl
gS\ ( = 22/7 ysa) (iii) Total Surface area/dqy i`"Bh; {ks=kiQy
SSC CGL TIER II (12/09/2019) = r(r + l)
 Mensuration-3D

57. If the height of a cone is 7 cm and the 62. A heap of wheat is in the form of a cone
diameter of the circular base is 12 cm, whose base diameter is 8.4 m and height
then its volume is (nearest to integer):
is 1.75 m. The heap is to be covered by
;fn ,d 'kadq dh ÅapkbZ 7 lseh- gS] vkSj blds
canvass. What is the area (in m²) of the
o`Ùkkdkj vk/kj dk O;kl 12 lseh- gS] rks bldk
canvas required? (Use =22/7)
vk;ru D?;k gksxk\ (fudVre iw.kkZad rd)
SSC CPO 03/10/2023 (Shift-01) xsgwa dk ,d <sj ,d 'kadq ds vkdkj dk gS ft
(a) 264 cm3 (b) 284 cm3 vk/kj O;kl 8-4 ehVj vkSj ÅapkbZ 1-75 ehVj
dks dSuokl ls <duk gSA dSuokl dk {ks=kiQy)
3
(c) 274 cm (d) 254 cm3 2
58. If the slant height of a cone is 29 cm and
Kkr djsaA
(=22/7 dk iz;ksx djsa
)
its height is 20 cm, find the ratio between
the magnitudes of total surface area and SSC CGL 16/08/2021 (Shift 02)
the volume. (a) 60.06 (b) 115.05
;fn ,d 'kadq dh frjNh ÅapkbZ 29 lseh gS vkSj bldh
(c) 60.6 (d) 115.5
ÅapkbZ 20 lseh gS] rks dqy lrg {ks=k vkSj vk;ru ds
63. A conical shape vessel has a radius of 21
ifjek.k ds chp vuqikr Kkr djsaA

r
cm and has a slant height of 25 cm. If the
SSC CPO 05/10/2023 (Shift-03) curved part of the vessel is to be painted

si
(a) 3 : 7 (b) 3 : 14 white, find the cost (in `) of painting at
(c) 5 : 14 an by (d) 7 : 15 t he r at e of `1.5 per cm2.
59. If the ratio of the base radius to the height ,d 'kadkdkj crZu dh f=kT;k 21 cm vkSj fr;Z
of a cone is 7 : 9, and the area of the base Å¡pkbZ 25 cm gSA ;fn crZu ds oØh; Hkkx

n
of the cone is 1386 cm2. then the volume lisQn jax ls isaV fd;k tkrk gS] `1.5
rks izfr cm2
of the cone (in cm3) is:
dh nj ls isafVax dh ykxr `( esa) Kkr dhft,A
ja
;fn ,d 'kadq ds vk/kj dh f=kT;k dk mldh Å¡pkbZ ls
R s
SSC CPO 03/10/2023 (Shift-02)
vuqikr7 : 9 gS vkSj 'kadq ds vk/kj dk {ks=kiQy
1386
a th

(a) 2475 (b) 825

lseh2 gS] rks 'kadq dk vk;ru (lseh
3
esa) fdruk gksxk\
(c) 1680 (d) 1250
ICAR Mains, 10/07/2023 (Shift-1)
64. The radius of the base of a conical tent is
(a) 12,474 (b) 13,652
8 m and its height is 15 m, what is the cost
ty a

(c) 12,768 (d) 13,125 of the material needed to make it if it costs

60. The height and the radius of the base of a Rs 54 per  m2?
di M

right circular cone are in the ratio of 12:

5. If its volume is 314 cm³, then what is ,d 'kaDokdkj racw ds vk/kj dh f=kT;k 8 eh
the slant height of the cone? (Use  = 3.14) vkSj bldh ÅapkbZ 15 ehVj gS] ;fn bldh ykx
,d yac o`Ùkh; 'kadq dh ÅapkbZ vkSj mlds vk/kj #i;s çfr ehVj gS rks bls cukus ds fy, vko';
2

dh f=kT;k dk vuqikr 12% 5 gSA ;fn bldk vk;ru lkexzh dh ykxr D;k gS\
314 lseh gS] rks 'kadq dh fr;Zd ÅapkbZ D;k gS \
3 SSC MTS 08/05/2023 (Shift-02)
(  ¾ 3-14 dk ç;ksx djsa) (a) Rs.6454 (b) Rs.7344
SSC CPO 05/10/2023 (Shift-01) (c) Rs.8678 (d) Rs.7454
(a) 11 cm (b) 14 cm 65. How many metres of 2-m-wide cloth will
(c) 12 cm (d) 13 cm be required to make a conical tent with
61. The circumference of the base of a right the diameter of the base as 14 m and slant
A

circular cone is 44 cm and its height is 24 height as 9 m ignore wastage?

cm. The curved surface area ( in cm2) of ds vk/kj&O;kl vkSj9 m dh fr;Zd Å¡pk
14 m
the cone is : okys fdlh 'kaDokdkj VasV dks cukus ds 2mf
,d yac o`Ùkh; 'kadq ds vk/kj dh ifjf/ 44 lseh dh pkSM+kbZ okys fdrus ehVj diM+s dh
gS vkSj bldh ÅapkbZ 24 lseh gSA 'kadq dk oØ i`"Bh;
gksxh] ;fn viO;; dks ux.; eku fy;k tk,\
{ks=kiQy (lseh
2
esa) gS %
SSC CGL 12/12/2022 (Shift- 01)
 Mensuration-3D

66. The volume of conical tent is 1232cm3 and 70. The height of a cone is three times the
area of base is 154cm2. Find the length of radius of its base and its total surface area
canvas required to build the tent whose
is 36  10  1 cm2. What is the volume
width is 2 m?
fdlh 'kaDokdkj rEcw dk vk;ru 1232 lseh
3
gS] vkSj (in cm3) of the cone?
blds vk/kj dk {ks=kiQy 154 lseh
2
gS] rc rECkw ds ,d 'kadq dh ÅapkbZ mlds vk/kj dh f=kT;k ls rh
dSuokl dh yEckbZ D;k gksxh ;fn bldh pkSM+kbZ gS2 vkSj mldk laiw.kZ i`"Bh; {ks=kiQy
36  10  1
ehVj gS\ oxZ lseh- gSA 'kadq dk vk;ru (lseh
3
esa) fdruk gks
(a) 255 m (b) 265 m
(c) 275 m (d) 225 m ICAR Mains, 07/07/2023 (Shift-2)
67. The volume of a right circular cone is 308 (a) 288  (b) 125 
cm³ and the radius of its base is 7 cm. (c) 216  (d) 144 
What is the curved surface area (in cm²) of
71. A semi-vertical angle of a right circular
 22  cone is 60º, and its slant height is 3 cm.
the cone?  Take = 
7  Find the ratio of the height of cone and
,d yac o`Ùkh; 'kadq dk vk;ru 308 lseh3
gS vkSj radius of the base of cone?

r
blds vk/kj dh f=kT;k 7 lseh gSA 'kadq dk oØ i`"Bh; ,d yEc o`Ùkh; 'kadq dk v/Z&ÅèokZ/j dks.k º gS

si
{ks=kiQy (lseh
2
esa) fdruk gSaA vkSj bldh frjNh Å¡pkbZ3 lseh gSA 'kadq dh Å
SSC CGL MAINS 03/02/2022
vkSj 'kadq ds vk/kj dh f=kT;k dk vuqikr Kkr d
(a) 22 21
an by (b) 44 21
CRPF HCM 01/03/2023 (Shift - 02)

n
(c) 22 85 (d) 11 85
(a) 3 : 1 (b) 1 : 3
68. Volume of a cone whose radius of base and
ja
height are r and h respectively, is 400 cm3. (c) 2 : 3 (d) 3 : 2
R s
What will be the volume of a cone whose 72. The radii of the base of a cylinder and a
cone are in the ratio 3 : 2 and their
a th

radius of base and height are 2r cm and h

cm respectively? heights are in the ratio 2 : 3. Their
,d 'kadq] ftlds vk/kj dh f=kT;k rFkk Å¡pkbZ Øe'k%
r volumes are in the ratio of
rFkkh gS] dk vk;ru 400 lseh3 gSA ,d 'kadq] ftlds ,d csyu vkSj 'kadq ds vk/kj dh f=kT;k
3 : 2 ds
ty a

vk/kj dh f=kT;k rFkk Å¡pkbZ Øe'k%

2r lseh rFkkh lseh vuqikr esa gSa vkSj mudh 2Å¡pkbZ
: 3 ds vuqikr e
gS] dk vk;ru D;k gksxk\ gSaA vk;ruksa dk vuqikr gSa\
di M

SSC CGL MAINS (08/08/2022) (a) 3 : 2

(a) 1000 cm³ (b) 1200 cm³ (b) 33 : 2
(c) 1600 cm³ (d) 800 cm³ (c) 3 : 22
69. The curved surface area of a right circular
(d) 2 : 6
cone is 2310cm² and its radius is 21 cm.
If its radius is increased by 100% and 73. If the radius of the base of a cone is
height is reduced by 50%, then its capacity doubled, and the volume of the new cone
is three times the volume of the original
(in litres) will be correct to one decimal
cone, then what will be the ratio of the
 22  height of the original cone to that of the
place)  Take = 7 
new cone?
,d yEc o`Ùkh; 'kadq dk oØ i`"Bh; {ks=kiQy 2310
A

;fn 'kadq ds vk/kj dh f=kT;k nksxquh gks tkrh

lseh2 gS vkSj bldh f=kT;k 21 lseh gSA ;fn bldh u, 'kadq dk vk;ru ewy 'kadq ds vk;ru ls rhu x
f=kT;k 100» c<+k nh tk, vkSj ÅapkbZ 50» de dj gS] rks u, 'kadq ds ewy 'kadq dh Å¡pkbZ d
nh tk,] rks bldh {kerk (yhVj esa) ,d n'keyo D;k gksxk\
LFkku rd lgh gksxh\ (a) 9 : 4
SSC CGL MAINS 03/02/2022 (b) 4 : 3
 Mensuration-3D

74. The numerical values of the volume and 'kadq dk oØ i`"Bh; {ks=kiQy
= f=kT;•aM dk {ks
the area of the lateral surface of a right
circular cone are equal. If the height of the r 2
=
cone be h and radius be r, the value of 4
1 1 Volume of cone/'kadq dk vk;ru
2
 2
h r
2
fdlh 'kadq ds oØ i`"B ds {ks=kiQy vkSj blds vk;ru 1 r 15r 15r 3
    
3 4 4 192
dk la[;kRed eku cjkcj gS] ;fn 'kadq dh Å¡pkbZ
h
1 1 2. A semicircular sector of radius r cm is rolled
vkSj f=kT;k
r gS] rc  dk eku Kkr djs\ into a cone.
h2 r 2
( r lseh f=kT;k ds fdlh v/Zo`Ùkkdkj f=kT;[kaM dks ?
4 3
(a) (b) 'kadq)
15 1
1 1 r
(c) (d)
6 9 l

r
h
Rolled Up
 Cone formed by rolling up a sector

si
f=kT;[kaM dks ?kqekus ijdqcuk 'ka r1
an by 2r
When a sector is rolled up in such a way = r
2
that the two binding radii are joined
together then a cone generates.
Height of cone 'kadq dh ÅapkbZ
(h)

n
tc fdlh f=kT;[kaM dks bl izdkj ?kqek;k tkrk gS fd r2 3r
= l 2 – r1 ² =
bldh nks f=kT;k,a tqM+ tkrh gS] rks ,d 'kadq dk fuekZ.k r2 – =
ja 4 2
R s
gksrk gSA
Curved surface area of cone = Area of sector
a th

1. A right angled sector of radius a cm is rolled

up into a cone r 2
=
a lseh f=kT;k ds ,d ledks.kh; f=kT;[kaM dks ?kqekdj 2
cuk 'kadq
ty a

r 2
'kadq dk oØ i`"Bh; {ks=kiQy ¾ f=kT;•aM dk={ks=kiQ
2
di M

r
Volume of cone ('kadq dk vk;ru)
90º
l
1 1 r2 3r r 3
h = r1 ² h =  × × =
Rolled Up 3 3 4 2 8 3
r
4 = 
2
r

Note: Canvas required to construct a

2

r'
conical tent
= Curved surface area of cone
75. A semicircular sheet of diameter 32cm is
r r bent into a conical cup. Find the depth of
 2r'   r'  & l  r
2 4 cup?
A

Height of cone ('akdq dh ÅapkbZ)

(h) 32 lseh O;kl okys v¼Zo`Ùkkdkj dkxt dks
,d 'kaqd cuk;k tkrk gS rc bl 'kadq dh xgjkbZ
 l 2 – r'²
gksxh\
r2 15r (a) 85
 r2 –
16 4
(b) 83
Curved surface area of cone = Area of sector
 Mensuration-3D

76. A sector of radius 10.5 cm with the central (i) Volume of frustum (fNUud dk vk;ru)
angle 120º is formed to form a cone by
joining the two bounding radii of the 1
sector. What is the volume (in cm3) of the = (R² + r² + Rr) h
3
cone so formed?
(ii) Curved surface area (ik'oZ i`"Bh; {ks=k
10-5 lseh f=kT;k vkSjº120
dsaæh; dks.k okys ,d f=kT;k&
= (R + r)l
[kaM (lsDVj) dks ,d 'kadq cukus ds fy,] bldh nks
(iii)Total surface area (dqy i`"Bh; {ks=kiQy)
lhed f=kT;kvksa dks feykdj eksM+k tkrk gSA cus gq,
= l(R + r) +  (R² + r²) Where l
'kadq dk vk;ru D;k gS\ (lseh
3
esa)
= h2  (R – r)2
343 3 343 3
(a)  (b)  79. The radii of the ends of a frustum of a cone
12 6
7 cm height are 5 cm and 3 cm. Find its
343 2 343 2 volume correct to one decimal place.
(c)  (d) 
12 6
77. From a circular sheet of paper of radius
 22 
 Use = 
25cm, a sector area 60% is removed. If the  7 

r
remaining part is wood to make a conical
7 cm Å¡pkbZ okys ,d 'kadq ds fN=kd ds fl
cup. Then find the ratio of height and

si
radius of cone? f=kT;k,¡5 cm vkSj 3 cm gSaA bldk vk;ru K
f=kT;k 25 lseh ds dkxt dh ,d o`Ùkkdkj 'khV ls
an by dhft, tks n'eyo ds ,d LFkku rd lgh gksA
,d o`Ùk[kaM ftldk {ks=kiQy 60» gS] dkV dj fudky  22 
fn;k x;k gS ;fn 'ks"k Hkkx ls ,d 'kaDokdkj di  = dk iz;ksx dhft,

n
 7 
cuk;k x;k gSA di dh Å¡pkbZ vkSj f=kT;k dk vuqikr
SSC CGL 12/12/2022 (Shift- 02)
Kkr dhft,\
ja (a) 345.6 cm³ (b) 359.3 cm³
R s
(a) 19 : 6 (b) 24 : 7
(c) 379.3 cm³ (d) 369.3 cm³
(c) 4 : 3 (d) 21 : 2
a th

80. The lateral surface area of frustum of a right

78. A right angle triangle whose sides are 15cm
and 20cm (other than hypotenuse) is made circular cone, if the area of its base is 16
to revolve about its hypotenuse. Find the cm2 and the diameter of circular upper
volume and surface area of double cone so surface is 4 cm and slant height 6 cm, will be
ty a

formed? ,d yEc o`Ùkh; 'kadq ds fNUud ds vk/kj dk {k

,d ledks.k f=kHkqt ftldh Hkqtk,¡ 15 lseh vkSj 20 16 lseh2 gS vkSj o`Ùkkdkj Åijh lrg dk O;kl 4
di M

lseh (d.kZ ds vykok) gSa] dks mlds d.kZ ds pkjks vkSj frjNh Å¡pkbZ 6 lseh gS] rks ik'oZ i`"Bh; {k
vksj ?kqek;k tkrk gSA bl izdkj cus nksuksa 'kadqvksa
(a) ds
30 cm2 (b) 48 cm2
vk;ru vkSj i`"B {ks=kiQy Kkr dhft,\ (c) 36 cm 2 (d) 60 cm2
(a) 1200, 1320 (b) 1600, 1320  All triangles formed by cutting cone are
(c) 1250, 1444 (d) 1000, 1260 similar to each other.
'kadq dks dkVus ls cus lHkh f=kHkqt ,d nwljs
Frustum/fNUud
gksrs gSaA
If a cone is cut by a plane parallel to its base, O
the portion of solid between this plane and
the base is known as frustum of the cone. l
h
(tc ,d 'kadq dks blds vk/kj ds lekUrj dkVk tkrk gS rks
A

L
uhps okyk Hkkx fNUud dgykrk A gS) H
r
B
A

h1 l1
h1 D
h2 rr11 l2 r C
Frustum R
(i) OCD  OAB  A  C  90, O  O 
 Mensuration-3D

 Let V is volume of larger cone and v is Surface area (i`"Bh; {ks=kiQy)

= 4r²
volume of smaller cone
4
(ekuk] V cMs+ 'kadq dk vk;ru rFkk
v NksVs 'kadq dk Volume (vk;ru) = r³
3
vk;ru gS)
1
 Let v1 and v2 be volume and S1 and S2 be
R 2 H area of two sphere then
V 3 R2H
(ii)  
v 1 2 r2h ekukv1 rFkkv2 vkSjS1 rFkkS2 nks xksys ds vk
r h
3 vkSj {ks=kiQy gSaA
V H3 R3 L3 3/2 2/3
(iii)    v1  S1  S1  v1 
v h3 r3 l3  or 
81. The height of a cone is 40 cm. If a small v 2  S2  S2  v 2 
cone is cut off at the top by a plane parallel
to the base of the cone, the volume of the 83. If the volume of a sphere is 38808 cm3,
then its surface area is:
1
smaller con is
64
the volume of the larger ;fn fdlh xksys dk vk;ru 38808 lseh3 gS] rks ml
i`"Bh; {ks=kiQy gS%

r
cone. Find the height of the frustum.
,d 'kadq dh Å¡pkbZ 40 lseh gSA ;fn 'kadq ds vk/kj

si
SSC CPO 04/10/2023 (Shift-3)
ds lekukarj ,d lery }kjk 'kh"kZ ij ,d NksVk 'kadq
an by (a) 5564 cm2 (b) 5544 cm2
dkVk tkrk gS] rks NksVs 'kadq dk vk;ru cM+s 'kadq(c)
dk 5554 cm2 (d) 5574 cm2

n
1
vk;ru gSA fNUud dh Å¡pkbZ Kkr djsaA 84. If the diameter of a sphere is reduced to
64 ja its half, then the volume would be:
CRPF HCM 23/02/2023 (Shift-03)
;fn fdlh xksys dk O;kl ?kVkdj vk/k dj fn;
R s
(a) 5 (b) 15
(c) 10 (d) 30
tk,] rks mlds vk;ru ij D;k çHkko iM+sxk\
a th

82. If a right circular cone is separated into SSC CPO 04/10/2023 (Shift-02)
solids of volumes V1, V2, V3 by two planes
parallel to the base which also trisect the 1
(a) Reduced by of the former volume
altitude, then V1 : V2 : V3 is- 8
ty a

vxj ,d yac o`Ùkh; 'kadq dks vk/kj ds lekarj nks

1
leryksa }kjk rhu Bksl Hkkxksa esa ck¡V fn;k tkrk gS ftuds
di M

(b) Increased by of the former volume

4
vk;ru Øe'k% V1, V2, V3 gS] tks mlds yac (Å¡pkbZ)
dks Hkh rhu cjkcj Hkkxksa esa ck¡Vrk gS
V1 : V 2
:rc
V3 1
(c) Reduced by of the former volume
dk eku gksxk& 4
(a) 1 : 2 : 3 (b) 1 : 4 : 6
1
(c) 1 : 6 : 9 (d) 1 : 7 : 19 (d) Increased by of the former volume
8
Sphere/xksyk 85. The sum of the radii of spheres A and B is
A sphere is a solid obtained on revolving a 14 cm, the radius of A being larger than
circle about any diameter of it. that of B. The difference between their
xksyk vius fdlh Hkh O;kl ds pkjksa vksj ,d o`Ùk dks surface area is 112 . What is the ratio of
volumes of A and B?
A

?kqekus ij çkIr ,d Bksl gSA

xksykA vkSjB ds f=kT;k dk ;ksx 14 lseh gS]
A dh
r f=kT;k
B dh rqyuk esa cM+h gSA muds lrg {k
chp dk varj 112 gSAA vkSj B ds vk;ru dk
vuqikr D;k gS\
d
CGL MAINS 15/10/2020
 Mensuration-3D

86. The sum of radii of two spheres is 10 cm

4
and the sum of their volumes is 880cm3. its volume = (R³ – r³) = Volume of
What will be the product of their radii ? 3
material in spherical shell
nks xksyks dh f=kT;kvksa dk ;ksx 10 lseh gS] vkSj muds
vk;ruksa dk ;ksxiQy 880 lseh
3
gS] rc mudh f=kT;kvksa 4
dk xq.kuiQy Kkr djsA bldk vk;ru = (R³ – r³) = xksykdkj •ksy
3
2 1
(a) 25 cm² (b) 26 cm² inkFkZ dk vk;ru
3 3
1 1 Total surface area = 4(R² – r²)
(c) 33 cm² (d) 27 cm²
3 3
87. The cost of whitewashing the surface area of dqy i`"Bh; {ks=kiQy
= 4(R² – r²)
a sphere is ` 8,393 at the rate of ` 54.5/cm2.
89. What is the volume (in cm³) of a spherical
What is the volume of the sphere (in cm3)? shell whose inner and outer radii are
(Round off your answer to the nearest respectively 2 cm and 3 cm?
22
intege) [Use  = ]
7 ml xksydkj 'ksy dk vk;ru (lseh3 esa) Kkr dj

r
,d xksys ds i`"Bh; {ks=kiQy ij lisQnh djkus dh`ykxr ftldh vkarfjd vkSj ckgjh f=kT;k Øe'k% 2 lseh
54.5 izfr oxZ ehVj dh nj ls ` 8,393 gSA xksys dk

si
3 lseh gSa\
vk;ru (?ku lseh esa)fdruk gksxk\ (vius mÙkj dks fudVre
an by 22 76 
iw.kk±d rd iw.kk±fdr djsa)
 = ( iz;ksx djsaA)
7
(a)
3

n
ICAR Mains, 08/07/2023 (Shift-1)
(a) 192 ja (b) 180 106 
(c) 185 (d) 174 (b)
R s
3
88. When the radius of a sphere is increased by
5cm, its surface area increases by 704 cm2.
a th

The diameter of the original sphere is 56 

(c)
tc ,d xksys dh f=kT;k esa 5 lseh dh o`f¼ dh tkrh 3
gS] rks blds i`"Bh; {ks=kiQy esa 2704
dh o`f¼
lseh gks
tkrh gSA okLrfod xksys dk O;kl fdruk gS\ (eku 86 
ty a

(d)
22 3
yhft, = )
di M

7 90. A hollow spherical shell is made of a metal

(a) 8.2 cm (b) 6.8 cm of density 2.5 g/cm3. If the external and
(c) 5.2 cm (d) 6.2 cm
the internal radii of the given sphere are
Spherical Shell/[kks[kyk xksyk 35 cm and 14 cm, respectively, find the

It is solid encosed between two concentric  22 

mass of the shell. Use =
spheres.  7 
;g nks ladsafær xksyksa ds chp f?kjk gqvk Bksl gSA
,d [kks[kyk xksykdkj lsy 2-5 g/cm3 ?kuRo okyh
Let R be the external radius and r be the
krq ls cuk gSA ;fn fn, x, lsy dh okÞ; vkSj vkar
internal radius of a spherical shell then
f=kT;k,a Øe'k% 35 lseh vkSj 14 lseh gSa] rk
ekukR ckgjh f=kT;k gS vkSj
r ,d xksykdkj dks'k dh æO;eku Kkr dhft,A
A

vkarfjd f=kT;k gS] rks

SSC MTS 25/07/2022 (Shift- 3)

(a) 398567 g
r
(b) 324120 g
R (c) 440040 g
 Mensuration-3D

Hemisphere/v¼Zxksyk yksgs ls cus ,d v/Zxksyh; dVksjs dk vkarfjd

 22 
When a solid sphere is cut through its 84 lseh gSA 21 #i;s çfr 100 lseh
2  =
 y sa


center into two equal (identical) piece, each  7 
piece is called a hemisphere. dh nj ls n'keyo ds nks LFkkuksa rd vanj dh
tc ,d Bksl xksys dks mlds dsaæ ls nks cjkcj (leku) fVu p<+kus dh ykxr Kkr dhft,A
VqdM+ksa esa dkVk tkrk gS] rks çR;sd VqdM+s dks xksyk¼Z SSC CGL 02/12/2022 (Shift- 01)
vFkok v¼Zxksyk dgk tkrk gSA (a) Rs.2,328.48 (b) Rs.2,425.48
(c) Rs.2,425.60 (d) Rs.2,355.48
r
94. The diameter of a hemisphere is equal to
the diagonal of a rectangle of length 4 cm
and breadth 3 cm. Find the total surface
area (in cm2) of the hemisphere.
,d v/Zxksys dk O;kl 4 lseh yackbZ vkSj 3 lseh
okyk ,d vk;r dk fod.kZ ds cjkcj gSA xksyk/Z

r
(i) Curved Surface area (ik'oZ i`"Bh; {ks=kiQy) dqy i`"Bh; fiQYe (lseh
2
esa) Kkr fdvksA

si
= 2r² SSC CPO 04/10/2023 (Shift-3)
(ii) Tot al sur face ar ea (dqy i`"Bh; {ks=kiQy)
=
3r²
an by (a) 25 (b)
50 
4

n
2
(iii) Volume (vk;ru) = r³
3 25 75
ja (c) (d)
91. The total surface area of a solid 4 4
R s
hemisphere is 4158 cm2. Find its volume 95. If the radius of a hemispherical balloon
(in cm3).
a th

increases from 4 cm to 7 cm as air is

,d Bksl v¼Zxksys dk laiw.kZ i`"Bh; {ks=kiQy 4158
pumped into it, find the ratio of the
lseh2 gSA bldk vk;ru (lseh3 esa) Kkr dhft, surface area of the new balloon to its
SSC CPO 05/10/2023 (Shift-01) original.
ty a

(a) 9702 (b) 19404 ;fn ,d v/Zxksykdkj xqCckjs esa gok Hkjus i
(c) 1848 (d) 462 f=kT;k 4 lseh ls 7 lseh rd c<+ tkrh gS] rk
di M

92. The radius of hemisphere is 14cm. What is xqCckjs ds lrg {ks=k dk mlds ewy ls vuq
the cost of painting the outer curved surface dhft,A
of the hemisphere at the rate of Rs. 45 per SSC CPO 04/10/2023 (Shift-01)
22  (a) 20 : 49 (b) 16 : 21

sq.cm?   =  (c) 49 : 16 (d) 21 : 12
7
Hemispherical Shell/v¼Zxksyh; 'k
,d v/Zxksys dh f=kT;k 14 lseh gSA bl v/Zxksys ds
ckgjh oØ i`"B dks 45 :i;s izfr oxZ lseh dh nj ls
R r
jaxus dh ykxr Kkr djsaA
SSC MTS 8/08/ 2019 (Shift-01)
A

(a) Rs. 53160 (b) Rs. 55440

(c) Rs. 56820 (d) Rs. 58280
93. A hemispherical bowl made of iron has (i) Curved Surface area (ik'oZ i`"Bh; {ks=k
inner diameter 84 cm. Find the cost of
= 2(R2 – r2)
tin plating it on the inside at the rate of
(ii) Total surface area (dqy i`"Bh; {ks=kiQ=
 22  3(R² – r²)
 Mensuration-3D

96. The internal diameter of a hollow  When we transform one object into another
hemispherical vessel is 24 cm. It is made object, volume remains constant.
of a steel sheet which is 0.5 cm thick.
What is the total surface area (in cm2) of tc ge ,d oLrq dks nwljh oLrq esa cnyrs gSa] rks
the vessel ? fLFkj jgrk gSA
,d •ks•ys v¼Zxksyh; crZu dk vkrafjd O;kl 24 lseh(i) Concept of Melting & Recasting
gSA ;g bLikr dh pknj ls cuk gqvk gS ftldh eksVkbZ fi?kyus vkSj iqujZpuk dh vo/kj.kk
0-5 lseh gSA bl crZu dk dqy i`"B {ks=kiQy (oxZ lseh
esa) Kkr djsaA Volume of Melted object/fi?kyh gqbZ oLr
vk;ru = Volume of recasted object/iqujZp
SSC CGL TIER II (13/09/2019)
ds ckn oLrq dk vk;ruA
(a) 612.75  (b) 468.75 
100. A cylindrical metallic rod of diameter 2 cm
(c) 600.2  (d) 600.5  and length 45 cm is melted and converted
97. The internal and external radii of a into wire of uniform thickness and length
hollow hemispherical vessel are 6 cm and 5 m. The diameter of the wire is: O;kl 2 lse

r
7 cm respectively. What is the total vkSj yackbZ 45 lseh dh ,d csyukdkj /krq dh
surface are (in) of the vessel cm2 ? dks fi?kyk;k tkrk gS vkSj ,d leku eksVkbZ

si
,d [kks[kys v¼Zxksyh; crZu dh vkarfjd vkSj oká ehVj yackbZ ds rkj esa ifjofrZr fd;k tkrk gSA
an by
f=kT;k Øe'k% 6 lseh vkSj 7 lseh gSA bl crZu dk O;kl gS%
dqy i`"B {ks=kiQy (oxZ lseh esa) Kkr djsaA SSC CPO 04/10/2023 (Shift-3)

n
SSC CGL TIER II (11/09/2019)
ja (a) 5 mm (b) 3 mm
R s
(a) 183  (b) 189  (c) 2 mm (d) 6 mm
(c) 177  (d) 174  101. A solid sphere made of wax of radius 12
a th

cm is melted and cast into solid

98. A metallic hemispherical bowl is made up
hemispheres of radius 4 cm each. Find the
of steel. The total steel used in making the
bowl is 342cm3. The bowl can hold 144 number of such solid hemispheres.
ty a

cm3 water. What is the thickness (in cm3) 12 lseh f=kT;k okys ekse ls cus ,d Bksl xk
of bowl and the curved surface area (in
fi?kykdj 4 lseh f=kT;k okys Bksl xksyk/ks± es
di M

cm2) of outer side?

gSA ,sls Bksl xksyk/ks± dh la[;k Kkr dhft,A
,d v¼Z xksykdkj ckmy LVhy dk cuk gqvk gS] bl
SSC CPO 03/10/2023 (Shift-3)
ckmy dks cukus esa342
lseh3 LVhy dk mi;ksx gqvk
gS] nl ckmy esa 144
 lseh3 ikuh vk ldrk gS] bl (a) 14 (b) 54
ckmy dk eksVkbZ vkSj ckgjh i`"B dk oØ i`"B {ks=kiQy(c) 28 (d) 27
Kkr dhft,\ 102. Three solid iron cubes of edges 4 cm, 5 cm
(a) 6162 (b) 3162 and 6 cm are melted together to make a
new cube. 62 cm3 of the melted material
(c) 681 (d) 381 is lost due to improper handling. The area
99. A sphere and another solid hemisphere (in cm2) of the whole surface of the newly
have the same surface area. The ratio of formed cube is
A

their volumes is:

4 lseh] 5 lseh vkSj 6 lseh fdukjksa ds rhu Bh
,d xksys vkSj nwljs Bksl xksyk/Z dk i`"Bh; {ks=kiQy
ds ?ku ,d u;k ?ku cuku ds fy, ,d lkFk fi?kyk,
leku gSA muds vk;ru dk vuqikr gS% tkrs gSaA fi?kyh lkexzh dk 62 3
vuqfpr
lseh gSaM
SSC CPO 05/10/2023 (Shift-2) ds dkj.k [kjkc tkrk gSA uoxfBr ?ku dh ijh l
dk {ks=k (lseh
2
esa) gS
(a) (b)
 Mensuration-3D

103. If a cuboid of dimensions 32 cm × 12cm × Bksl /krq dh uDdk'khnkj lrg dk {ks=kiQy  lseh
262
9cm is melted into two cubes of same size, gS vkSj bldh frjNh ÅapkbZ 26 lseh gSA bls
what will be the ratio of the surface area of
the cuboid to the total surface area of the
tkrk gS vkSj vk/kj f=kT;k 5 lseh ds ,d Bksl yEc
two cubes ? csyu esa <kyk tkrk gSA csyu dk dqy i`"Bh;
(lseh2 esa) D;k gS\
;fn 32 lseh × 12 lseh × 9 lseh foekvksa okys ,d
CRPF HCM 22/02/2023 (Shift - 02)
?kukHk dks cjkcj vkdkj ds nks ?kuksa esa fi?kyk;k tk,] rks
(a) 370 (b) 300
?kukHk ds i`"B {ks=kiQy vkSj nksuksa ?kuksa ds dqy i`"B
(c) 350 (d) 320
{ks=kiQy esa D;k vuqikr gksxk\
107. The base radius of a solid right circular
SSC CGL TIER II (11 /09/2019) cylinder is 15 cm and its total surface area
is 770 cm2. It is melted and recast into
(a) 65:72 (b) 37:48 another solid cylinder of height 24 cm.
What is the curved surface area (in cm2)
(c) 24:35 (d) 32:39
of the new cylinder so formed?
104. A solid metallic cube of side 9 cm and a ,d Bksl yEco`Ùkh; csyu dh vk/kj f=kT;k 15 ls
solid metallic cuboid having dimensions 5 vkSj bldk laiw.kZ i`"Bh; {ks=kiQy
 oxZ770
lseh gSA

r
cm, 13 cm, 31 cm are melted to form a fi?kyk;k x;k gS vkSj 24 lseh ÅapkbZ okys ,d v

si
single cube. How much (in Rs) is the cost
csyu esa <kyk x;k gsA bl izdkj fufeZr u, csyu d
to polish the new cube at a rate of Rs10
i`"Bh; {ks=kiQy (oxZ lseh esa) fdruk gksxk\
per cm²?
an by ICAR Mains, 07/07/2023 (Shift-2)
9 lseh Hkqtk okys ,d Bksl /kfRod ?ku rFkk 5 lseh] (a) 240

n
(b) 432
13 lseh] 31 lseh foekvksa okys ,d Bksl /kfRod ?kukHk (c) 480
ja (d) 384
dks fi?kykdj ,d ,dy ?ku cuk;k tkrk gSA bl u, 108. A steel cylinder of radius 3.5 cm and
R s
?ku dks ikWfy'k djus esa :i;s 10 çfr 2lseh
dh nj ls height 7 cm is melted to form bearings of
a th

radius 1 cm. How many such bearings can

fdruh ykxr (:- esa) vk,xh\
be made, assuming that 9.5 cm³ of steel
(a) 8,650 (b) 27,440 goes waste in manufacturing?
(c) 13,620 (d) 11,760 f=kT;k 3-5 lseh vkSj ÅapkbZ 7 lseh okys ,d
ty a

105. Two solid metallic right circular cones each cuk gqvk yEco`Ùkh; csyu dks 1 lseh f=kT;k ok
of base radius 4.5 cm and heights 10 cm cukus ds fy, fi?kyk;k tkrk gSA bl rjg ds fd
di M

and 8 cm, are melted and recast into a

ch;fjax cuk, tk ldrs gSa] ;g ekurs gq, fd 9
solid sphere. What is the cost of polishing
the surface area of the sphere at the rate lseh3 LVhy fofuekZ.k esa cckZn gks tkrk gS
of ` 3.50 per 10 cm2? (nearest to a `)? (a) 57 (b) 62
4.5 vk/kj f=kT;k vkSj10 lseh rFkk8 lseh Å¡pkb;ksa (c) 65 (d) 64
okys nks Bksl /kfRod yEco`Ùklh; 'kadqvksa dks fi?kykdj
109. A solid cylinder of diameter 12cm and
,d Bksl xksys esa <kyk x;k` gSA
3.50 izfr 10 oxZ lseh height 15cm is melted and recast into toys
with the shape of a right circular cone
dh nj ls xksys ds i`"Bh; {ks=kiQy ij ikWfy'k djus dh mounted on a hemisphere of radius 3cm.
ykxr fdruh gksxh (fudVre` rd iw.kk±fdr)\ If the height of the toy is 12cm find the
A

ICAR Mains, 07/07/2023 (Shift-3) number of toys so formed.

(a) ` 90 (b) ` 88 O;kl 12 lseh vkSj ÅapkbZ 15 lseh ds ,d Bksl

(c) ` 92 (d) ` 89 dks fi?kykdj f•ykSus cuk, tkrs gS] tks 'kadq ds
106. The carved surface area of solid metallic
dk gS ftl ij 3 lseh ds f=kT;k dk v/Z xksyk fL
cone is 260 cm2 and its slant height is 26 gSA ;fn f•ykSus dh ÅapkbZ 12 lseh gS rks
cm. It is melted and recast into a solid la[;k Kkr djsa\
 Mensuration-3D

110. A solid metallic sphere of radius 15 cm is (ii) Concept of digging & earth taken out.
melted and recast into spherical balls of
•qnkbZ vkSj fudkyh xbZ feêðh dh vo/kj.kkA
radius 3 cm each. What is the ratio of the
surface area of the original sphere and the Volume of earth taken out/fudkyh xbZ fe
sum of the surface areas of all the balls? dk vk;ru
15 lseh f=kT;k okys Bksl /kfRod xksys dks fi?kyk;k= Volume of ambankment or volume of
tkrk gS vkSj fiQj çR;sd 3 lseh f=kT;k okyh xksykdkjcuboid cylinder hemisphere/pcwrjs dk vk;ru
xsan cukbZ tkrh gSaA ewy xksys ds i`"Bh; {ks=kiQy vFkok
vkSj?ku @csyu@ v¼Zxksys dk vk;ruA
lHkh xsanksa ds i`"Bh; {ks=kiQy ds ;ksxiQy dk vuqikr w
Kkr dhft,A
(a) 1 : 5 (b) 1 : 10 x
(c) 5 : 27 (d) 3 : 40
111. The radius of base of solid cylinder is 7 cm
and its height is 21 cm. It melted and
converted into small bullets. Each bullet is

r
of same size. Each bullet consisted of two
parts viz. a cylinder and a hemisphere on

si
one of its base. The total height of bullet r
is 3.5 cm and radius of base is 2.1 cm.
an by
Approximately how many complete bullets
r 2 h =   (r  w)2 – r 2  × x
can be obtained? Note: Generally, the soil taken out from a

n
Bksl csyu ds vk/kj dh f=kT;k 7 lseh gS vkSj bldh cuboidal or cylindrical well is used to make
ÅapkbZ 21 lseh gSA ;g fi?ky x;k vkSj NksVh xksfy;ksa
a platform around the same well or to fill
ja
R s
another cuboidal / cuboidal / cylindrical /
esa cny x;kA çR;sd xksyh ,d gh vkdkj dh gSA çR;sd hemispherical vessel. The volume of the
xksyh esa nks Hkkx gS] blds ,d gh vk/kj ij ,d
a th

removed soil figure is made equal to the

csyu vkSj ,d xksyk/Z gSA cqysV dh dqy ÅapkbZ 3-5 volume of the used soil figure.
lseh vkSj vk/kj dh f=kT;k 2-1 lseh gSA rks Kkr djs uksV% lkekU;r% feV~Vh ?kukHkkdkj vFkok csy
fdruh xksfy;ka çkIr dh tk ldrh gSa\ fudky dj mlh dq,a ds pkjks vksj pcwrjk cukus ;
ty a

(a) 83 (b) 89 fdlh nwljs ?kukdkj @ ?kukHkkdkj @ csyukdkj @

ik=k dks Hkjus esa mi;ksx dhfudkyh
tkrh gSA
xbZ feV~V
di M

(c) 74 (d) 79
112. The radii of the ends of a frustum of a solid
vkÑfr ds vk;ru dks mi;ksfxr feV~Vh dh vkÑf
right-circular cone 45 cm high are 28 cm
and 7 cm. If this frustum is melted and vk;ru ds cjkcj dj nsrs gSA
reconstructed into a solid right circular 113. A well 20 m in diameter is dug 14 m deep
cylinder whose radius of base and height and the earth taken out is spread all around
are in the ratio 3: 5, find the curved it to a width of 5 m to form an embankment.
surface area (in cm²) of this cylinder. The height of the embankment is:
 22  20 ehVj O;kl okyk ,d dqvk¡
14 ehVj xgjk •ksnk tk
 Use   7  gS vkSj fudkyh xbZ feV~Vh dks pcwrjk cukus
 
45 lseh Åaps ,d Bksl yac&o`Ùkh; 'kadq ds fNUud dspkjksa vksj 5 ehVj dh pkSM+kbZ esa iQSyk fn;k tk
fljksa dh f=kT;k,a 28 lseh vkSj 7 lseh gSaA ;fn bl dh ÅapkbZ gS%
A

fNUud dks fi?kykdj ,d Bksl yac o`Ùkh; csyu cuk;k (a) 12.4 m (b) 9.5 m
tkrk gS] ftlds vk/kj dh f=kT;k vkSj ÅapkbZ dk vuqikr (c) 11.2 m (d) 8.4 m
114.
3% 5 gS rks bl 22 csyu dk oØ i`"Bh; {ks=kiQy (lseh
2 A cylindrical tank of radius 5.6 m and
depth of 'h' m is built by digging out earth.
22 The sand taken out is spread all around the
esa) Kkr djsaA
 ¹ dk ç;ksx djsaAº
7 tank to form a circular embankment to a
 Mensuration-3D

i`Foh dh •qnkbZ djds 5-6 ^ehVj f=kT;kh vkSj

ehVj (a) 29.8cm³ (b) 31m³
xgjkbZ dk ,d csyukdkj VSad cuk;k x;k gSA ckgj fudkyh (c) 30.2m³ (d) 33.6m³
xbZ jsr VSad ds pkjksa vksj tehu ij iQSykdj 7 (iii)
ehVj Filling a container by another shape
pkSM+k ,d o`Ùkkdkj pcwrjk cuk;k x;k gSA VSad dh xgjkbZcontainer.
fdruh gksxh] ;fn pcwrjs dh ÅapkbZ 1-96 ehVj gS\ ,d ik=k dks nwljs vkdkj ds ik=k ls HkjukA
(a) 7.2 m (b) 7 m
(c) 8 m (d) 9.5 m Volume of filling container/Hkjus okys ik=
115. A field is 125m long and 15m wide. A tank vk;ru = Volume of to be filled container/Hk
10 m × 7.5 m × 6 m was dug in it and the tkus okys ik=k dk vk;ru
Earth thus dug out was spread equally on
the remaining field. The level of the field 118. A hemishphere bowl of internal radius 15cm
thus raised is equal to which one of the contains a liquid. The liquid is to be filted
following? into cylinderical shaped bottles of diameter
,d eSnku 125 ehVj yack vkSj 15 ehVj pkSM+k gSA blesa 5 cm and height 6cm The number of bottles
required to empty the bowl is:
10 ehVj× 7.5 ehVj × 6 ehVj dk ,d VSad •ksnk x;k
vkSj bl rjg •ksnh xbZ feêðh dks 'ks"k eSnku ij leku 15 lseh vkarfjd f=kT;k okys ,d v/Zxksys dVksjs es

r
:i ls iQSyk fn;k x;kA bl çdkj eSnku dh Å¡pkbZ esa inkFkZ gSA rjy dks 5 lseh O;kl vkSj 6 lseh Åa

si
o`f¼ fuEufyf•r esa ls fdlds cjkcj gS\ csyukdkj vkdkj dh cksryksa esa fiQYVj fd;k tkuk
dks •kyh djus ds fy, vko';d cksryksa dh la[;k gS%
(a) 20 cm
(c) 28 cm
an by (b) 30 cm
(d) 25 cm
(a) 72 (b) 54

n
116. A field is 119m × 18 m in dimension. A tank
(c) 66 (d) 60
17m × 6m × 3m is dug out in the middle
ja
and the soil removed is evenly spread over 119. The diameter of the base of cylindrical drum
R s
is 35dm. and the height is 24 dm. It is full
the remaining part of the field. The increase
of kerosane. How many tins each of size 25
a th

in the level on the remaining part of the cm × 22 cm × 35 can be filled with kerosene
field is:
 22 
,d eSnku dh yackbZ&pkSM+kbZ × 18119
eh gSA
eh chp esa from the drum?  Use  = 
7 

17 eh × 6 eh × 3 eh vkdkj okyk ,d rkykc [kksnk
ty a

csyukdkj Mªe ds vk/kj dk O;kl35 dm gSA v

tkrk gS rFkk blls fudyh feV~Vh dks eSnku ds 'ks"k Hkkx
di M

esa iQSyk fn;k tkrk gSA eSnku ds 'ks"k Hkkx ds Lrj esa
ÅapkbZ
gqbZ
24 dm gSA blesa dsjksflu Hkjk gqvk gS
o`f¼ Kkr djsaA lseh × 22 lseh × 35 vkdkj ds fdrus fVu feV~V
SSC CPO 16/03/2019 (Shift -02) rsy ls Hkjs tk ldrs gSa\
(a) 14 cm (b) 13 cm
(a) 900 (b) 1000
(c) 15 cm (d) 12 cm
(c) 1280 (d) 1200
117. A field is in the form of a rectangle of length
20m and width 14m. a pit 6m long, 3m 120. A hemispherical bowl of internal radius 18
wide and 2.5 m deep is dug in a corner of cm is full of liquid. This liquid is to be filled
the field and the earth taken out of the pit in cylindrical bottles each of radius 3 cm
is spread uniformly over the remaining part and height 6 cm. How many bottles are
of field in order to raise the height of field required to empty the bowl?
A

by 30cm. How much more soil is required?

18 cm dh vkarfjd f=kT;k dk ,d v¼Zxksyh; dV
,d eSnku ftldh yackbZ 20 ehVj vkSj pkSM+kbZ 14 æo ls Hkjk gqvk gSA bl æo3dks cm f=kT;k vkS 6
ehVj dh vk;r ds :i esa gSA ,d xîôk 6 ehVj yack] cm Å¡pkbZ okyh csyukdkj cksryksa esa Hkjk t
3 ehVj pkSM+k vkSj 2-5 ehVj xgjk •sr ds ,d dksus •kyh djus ds fy, fdruh cksryksa dh vko';drk gks
esa •ksnk x;k gS vkSj xîôs ls fudkyh xbZ /jrh 30
SSC CGL TIER- II 03/03/2023
lseh rd eSnku dh ÅapkbZ c<+kus ds fy, eSnku ds 'ks"k
 Mensuration-3D

(iv) Filling or emptying a tank. (a) 1 hour, 30 min. (b) 1 hour, 20 min.
fdlh Vadh dks Hkjuk ;k •kyh djukA (c) 1 hour, 50 min. (d) 1 hour, 40 min.
Volume of water flowing through pipe 124. Water flows into a tank which is 200m long
(Cylindrical / Cuboidal) in t time/t le; esa and 150m wide, through a pipe of cross-
ikbi (csyukdkj@?kukdkj) ls cgus okys ikuh dk vk;ru section 0.3m × 2m at 20 km/hour. Then
the time (in hours) for the water level in
= Volume of the tank (Cylindrical /
the tank to reach 8m is:
Cuboidal)/VSad (csyukdkj@?kukdkj) dk vk;ruA
ikuh ,d VSad esa cgrk gS tks 200 ehVj yack v
Area of base × speed of flow × time/vk/kj dk
ehVj pkSM+k gS] 20 fdeh@?kaVk dh xfr ls× 0- 2
{ks=kiQy
× çokg dh xfr × le; = Volume of the tank
ehVj ØkWl&lsD'ku ds ikbi ds ekè;e lsA fiQj VS
(Cylindrical/Cuboidal)/VSad (csyukdkj@?kukdkj) dk
dk Lrj 8 ehVj rd igqapus dk le; (?kaVksa esa)
vk;ruA (a) 225 hours (b) 180 hours
1 (c) 196 hours (d) 200 hours
(a) r² × v × t = R²h or LBH or R2H
3 125. Water flows into a rank 200m × 150 m
(b) l × b × v × t = LBH or R²h through a rectangular pipe 1.5 m × 1.2 m
121. Water flows through a cylinderical pipe, at the rate 15 km/h in what time (in

r
whose radius is 7cm, at 5 metre per second. minutes) will the water rise 3 metres?
The time, it takes to fill an empty water 200m × 150 m ds ,d VSad esa 1.5 m × 1.2 dh

si
tank, with height 1.54 metres and area of
vk;rkdkj ikbi ds ekè;e ls ikuh 15 fdeh@?kaVs
the base (3 × 5) square metres is
 22 
an by ls izokfgr gksrk gSA tyLrj fdrus le; (feuVksa3
 Take  = ehVj rd Åij mBsxk\

n

 7 
ICAR Mains, 10/07/2023 (Shift-2)
ikuh ,d csyukdkj ikbi ls ikuh] ftldh f=kT;k 7 lseh
ja (a) 160 (b) 150
gS] 5 ehVj çfr lsdaM dh xfr ls cgrk gSA 1-54 ehVj
R s
(c) 180 (d) 200
ÅapkbZ vkSj vk/kj {ks=kiQy
(3 × 5) oxZ ehVj okys ,d 126. Water in a canal 40 decimetre wide and 16
a th

decimetre deep is flowing with a velocity

•kyh ikuh VSad dks Hkjus esa yxus okyk le; gS of 15 km/h. How much area (in m2) will it
 22  irrigate in 30 minutes if 12 cm of standing
 Take  =  water is required for irrigation?
 7 
ty a

(a) 5 min. (b) 6 min.

40 MslhehVj pkSM+h vkSj 16 MslhehVj xg
(c) 8 min. (d) 4 min. esa ikuh 15 fdeh@?k.Vk ds osx ls cg jgk g
di M

122. Water flows at the rate of 10 metres per flapkbZ ds fy, 12 lseh •M+s ikuh dh vko';
minute from a cylinderical pipe 5 mm in gksrh gS rks ;g 30 feuV esa fdrus {ks=k
2
esa)(eh
d
diameter. How long it take to fill up a flapkbZ djsxk\
conical vessel whose diameter at the base CRPF HCM 24/02/2023 (Shift - 02)
is 30 cm and depth 24cm? (a) 2,40,000 (b) 3,20,000
5 feeh O;kl okys ,d csyukdkj ikbi ls 10 ehVj çfr (c) 4,80,000 (d) 4,00,000
feuV dh nj ls ikuh cgrk gSA ,d 'kaDokdkj crZu] 127. Water flows out through a pipe with internal
ftlds vk/kj dk O;kl 30 lseh vkSj xgjkbZ 24 lseh gS] diameter 3 cm at the speed of 6.3 km/h
dks Hkjus esa fdruk le; yxsxk\ into a cylindrical tank whose internal base
(a) 25 min 24 sec (b) 24 min 24 sec radius is 1.5m. In 30 minutes, the water
(c) 28 min 48 sec (d) 30 min 36 sec level in the tank will rise by (assuming no
A

123. Water is flowing at the rate of 3km/hr overflow):

through a circular pipe of 20cm internal ikuh3 lseh vkarfjd O;kl okys ,d ikbi ds ekè;e ls
diameter into a circular cistern of diameter 6.3 fdeh@?kaVk dh xfr ls ,d csyukdkj VSad e
10m and depth 2m. In how much time will gS ftldh vkarfjd vk/kj f=kT;k1.5 ehVj gSA 30
the cistern be filled?
feuV esa] VSad esa ikuh dk Lrj c<+ tk,xk (;g e
20 lseh vkarfjd O;kl okys ,d xksykdkj ikbi ds fd dksbZ vfrçokg ugha gksxk)%
ekè;e ls 10 ehVj O;kl vkSj 2 ehVj xgjkbZ okys ,d
 Mensuration-3D

(v) Concept of inserting an object into another 132. A rectangular tank whose length and
object. breadth are 2.5 m and 1.5 m, respectively
,d oLrq dks nwljh oLrq esa Mkyus dh vo/kj.kkA is half fill of water. If 750 L more L more
Volume of inserted object/Mkyh xbZ oLrq dk warter is poured into the tank, then what
is the height through which water level
vk;ru = Volume of water/liquid displaced further goes up?
by inserted object/Mkyh xbZ oLrq }kjk foLFkkfir
ikuh@æo dk vk;ruA ,d vk;rkdkj VSad ftldh yackbZ vkSj pkSM+kb
128. A cylindrical vessel of base radius 14 cm is 5 ehVj vkSj 1-5 ehVj gS] mlesa vk/k ikuh gSA ;
filled with water to some height. If a 750 yhVj vkSj ikuh Mkyk tkrk gS] rks og Å¡pk
rectangular solid of dimensions 22 cm × 7 ftlls ikuh dk Lrj vkSj Åij pyk tkrk gS\
cm × 5 cm is immersed in it what is the
(a) 20 cm (b) 24 cm
rise in water level?
vk/kj f=kT;k 14 lseh dk ,d csyukdkj crZu dqN (c) 18 cm (d) 22 cm
ÅapkbZ rd ikuh ls Hkjk gqvk gSA ;fn 22× 7lseh
lseh 133. The base radius and slant height of a
× 2 lseh vk;ke okys ,d vk;rkdkj Bksl dks blesa conical vessel is 3 cm and 6 cm
respectively. Find the volume of sufficient
Mqcks;k tk, rks ty Lrj esa fdruh o`f¼ gksxh\

r
water in the vessel such that a sphere of
(a) 1.48 m (b) 1.50 m
radius 1 cm is placed into it water just

si
(c) 1.25 m (d) 2.25 m
129. Two irons sphere each of diameter 6cm are
an by immerse it?
immeresed in the water contained in a ,d 'kaDokdkj crZu dh f=kT;k 3lseh vkSj fr;Zd
cylindrical vessel of radius 6cm. The level
6 lseh gS] bl 'kadq esa fdrus ikuh dh vko'

n
of the water in the vessel will be raised by.
gksxh rkfd blds vanj j•k gqvk 1 lseh f=kT;
6 lseh O;kl okys nks yksgs ds xksys 6 lseh f=kT;k okys
,d csyukdkj crZu esa j•s ikuh esa Mkys tkrs gSaA crZu xksyk
esa iw.kZr% Mwc lds\
ja
R s
ikuh dk Lrj fdruk c<+ tk,xk\ 4 5
(a) (b)
a th

(a) 1 m (b) 2 m 3 3
(c) 2.5 m (d) 3 m
130. A cylindrical vessel of diameter 32 cm is 7 2
partially filled with water. A solid metallic (c) (d)
3 3
ty a

sphere of radius 12 cm is dropped into it.

What will be the increase in the level of 134. A sphere of diameter 18 cm is dropped in a
di M

water in the vessel (in cm)? right circular cylindrical vessel partly filled
32 lseh- O;kl okyk ,d csyukdkj crZu vkaf'kd :i with water. The radius of the base of the
cylindrical vessel is twice the radius of the
ls ikuh ls Hkjk gSA 12 lseh- f=kT;k okyk ,d Bksl /krqsphere. If the sphere is completely
dk Xkksyk blesa fXkjk;k tkrk gSA crZu esa ikuh dk Lrjsubmerged in water, by how much will the
(lseh- esa) fdruk Åij mBsXkk\ level of water rise in the cylindrical vessel?
SSC CGL 13/04/2022 (Shift-01)  22 
(a) 9 (b) 72  use   
7 
(c) 27 (d) 2.25
131. A cylindrical tank of diameter 35 cm is full 18 lseh O;kl dk ,d xksyk vkaf'kd :i ls ikuh l
of water. If 11 litres of water is drawn of Hkjs ,d yEc o`Ùkh; csyukdkj crZu esa fxjk
the water level in the tank will drop by:
gSA csyukdkj crZu ds vk/kj dh f=kT;k xksys
A

 22 
 use  =  dh nksxquh gSA ;fn xksyk iwjh rjg ls ikuh es
 7 
gS] rks csyukdkj crZu esa ikuh dk Lrj fdru
35 lseh O;kl okyk ,d csyukdkj VSad ikuh ls Hkjk gSA
;fn 11 yhVj ikuh fudkyk tk, rks Vadh esa ikuh dk  22 
mBsxk\ use   7 
Lrj fxj tk,xk%
4 2 CRPF HCM 28/02/2023 (Shift - 01)
(a) 9 cm (b) 10 cm
 Mensuration-3D

135. Some marbles each of diameter 4.2 cm, are 139. A solid cube of volume 13824 cm3 is cut
dropped into a cylindrical beaker containing into 8 cubes of equal volumes. The ratio
some water and are fully submerged. The of the surface area of the original cube to
diameter of the beaker is 28 cm. Find how the sum of the surface areas of three of the
many marbles have been dropped in it if smaller cubes is :
the water rises by 15.75 cm.
13824 lseh vk;ru dk ,d Bksl ?ku cjkcj vk;ruksa
dqN daps ls izR;sd dk O;kl
4.2 lseh gS] ,d csyukdkj 8 NksVs ?kuks esa dkVk tkrk gSA ewy ?ku d
chdj esa Mkys x, gSa ftlesa dqN ikuh gS vkSj os iwjh rjg
rhu NksVs ?kuks ds lrg {ks=kksa ds ;ksx ls D;k
Mwc x, gSaA chdj dk O;kl
28 lseh gSA ;fn ikuh ds Lrj
esa15.75 lseh dh o`f¼ gks tkrh gS rks blesa Mkys x, (a) 2 : 3 (b) 4 : 3

dapksa dh la[;k Kkr dhft,A (c) 8 : 3 (d) 2 : 1

ICAR Mains, 08/07/2023 (Shift-2) 140. A solid cylinder having radius of base as
(a) 225 (b) 275 28 cm and height as 24 cm is bisected
(c) 250 (d) 290 from its height to get two identical
cylinders. What will be the percentage

r
(vi) Concept of cutting a solid
increase in the total surface area?
Bksl dks dkVus dh vo/kj.kk

si
(a) Volume of solid does'nt change ,d Bksl flysaMj ftldk vk/kj f=kT;k 28 lseh v
Bksl dk vk;ru ugha cnyrk gS ÅapkbZ 24 lseh gS] nks leku flysaMj çkIr
an by
136. A sphere is cut into hemisphere. One of them fy, bldh ÅapkbZ ls lef}Hkkftr fd;k tkrk gSA
lrg {ks=k esa fdrus çfr'kr dh o`f¼ gksxh\

n
is used as bowl. It takes 8 bowlfuls of this to
fill a conical vessel of height 12cm and radius
6cm. The radius of the cylinder is:
ja SSC MTS 17/05/2023 (Shift-01)
,d xksys dks v/Zxksys esa dkVk tkrk gSA muesa ls ,d dk
R s
(a) 61.72 percent (b) 41.92 percent
mi;ksx dVksjs ds :i esa fd;k tkrk gSA 12 lseh ÅapkbZ vkSj
a th

(c) 53.85 percent (d) 48.64 percent

6 lseh f=kT;k okys ,d 'kaDokdkj crZu dks Hkjus esa bldh
8 dVksfj;ka yxrh gSaA csyu dh f=kT;k gS% 141. A right circular cylinder has height as 18
(a) 3 cm (b) 4 cm cm and radius as 7 cm. The cylinder is cut
ty a

in three equal parts (by 2 cuts parallel to

(c) 2 cm (d) 3.5 cm base). What is the percentage increase in
(b) Surface area of solid increases. total surface area?
di M

Bksl dk lrg {ks=k c<+ tkrk gSA

137. A cuboid of size 50 cm x 40 cm x 30 cm is
,d yEc o`Ùkh; csyu dh Å¡pkbZ 18 lseh vkSj f=
cut into 8 identical parts by 3 cuts. What is lseh gSA csyu dks rhu cjkcj Hkkxksa esa d
the total surface area (in cm2) of all the 8 (vk/kj ds lekukarj 2 dVksa ls)A dqy lrg {ks
parts? çfr'kr o`f¼ D;k gS\
50 lsaeh× 40 lsaeh× 30 lsaeh vkdkj ds ,d ?kukHk dks
(a) 62% (b) 56%
3 dkV }kjk 8 leku Hkkxksa esa dkVk tkrk gSA lHkh 8 Hkkxksa
ds laiw.kZ i`"B dk {ks=kiQy (oxZ lseh esa) D;k gS\ (c) 48% (d) 52%
(a) 18,800 cm² (b) 16,400 cm² 142. A solid sphere has a surface area of 616 cm².
(c) 20,800 cm² (d) 21,600 cm² This sphere is now cut into two
hemispheres. What is the total surface area
A

138. A solid cube has side 8 cm. It is cut along

diagonals of top face to get 4 equal parts. of one of the hemispheres?
What is the total surface area (in cm2) of ,d Bksl xksys dk i`"Bh; {ks=kiQy 616
2
gSA
lsehbl x
each part.
dks nks xksyk¼ (v¼Zxksyksa) esa dkVk tkrk
,d Bksl ?ku dh Hkqtk 8 lseh gSA bls 'kh"kZ iQyd ds fod.kZ
xksyk¼Z dk laiw.kZ i`"Bh; {ks=kiQy Kkr dj
ds lekarj dkVdj 4 cjkcj Hkkx çkIr fd, x, gSaA çR;sd
Hkkx ds laiw.kZ i`"B dk {ks=kiQy (oxZ lseh esa) D;k gS\ SSC PHASE XI 28/06/2023 (Shift-01)
 Mensuration-3D

143. A solid sphere of diameter 17.5cm is cut (ii) A cylinder just encloses a sphere then
into two equal halves. What will be the csyu ds vUnj xksyk
increase (in cm²) in the total surface area?
17-5 lseh O;kl okys ,d Bksl xksys dks nks cjkcj Hkkxksa
esa dkVk tkrk gSA dqy i`"Bh; {ks=kiQy esa fdruh o`f¼
(lseh2 esa) gksxh\
(a) 289 (b) 481.25 h = 2r
(c) 361.5 (d) 962.5
144. A spherical ball of diameter 8 cm is cut into
two equal parts. The curved area of one such
part has to be painted with green colour, r
while the other part has to be painted with Height of cylinder/csyu dh Å¡pkbZ
red colour. The bases of both the = Diameter of sphere/xksys dk O;kl
= 2r
hemispheres are to be painted with blue Volume of Cylinder/csyu dk vk;ru : Volume
colour. The cost of painting with blue is Rs of Sphere/xksys dk vk;ru
2/cm², while the cost of painting the curved
4 3 4

r
area is Rs 3/cm². What will be the cost (in = r 2 (2r) :
r = 2 : = 3 : 2
Rs) of painting the hemispheres? 3 3

si
Take  = 3.14 (iii) A maximum cylinder inside cube then
8 lseh O;kl okyh ,d xksykdkj xsan dks nks cjkcj Hkkxksa
an by ?ku ds Hkhrj
vf/dre vk;ru dk csyu
esa dkVk tkrk gSA bl rjg ds ,d fgLls ds ofØr {ks=kiQy
dks gjs jax ls jaxuk gS] tcfd nwljs fgLls dks yky jax ls

n
jaxuk gSA nksuksa xksyk¼ksZa ds vk/kjksa dks uhys jax ls jaxuk
gSA uhys jax ls jaxus dh ykxr :i;s 2@lseh
2
gS] tcfd
ja h
R s
OkfØr {ks=kiQy dks jaxus dh ykxr :i;s 3@lseh 2
gSA
xksyk¼ksZa dks jaxus dh ykxr (:i;s esa) D;k gksxk\
a th

 ¾ 3-14 yhft,
SSC PHASE XI 27/06/2023 (Shift-04) a
(a) Rs 451.92 (b) Rs 492.92 Radius of cylinder/csyu dh f=kT;k
ty a

(c) Rs 803.84 (d) Rs 401.92 1 a

= × edge of cube =
Combination of 3-D Objects 2 2
di M

Height of cylinder/csyu dh Å¡pkbZ

f=kfoeh; vkÑfr;ksa dk la;kstu = edge of
cube/?ku dk fdukjk= a
 A solid maximum 3-D object inside another Volume of cube/?ku dk vk;ru : Volume of
3-D object 2

fdlh f=kfoeh; vkÑfr ds Hkhrj vf/dre vk;ru dh cylinder /csyu dk vk;ru = a3 :   a  a

2
nwljh f=kfoeh; vkÑfr
22 1
(i) A maximum cone inside a cylinder =1: × =14:11
7 4
csyu ds Hkhrj vf/dre vk;ru dk 'kadq 145. A right circular cylinder of maximum
volume is cut out from a solid wooden
cube. The material left is what percent of
the volume (nearest to an integer) of the
A

original cube ?
h ,d Bksl ydM+h ds ?ku ls vf/dre vk;ru okyk
,d yEc o`Ùkh; csyu dkVk tkrk gSA cph gqb
r vkjafHkd ?ku ds vk;ru (,d iw.kk±d ds fudV
Volume of cylinder/csyu dk vk;ru : Volume dk fdruk çfr'kr gS \
 Mensuration-3D

(iv) A maximum sphere inside a cube then (v) A maximum cube inside a sphere
?ku ds Hkhrj vf/dre vk;ru dk xksyk fdlh xksys ds Hkhrj vf/dre vk;ru dk ?ku
Diagonal of cube/?ku dk fod.kZ = Diameter of
sphere/xksys dk O;kl

a 3 a = 2r  a =
2r
Diameter of sphere/xksys dkO;kl
(2r) = edge 3
of cube/?kudh dksj= a Volume of sphere/xksys dk vk;ru: Volume of
Volume of cube/?ku dk vk;ru : Volume of cube/?ku dk vk;ru
3
3 4 3  2r  4 22 8

r
4 a   r :     :  11 3 : 7
sphere/xksys dk vk;ru= a : 3   2 
3
= 21: 3  3 3 7 3 3
 

si
(vi) A maximum sphere inside a cone
11 an by fdlh 'kadq ds Hkhrj vf/dre vk;ru dk xksyk
146. What is the volume of the largest sphere O
that can be carved out of a wooden cube of

n
 22 
 =
sides 21 cm?   h l
 7 
ja r B
R s
ydM+h ds 21cm dh Hkqtk okys ?ku ls dkVs tk A
r
ldus okys lcls cM+s xksys dk vk;ru fdruk gS\
a th

D
C R
SSC CGL TIER - II 02/03/2023
then, OCD ~ OBA
(a) 3851cm³ (b) 6858cm³ OD CD
ty a

 
(c) 4851cm³ (d) 5821cm³ OA AB
147. From the body of a solid cube of edge 7 l R
di M


cm. a solid sphere is removed. The h–r r
l × r = hR – Rr
volume of the remaining solid was found
 r = hR
1 l+R
to be 163 cm³. What is the diameter (in
3 (vii) Maximum cylinder inside a cone
cm) of the sphere? fdlh 'kadq ds Hkhrj vf/dre vk;ru dk csyu
O
 22 
 Take   7 
 
H
A B
7cm dksj okys fdlh Bksl ?ku ls ,d Bksl xksyk dkVk rr
A

h
1
tkrk gS 'ks"k Bksl dk vk;ru
163 cm³ ik;k x;kA R
3 C D

22
xksys dk O;kl
(cm esa
) D;k gS\ (π = yhft,) OCD ~ OAB
7 (A = C = 90º, O common)
SSC CGL 19/04/2022 (Shift- 01) OC CD

OA AB
 Mensuration-3D

(viii) A maximum cube inside a cone 148. A solid cone of height 42 cm with diameter
'kadq ds Hkhrj vf/dre vk;ru dk ?ku of its base 42 cm is cut out from a wooden
solid sphere of radius 24 cm. Find the
O percentage of wood wasted correct to two
places of decimal.
42 cm Å¡pkbZ dk ,d Bksl 'kadq ftlds vk
dk O;kl 42 cm gS]24 cm f=kT;k ds ydM+
Bksl xkssys ls dkVk tkrk gSA cckZn gqb
H izfr'kr Kkr dhft, tks n'keyo ds nks LFkkuk
lgh gksA
B
A SSC CGL 09/12/2022 (Shift- 01)
a a (a) 75.56% (b) 56.65%
a (c) 66.50% (d) 67.50%
149. A hemispherical depression of diameter 4
R cm is cut out from each face of a cubical
D block of sides 10 cm. Find the surface area

r
C a of the remaining solid (in cm2).
a

si
OCD ~ OAB 22
(Use π = )
an by 7
OC CD
 10cm Hkqtkvksa okys ,d ?ku ds izR;sd iQyd
4cml
OA AB
O;kl dk ,d v/Zxksykdkj xM~<k dkVk tkrk g

n
H R  2a a  Bksl dk i`"Bh; {ks=kiQy
(cm2 esa) Kkr dhft,A
  AB   
H– a a/ 2
ja 2 2
(π=
22
yhft,)
R s

7
(ix) Largest cube inside a hemisphere
a th

SSC CGL 18/04/2022 (Shift- 01)

xksyk¼Z ds vanj lcls cM+k ?ku
4 1
(a) 900 (b) 700
7 7
ty a

A 3 4
(c) 675 (d) 112
7 7
150. A sphere of maximum volume is cut out
di M

from a solid hemisphere. What is the ratio

of the volume of the sphere to that of the
remaining solid ?
C fdlh Bksl v¼Zxksys ls vf/dre vk;ru okyk ,d
B
E D xksyk dkVk tkrk gSA xksys ds vk;ru ,oa '
Bksl ds vk;ru ds chp vuqikr Kkr djsaA
Let R be the radius of hemisphere and x be SSC CGL TIER II (13/09/2019)
the side of cube. C is the centre of
(a) 1 : 4 (b) 1 : 2
hemisphere.
(c) 1 : 3 (d) 1 : 1
ekukR v¼Zxksys dh f=kT;k gS
x ?ku
vkSjdh Hkqtk C
gSA151. A sphere of maximum volume is cut out
A

xksyk¼Z dk dsaæ gSA from a solid hemisphere of radius r. The

ratio of the volume of the hemisphere to
BD = 2x that of the cut out sphere is :
x ,d r f=kT;k okys Bksl v¼Zxksys ls vf/dre v
 BC =
2 okyk ,d xksyk dkVk tkrk gSA v¼Zxksys d
In ABC, AC² = AB² + BC² dk dkVs x;s xksys ds vk;ru ls vuqikr D;k g
 Mensuration-3D

152. Radius of base of a hollow cone is 8 cm and 156. A 22.5 m high tent is in the shape of a
its height is 15 cm. A sphere of largest frustum of a cone surmounted by a
radius is put inside the cone. What is the hemisphere. If the diameters of the upper
ratio of radius of base of cone to the radius and the lower circular ends of the frustum
of sphere? are 21 m and 39m, respectively, then find
the area of the cloth (in m²) used to make
,d •ks•ys 'kadq ds vk/kj dh f=kT;k 8 lseh gS vkSj the tent (ignoring the wastage).
bldh ÅapkbZ 15 lseh gSA lcls cM+s f=kT;k dk ,d xksyk
'kadq ds vanj j•k x;k gSA 'kadq ds vk/kj dh f=kT;k dk Use   22 
xksys dh f=kT;k ls vuqikr D;k gS\  7 

(a) 5:3 (b) 4:1 ,d 22-5 ehVj Åapk racw ,d v/Zxksys ds Åij ,d 'k
(c) 2:1 (d) 7:3 ds fNUud ds vkdkj dk gSA ;fn fNUud ds Åijh
fupys o`Ùkkdkj fljksa dk O;kl Øe'k% 21 ehVj
153. A right circular cylinder has height 28 cm
and radius of base 14 cm. Two hemispheres ehVj gS] rks racw
cukusds fy, bLrsekyfd, x, diM+s
of radius 7 cm each are cut from each of the dk {ks=kiQy (ehVj2 esa)Kkrdhft, (diM+s ds viO;;
two bases of the cylinder. What is the total dks utj vankt djrs gq, mÙkj Kkr djsa)A

r
surface area (in cm²) of the remaining part?
SSC CGL 21/04/2022 (Shift -03)

si
,d yEc o`Ùkh; csyu dh Å¡pkbZ 28 lseh vkSj vk/kj dh
f=kT;k 14 lseh gSA csyu ds nksuksa vk/kjksa esa ls çR;sd ls 2
(a) 787
an by
7 lseh f=kT;k okys nks xksyk¼ks± dks dkVk tkrk gSA 'ks"k 7
Hkkx dk dqy i`"Bh; {ks=kiQy 2
esa
(lseh
) fdruk gS\

n
2
(a) 3842 (b) 4312 (b) 2800
ja 7
(c) 3296 (d) 4436
R s
6
154. From a solid cylinder wooden block of (c) 1635
a th

height 18 cm and radius 7.5 cm, a conical 7

cavity of same radius and same height is
2
taken out. What is total surface area (in cm2) (d) 2107
of the remaining solid? 7
ty a

Å¡pkbZ 18 lseh vkSj f=kT;k 7-5 lseh ds ,d Bksl flysaMj

157. A solid toy is in the shape which is a
ydM+h ds CykWd ls] leku f=kT;k vkSj leku ÅapkbZ combination of a cylinder, cone and a
di M

hemispherical bow. The cylinder

dk ,d 'kaDokdkj xqgk ckgj fudkyk tkrk gSA 'ks"k
contributes to 50% of the total volume of
Bksl dk dqy {ks=kiQy (oxZ lseh esa) D;k gS\ the toy, the cone contributes to 20% of the
(CGL MAINS 16/10/2020) volume. Find the ratio of the contribution
(in terms of volume) of the cone, cylinder
(a) 326.25  (b) 416.25 
and hemisphere.
(c) 472.5  (d) 270 
,d Bksl f[kykSus dh vkÑfr ,d csyu] 'kdq vkSj ,
155. Aright circular cone is inscribed in a cube
v/Zxksykdkj I;kys ds l;kstu ds leku gSA f[ky
of side 9 cm occupying the maximum space
possible. What is the ratio of the volume of
csyukdkj Hkkx dk vk;ru f[kykSus ds dqy v
the cube to the volume of the cone? dk 50» gS] 'kDdkdkj Hkkx dk vk;ru f[kykS
dqy vk;ru dk 20» gSA f[kykSus ds 'kDd
A

,d yEc o`Ùkh; 'kadq 9 lseh Hkqtk okys ?ku esa vafdr csyukdkj
gS vkSj v/Zxksykdkj Hkkxksa ds vk;ruk
tks vf/dre laHko LFkku ?ksjrk gSA ?ku ds vk;ru dk Kkr dhft,A
'kadq ds vk;ru ls vuqikr D;k gS\ SSC CHSL 30/05/2022 (Shift- 2)
(Take  = 22/7) (a) 2 : 3 : 5
SSC MTS 18 /10/ 2021 (b) 5 : 2 : 3
 Mensuration-3D

C.S.A/oØ
i`"B dk {ks=kiQy
Prism/fizTe = 4ah
T.S.A/laiw.kZ i`"B dk {ks=kiQy
= 4ah + 2a²
 A prism is a solid that has two faces that are
Volume/vk;ru = a²h
parallel and congruent and their faces
(Polygon) join by vertex to vertex. A prism (iii) Hexagonal Prism:/"kV~dks.kh; fizTEk
has a polygon as its base and vertical side
perpendicular to the base.
fçTEk ,d Bksl gksrk gS ftlds nks iQyd lekukarj vkSj lokZaxle
gksrs gSa vkSj muds iQyd (cgqHkqt) 'kh"kZ ls tqM+rs gSaA fçTEk esa
vk/kj ds :i esa ,d cgqHkqt gksrk gS vkSj ÅèokZ/j Hkqtk vk/kj
ds yEcor gksrh A gS
h
(a) Curved surface area of a prism = Perimeter
of base × height
fçTEk
dk oØ i`"Bh; {ks=kiQy ¾ vk/kj dk ifjeki
× Å¡pkbZ
(b) Total surface area of a prism = curved

r
surface area + 2 × area of base
fçTEk dk dqy i`"Bh; {ks=kiQy ¾ oØ i`"Bh; {ks=kiQy

si
a
$ 2 × vk/kj dk {ks=kiQy
an by C.S.A/oØ i`"B dk {ks=kiQy
= 6ah
(c) Volume of a prism = area of base × height
T.S.A/laiw.kZ i`"B dk {ks=kiQy 2
= 6ah + 3 3a
fçTEk dk vk;ru ¾ vk/kj dk {ks=kiQy
× Å¡pkbZ

n
6 3
(i) Equilateral triangular prism: Volume/vk;ru = a²h
4
ja
leckgq f=kHkqtkdkj fçTEk%
R s
158. The base of a right prism is a triangle with
sides 16 cm, 30 cm and 34 cm. Its height
a th

is 32 cm. The lateral surface area (in cm2)

and the volume (in cm3) are, respectively:
,d yac fçTe dk vk/kj ,d f=kHkqt gS ftldh Hkq
ty a

h 16 lseh] 30 lseh vkSj 34 lseh gSaA bldh Åa

lseh gSA ik'oZ i`"Bh; {ks=kiQy
2
esa)
(lseh
vkSj vk;r
di M

(lseh esa) Øe'k% fdrus&fdrus gksaxs\

3

a SSC CGL MAINS 29/01/2022

C.S.A./oØ i`"B dk {ks=kiQy
= 3ah (a) 2688 and 7680 (b) 2624 and 7040
3 2 (c) 2560 and 6400 (d) 2560 and 7680
T.S.A./laiw.kZ i`"B dk {ks=kiQy
= 3ah + 2 × a
4 159. The base of a prism is a right angle triangle
whose sides are 9 cm, 12 cm and 15 cm. Volume
3 2
Volume/vk;ru = a h of this prism is 648 cm³. What will be the
4 height of prism?
(ii) Square Prism/oxkZdkj fizTEk ,d fizTe dk vk/kj ,d ledks.k f=kHkqt gS ftld
Hkqtk,¡
9 cm, 12 cm rFkk15 cm gSA bl fizTe
A

vk;ru 648 cm³ gSA bl fizTe dh Å¡pkbZ D;k

SSC CGL MAINS (08/08/2022)

h
(a) 14 cm (b) 9 cm
(c) 16 cm (d) 12 cm
160. The base of a right prism is a triangle with
 Mensuration-3D

,d yEc fçTe dk vk/kj ,d f=kHkqt gS ftldh Hkqtk,a164. The base of a right prism is a regular
20 lseh] 21 lseh vkSj 29 lseh dh gSaA ;fn bldk hexagon of side 5 cm. If its height is
vk;ru 7560 ?ku lseh gS] rks bldk ik'oZ i`"B {ks=kiQy 12 3 cm, then its volume ( in cm3) is :
(oxZ lseh esa) Kkr djsaA
,d fçTe dk vk/kj 5 lseh dk ,d fu;fer "kV~Hkq
SSC CGL TIER II (12/09/2019) gSA ;fn bldh Å¡pkbZ
12 3 lseh gS] rks bldh vk;r
(a) 2484 (b) 2556 (?ku lseh esa) gS %
(c) 2520 (d) 2448
(CGL MAINS 15/10/2020)
161. The base of a right prism is an equilateral (a) 900 (b) 1800
triangle whose side is 10 cm. If height of (c) 1350 (d) 675
this prism is 10 3 cm, then what is the total 165. The base of a right prism is a quadrilateral
surface area of prism? ABCD, given that AB = 9cm, BC = 14 cm,
CD = 13 cm, DA = 12 cm and DAB = 90°.
,d fizTe dk vk/kj ,d leckgq f=kHkqt gS ftldh Hkqtk If the volume of the prism is 2070 cm³,
10 cm gSA ;fn fizT; dh Å¡pkbZ 10 3 cm gS] rks then the area of the lateral surface is ?

r
fizT; dk laiw.kZ i`"Bh; {ks=kiQy D;k gS\ ,d yEco`rh; fçTe dk vk/kj ,d prqHkZqtABCD

si
SSC CGL MAINS (08/08/2022) gS] tgkaAB = 9 lseh, BC = 14 lseh, CD = 13
lseh, DA = 12 lseh vkSj DAB = 90° gSA ;
(a) 325 3 cm²
an by (b) 350 3 cm² fçTe dk vk;ru 2070 lseh3 gS] rks ik'oZ lrg d

n
{ks=kiQy gS&
(c) 125 3 cm² (d) 150 3 cm²
ja (a) 720 cm² (b) 810 cm²
162. The base of right prism is a trapezium (c) 1260 cm² (d) 2070 cm²
R s
whose parallel sides are 11cm and 15cm 166. Let ABCDEF is prism whose base is a right
and the distance between them is 9 cm. If
a th

triangle whose perpendicular sides are 9cm

the volume of the prism is 1731.6 cm3 ,
and 12 cm, if cost of painting the prism is
then the height (in cm) of the prism will
Rs.151.20 at the rate of 20 paise/cm²,
be :
then find the height of the prism?
ty a

,d yEc fizTe dk vk/kj leyac gS ftldh lekukarj ekuk fd ABCDEF ,d fçTe gS] ftldk vk/kj
Hkqtk,a 11 lseh vkSj 15 lseh gSa rFkk muds chp dhledks.k f=kHkqt gS] ftldh nks yEcor Hkqtk,
di M

nwjh 9 lseh gSA ;fn fizTe dk vk;ru 1731-6 ?ku vkSj 12 lseh gSA ;fn fçTe dks jaxus dh ykxr
lseh gS] rks fçTe dh Å¡pkbZ gksxh %
çfr oxZ lseh dh nj ls 151-20 #i;s gS rks fçTe
SSC CGL TIER II (11/09/2019) ÅapkbZ fdruh gS\
(a) 15.6 (b) 15.2 (a) 17 cm (b) 15 cm
(c) 16 cm (d) 18 cm
(c) 14.8 (d) 14.2
167. A prism has a square base whose side is
163. The base of a solid prism of height 10 cm 8cm. The height of prism is 80cm. The
is a square and its volume is 160 cm3, What prism is cut into 10 identical parts by 9
is its total surface area of the prism (in cm2 ) ? cuts which are parallel to base of prism.
Å¡pkbZ 10 lseh ds ,d Bksl fçTe dk vk/kj ,d oxZ What is the total surface area (in cm²) of
A

gS vkSj bldh vk;ru 160 ?ku lseh gS] rfçTe dh all the 10 parts together?
dqy lrg dk {ks=kiQy (oxZ lseh esa) D;k gS\ ,d fçTe dk vk/kj oxkZdkj gS ftldk çR;sd Hk
8 lseh gS] vkSj fçTe dh ÅapkbZ 80 lseh gSA
(CGL MAINS 16/10/2020)
dks blds vk/kj ds lekUrj 9 dV }kjk 10 lek
(a) 200
Hkkxksa esa ckaVk x;k gS] rc bl çdkj cus
(b) 192 Hkkxksa ds dqy i`"B {ks=kiQy D;k gksxk
 Mensuration-3D

Pyramid/fijkfeM
A pyramid is a three-dimensional shape. A
pyramid has a polygonal base and flat
triangular faces, which join at a common
point called the apex. A pyramid is formed
by connecting the bases to an apex. Each edge
of the base is connected to the apex, and
forms the triangular face, called the lateral
face. If a pyramid has an n-sided base, then 1
it has n + 1 faces, n + 1 vertices, and 2n edges. C.S.A./oØ i`"B dk {ks=kiQy
=  3a  l
2
fijkfeM ,d f=k&vk;keh vkÑfr gSA fijkfeM esa ,d cgqHkqt
1 3 2
vk/kj vkSj likV f=kdks.kh; iQyd gksrh gSa] tks ,d T.S.A./laiw.kZ i`"B dk {ks=kiQy =  3al 
2 4
a
mHk;fu"B fcanq ij tqM+rs gSa ftls 'kh"kZ dgk tkrk gSA vk/
kjksa dks 'kh"kZ ls tksM+dj fijkfeM dk fuekZ.k fd;k tkrkVolume/
gSA vk;ru = 
1 3 2
a h
vk/kj dk çR;sd fdukjk 'kh"kZ ls tqM+k gqvk gS] vkSj

r
3 4
f=kdks.kh; iQyd cukrk gS] ftls ik'oZ iQyd dgk tkrk

si
2
 a 
gSA ;fn fdlh fijkfeM dk vk/kjn&Hkqtk dk gS] rks blesa Slant height (l)  h2  r 2  h2   
2 3
n + 1 iQyd] n + 1 'kh"kZ vkSj
2n fdukjs gksrs gSaA
an by (Slant edge)/fr;Zd fdukjk

n
2
 a 
h2  R 2  h2  
th

ja  
ng

 3
R s
le
t

Square Pyramid/oxkZdkj fijkfeM

an

(ii)
Sl

a th

Height

(a) Lateral/Curved surface area of Pyramid/

ty a

h
fijkfeM dk oØ i`"Bh; {ks=kiQy
= Sum of areas of l
SE
all the lateral triangular faces.
di M

1
= × Perimeter of base/vkèkkj dk ifjeki ×
2
a
slant height/frjNh Å¡pkbZ 2
a
(b) Total surface area of Pyramid/fijkfeM dk dqy 2

i`"Bh; {ks=kiQy
= Sum of the areas of all lateral a
faces + Area of the base. 1
C.S.A./oØ i`"B dk {ks=kiQy
=  4a  l
= Curved surface area + area of base/oØ i`"Bh; 2
{ks=kiQy
+ vk/kj dk {ks=kiQy
1
T.S.A./laiw.kZ i`"B dk {ks=kiQy 2
1 =  4al  a
2
= × Perimeter of base/vkèkkj dk ifjeki ×
A

2
1
slant height/frjNh
Å¡pkbZ
+ Area of base Volume/vk;ru =  a2  h
3
(c) Volume of a Pyramid/fijkfeM dk vk;ru
2
1 area of base/vk/kj dk {ks=kiQy
× height/ a
Slant height/fr;Zd Å¡pkbZ
2
= × = h  
3 2
Å¡pkbZ
 Mensuration-3D

(iii) Rectangular Pyramid/vk;rkdkj fijkfeM 1

C.S.A./oØ i`"B dk {ks=kiQy
= ×6al
There are two slant height/nks frjNh Å¡pkbZ 2
gksrh gSA
1 3 2
T.S.A./laiw.kZ i`"B dk {ks=kiQy
= 6al  6  a
2 4

1 6 3 2
Volume/vk;ru =  a h
h 3 4
l2 Slant height/frjNh Å¡pkbZ
2
2
 3 
l1 (l) = h +  a 
 2 
b
l/2
Slant edge/frjNh fdukjk = h 2 + a 2
b/2 168. The base of right pyramid is an equilateral
l triangle, each side of which is 20 cm. Each

r
First slant height /igyh frjNh Å¡pkbZ
(l1) = slant edge is 30 cm. The vertical height (in

si
cm) of the pyramid is:
2
 b
h2   
an by ,d yac fijkfeM dk vk/kj ,d ,slk leckgq f=kHk
2 gS] ftldh Hkqtk dh yackbZ 20 lseh gSA izR
dksj 30 lseh gSA fijkfeM dh ÅèokZ/j Å¡pk
Second slant height/nwljh frjNh Å¡pkbZ
(l2) =
esa) fdruh gksxh\

n
2
l  SSC CGL MAINS 29/01/2022
h2   
2
ja (a) 5 3 (b) 10 3
R s

C.S.A./oØ i`"B dk {ks=kiQy

= 23 23
a th

(c) 5 (d) 10
1 1 3 3
2 l  l1  2   b  l 2
2 2 169. The base of a right pyramid is a square of

T.S.A./laiw.kZ i`"B dk {ks=kiQy side 8 2 cm and each of its slant edge is

ty a

= C.S.A + lb
of length 10 cm. What is the volume (in
1 cm³) of the pyramid?
di M

Volume/vk;ru = × lb × h
3
,d yac fijkfeM dk vk/kj 8 2 lseh Hkqtk okyk
(iv) Hexagonal pyramid/"kV~dks.kh; fijkfeM oxZ gS vkSj bldh izR;sd fr;Zd dksj dh yackbZ
gSA fijkfeM dk vk;ru (lseh
3
esa) fdruk gS\
Slant SSC CGL MAINS 03/2/2022
edge (a) 256 (b) 224
2
(c) 426 (d) 96 2
3
Slant 170. The total surface area of a right pyramid,
height (l) with base as a square of side 8 cm, is 208
A

cm². What is the slant height (in cm) of

the pyramid?
8cm Hkqtk ds oxkZdkj vk/kj okys ,d le fijkf
a dk laiw.kZ i`"Bh; {ks=kiQy
208cm2 gSA fijkfeM
fr;Zd Å¡pkbZcm( esa) Kkr dhft,A
a SSC CGL 20/04/2022 (Shift- 02)
 Mensuration-3D

171. What is the total surface area of a pyramid 175. The base of a right pyramid is an equilateral
whose base is a square with side 8 cm and triangle with area 16 3 cm2. If the area of
height of the pyramid is 3 cm?
one of its lateral faces is 30 cm2 , then its
,d fijkfeM dk laiw.kZ i`"Bh; {ks=kiQy D;k gS ftldk height (in cm) is :
vkèkkj8 cm Hkqtk okyk ,d oxZ gS vkSj fijkfeM dh ,d yEc fijkfeM dk vk/kj ,d leckgq f=kHkqt
Å¡pkbZ 3 cm gS\ ftldk {ks=kiQy16 3 oxZ lseh gSA ;fn blds
SSC CGL TIER- II 03/03/2023 ik'oZ iQyd dk {ks=kiQy 30 oxZ lseh gS] r
(a) 169 cm² (b) 121 cm² Å¡pkbZ Kkr djsaA
(c) 144 cm² (d) 184 cm² SSC CGL TIER II (13/09/2019)
172. The base of a pyramid is a rectangle whose
739 209
length and breadth are 16 cm and 12 cm, (a) (b)
respectively. If the length of all the lateral 12 12
edges passing through the vertex of the 611 643
right rectangular pyramid is 26 cm, then (c) (d)
12 12
find the volume of the pyramid in cubic
176. A prism and a pyramid have the same base

r
centimeter.
and the same height. Find the ratio of the
,d fijkfeM dk vk/kj ,d vk;r gS ftldh yackbZ

si
volumes of the prism and the pyramid.
vkSj pkSM+kbZ Øe'k% 16 lseh vkSj 12 lseh gSA,d;fnfçTe vkSj ,d fijkfeM dk leku vkèkkj vk
an by
ledks.k vk;rkdkj fijkfeM ds 'kh"kZ ls xqtjus okys leku ÅapkbZ gSA fçTe vkSj fijkfeM ds vk
vuqikr Kkr dhft,A
lHkh ik'oZ fdukjksa dh yackbZ 26 lseh gS] rks fijkfeM

n
dk vk;ru ?ku lsaVhehVj esa Kkr djsaA SSC CGL TIER- II 07/03/2023
ja (a) 2 : 3 (b) 3 : 1
(a) 1536 (b) 1024
R s
(c) 1 : 3 (d) 3 : 2
(c) 718 (d) 2072
177. The base of a pyramid is a regular polygon,
a th

173. The base of a pyramid is an equilateral whose total surface area is 340 cm2, and
triangle whose each side is 8 cm. Is. Its area of base is 100 cm2, if area of each
(slant edge) is 24 cm. What is the total lateral surface is 30 cm2, then find no. of
surface area (in cm²) of the pyramid?
lateral surfaces of pyramid?
ty a

,d fijkfeM dk vk/kj ,d leckgq f=kHkqt gS] ftldh ,d fijkfeM dk vk/kj ,d le cgqHkqt gS] ftldk
izR;sd Hkqtk dh yEckbZ
8 cm. gSA bldk fr;Zd dksj
dqy i`"B {ks=kiQy 3402 lsehgS] vkSj vk/kj {ks
di M

(Slant edge) 24 cm. gSa fijkfeM dk lEiw.kZ i`"Bh;

100 lseh gS] ;fn çR;sd ik'oZ lrg dk {ks=kiQy
2
{ks=kiQy(cm² esa
) fdruk gS\
lseh2 gS] rc fijkfeM esa dqy fdrus ik'oZ lrg gS
SSC CGL TIER- II 06/03/2023
(a) 8 (b) 9
(c) 7 (d) 10
(a) 24 3  36 35  (b) 16 3  48 35  178. A pyramid has square base. The side of square
is 12cm and height of pyramid is 21cm. The
(c) 24 3  24 35  (d) 12 3  24 35 
pyramid is cut into 3parts by 2 cuts parallel
to its base. The cuts are at height of 7cm and
174. The volume of a right pyramid is 45 3
14cm respectively from the base. What is the
and its base is an equilateral triangle with difference (in cm3) in the volume of top most
side 6 cm. What is the height (in cm) of
A

and bottom most part?

the pyramid ?
oxkZdkj vk/kj okys fijkfeM dh çR;sd Hkqtk
,d yEc fijkfeM dk vk;ru 45 3 ?ku lseh gS vkSj ÅapkbZ 21 lseh gSA bl fijkfeM dks bld
vkSj bldk vk/kj ,d leckgq f=kHkqt gS ftldh Hkqtk ds lekUrj nks dV }kjk rhu Hkkxksa esa ck
6 lseh gSA bl fijkfeM dh ÅapkbZ (lseh esa) Kkr djsaAftudh vk/kj ls ÅapkbZ Øe'k% 7 lseh vkSj 14
gSA rc lcls Åij okys Hkkx vkSj lcls uhps
SSC CGL TIER II (11/09/2019)
 Mensuration-3D

Tetrahedron/leprq"iQyd (i) C. S. A. = Area of 3 equilateral triangle

Pyramid on a triangular base is a rhu leckgq f=kHkqtksa dk {ks=kiQy

tetrahedron. When a solid is bounded by
3 2
four triangular faces then it is a =3× a
tetrahedron. A right tetrahedron is so 4
called when the base of a tetrahedron is an (ii) T.S.A. = Area of 4 equilateral triangle
equilateral triangle and other triangular
faces are isosceles triangles. When we pkj leckgq f=kHkqtksa dk {ks=kiQy
encounter a tetrahedron that has all its
3 2
four faces equilateral then it is regular  4 a  3 a2
tetrahedron. 4

f=kdks.kh; vk/kj ij fijkfeM ,d prq"iQyd gSA tc(iii) Height/Å¡pkbZ

dksbZ Bksl pkj f=kHkqtkdkj iQydksa ls f?kjk gksrk gS rks 2
 a  2
og prq"iQyd gksrk gSA ,d ledks.k prq"BiQyd rc (h)  a 2 –    a
 3  3
dgykrk gSA tc ,d prq"iQyd dk vk/kj ,d leckgq

r
f=kHkqt gksrk gS vkSj vU; f=kHkqtkdkj iQyd len~fockgq
(iv) Volume/vk;ru (V)
f=kHkqt gksrs gSaA tc gekjk lkeuk fdlh ,sls prq"iQyd

si
ls gksrk gS ftlds pkjksa iQyd leckgq gksa rks og le = 1 × Area of base/vkèkkj dk {ks=kiQy
an by × height/
prq"iQyd gksrk gSA 3

Å¡pkbZ

n
ja 1 3 2 2 2 3
  a  a a
R s
3 4 3 12

a a
a th

179. If the side of a tetrahedron is 8 3 cm then

find the volume and its height.
;fn ,d prq"iQyd dh Hkqtk8 3 gS] rks mldk vk;
ty a

vkSj mldh Å¡pkbZ Kkr dhft,A

a (a) 128 6, 8 2 (b) 64 6, 4 2
di M

a (c) 32 6, 4 2 (d) 32 6, 8 2
R=
3 180. The length of one side of a regular
(a) There are four equilateral faces. tetrahedron is 8 cm. What is the ratio of
its surface area to its volume?
pkj leckgq iQyd gSaA ,d fu;fer prq"iQyd dh Hkqtk dh yackbZ 8 lse
(b) All edge are equal in length blds i`"Bh; {ks=kiQy vkSj blds vk;ru ds chp
vuqikr gS\
lHkh fdukjs yackbZ esa cjkcj gSA
(a) 3 3 : 2 2 (b) 2 : 12
(c) Slant edge is same as side of base
(c) (d) 1 : 1
A

fr;Zd mQ¡pkbZ vk/kj dh Hkqtk ds cjkcj gSA 3 :8

 Mensuration-3D

Answer Key
1.(d) 2.(b) 3.(d) 4.(c) 5.(d) 6.(c) 7.(a) 8.(b) 9.(b) 10.(b)

11.(a) 12.(d) 13.(a) 14.(b) 15.(d) 16.(a) 17.(c) 18.(a) 19.(a) 20.(d)

21.(b) 22.(c) 23.(a) 24.(d) 25.(b) 26.(a) 27.(a) 28.(b) 29.(c) 30.(a)

31.(a) 32.(a) 33.(a) 34.(d) 35.(c) 36.(a) 37.(d) 38.(b) 39.(a) 40.(a)

41.(a) 42.(c) 43.(b) 44.(a) 45.(c) 46.(b) 47.(b) 48.(d) 49.(d) 50.(c)

51.(c) 52.(a) 53.(d) 54.(b) 55.(c) 56.(a) 57.(a) 58.(c) 59.(a) 60.(d)

61.(b) 62.(a) 63.(a) 64.(b) 65.(c) 66.(c) 67.(c) 68.(c) 69.(d) 70.(c)

r
71.(b) 72.(b) 73.(b) 74.(d) 75.(b) 76.(c) 77.(d) 78.(a) 79.(b) 80.(c)

si
81.(d) 82.(d) 83.(b) 84.(a) 85.(a) 86.(b) 87.(b) 88.(d) 89.(a) 90.(d)
an by
91.(b) 92.(b) 93.(a) 94.(d) 95.(c) 96.(a) 97.(a) 98.(b) 99.(d) 100.(d)

n
101.(b) 102.(a) 103.(a)
ja 104.(d) 105.(d) 106.(a) 107.(c) 108.(b) 109.(d) 110.(a)
R s

111.(a) 112.(b) 113.(c) 114.(c) 115.(d) 116.(c) 117.(d) 118.(d) 119.(d) 120.(a)
a th

121.(a) 122.(c) 123.(d) 124.(d) 125.(d) 126.(d) 127.(d) 128.(c) 129.(b) 130.(a)

131.(c) 132.(a) 133.(b) 134.(c) 135.(c) 136.(a) 137.(a) 138.(d) 139.(b) 140.(c)
ty a

141.(b) 142.(b) 143.(b) 144.(d) 145.(d) 146.(c) 147.(b) 148.(c) 149.(c) 150.(c)
di M

151.(b) 152.(a) 153.(b) 154.(c) 155.(b) 156.(d) 157.(c) 158.(d) 159.(d) 160.(c)

161.(b) 162.(c) 163.(b) 164.(c) 165.(a) 166.(d) 167.(c) 168.(d) 169.(a) 170.(b)

171.(c) 172.(a) 173.(b) 174.(a) 175.(c) 176.(b) 177.(a) 178.(c) 179.(a) 180.(a)
A
 MENSURATION-3D /f=kfoeh; {ks=kfefr
(Practice Sheet With Solution)
1. Three cubes of metal, whose edges are 3 6. A cube of metal, each edge of which
cm, 4 cm and 5 cm respectively are melted measures 4 cm, weight 400 kg. What is the
to form a new cube. What is the total length of each edge of a cube of the same
surface area of the new cube? metal which weight 3200 kg?
èkkrq ds rhu ?ku] ftudh Hkqtk,¡ Øe'k% 3 lseh] 4 /krq dk ,d ?ku] ftlds çR;sd fdukjs dh eki 4
lseh vkSj 5 lseh gSa] dks fi?kykdj ,d u;k ?ku lseh gS] dk otu 400 fdxzk gSA mlh /krq ds ?k
cuk;k tkrk gSA u, ?ku dk dqy i`"Bh; {ks=kiQy D;k çR;sd fdukjs dh yackbZ fdruh gS ftldk otu 3200
gS\ fdyksxzke gS\
(a) 216 cm2 (b) 56 cm2 (a) 64 cm (b) 8 cm
2
(c) 36 cm (d) none of these
(c) 2 cm (d) None of these
2. There is a cubical block of wood of side

r
2 cm. If the cylinder of the largest possible7. 64 small cubes of 1 cm3 are to be arranged
volume is carved out from it. Find the in a cuboidal shape in such a way that the

si
volume of the remaining wood. surface area will be minimum. What is the
2 lseh Hkqtk dk ydM+h dk ,d ?kukdkj CykWd gSA
length of diagonal of the larger cuboid ?

an by
;fn lcls cM+s laHko vk;ru dk flysaMj mlesa ls 1 lseh3 vkdkj ds 64 ?kuksa dks bl çdkj O;ofLF
fudkyk tkrk gSA 'ks"k ydM+h dk vk;ru Kkr dhft,A fd;k x;k fd mu ls cus cM+s ?ku dk lEiw.kZ i`

n
7 12 {ks=kiQy de ls de gksA ml cM+s ?ku dk fod.k
(a) cm3 (b) cm3 gksxk \
12

ja 7
R s
5 (a) 8 2 cm. (b) 273 cm.
(c) 5 cm3 (d) none of these
a th

7 (c) 4 3 cm. (d) 4 cm.

3. The cost of the paint is Rs.36.50 per kg.
if 1kg of paint covers 16sq.ft, how much 8.
125 identical cubes are cut from a big cube
will it cost to paint outside of a cube and all the smaller cubes are arranged in
having 8 feet each side.
ty a

a row to form a long cuboid. What is the

isaV dh dher 36-50 #i;s çfr fdyks gSA ;fn 1 percentage increase in the total surface
fdyks isaV 16 oxZ iQqV dks ?ksjrk gS] rks 8 iQhV
area of the cuboid over the total surface
di M

area of the cube?

Hkqtk okys ?ku ds lHkh Hkqtk ckgj ls isaV djus esa
fdruk [kpZ vk,xk\ ,d cM+s ?ku ls 125 cjkcj vkdkj ds ?ku dkVs tk
(a) Rs.962 (b) Rs.672 gS vkSj mUgsa ,d ykbu esa j•dj ,d ?kukHk cuk f
(c) Rs.546 (d) Rs.876 bl çfØ;k ds dkj.k ?kukHk dk i`"B {ks=kiQy ?
4. How many cubes each of surface area 24 lEiw.kZ i`"B {ks=kiQy dk fdruk çfr'kr c<+sxk \
sq. dm can be made out of a metre cube,
without any wastage: 1 2
(a) 234 % (b) 234 %
,d ehVj ?ku ls fcuk fdlh viO;; ds 24 oxZ 3 3
MslhehVj {ks=kiQy okys fdrus ?ku cuk, tk ldrs gSa%(c) 117% (d) None of these
(a) 75 (b) 250 9. A cistern from inside is 12.5 m long, 8.5
(c) 125 (d) 62 m broad and 4 m high and is open at top.
5. The three co-terminus edges of a Find the cost of cementing the inside of
A

rectangular solid are 36 cm, 75 cm and 80 a cistern at Rs.24 per sq. m:

cm respectively. Find the edge of a cube
which will be of the same capacity: vanj ls ,d dqaM 12-5 ehVj yack] 8-5 ehVj pkS
,d vk;rkdkj Bksl ds rhu lg&VfeZul fdukjs Øe'k% vkSj 4 ehVj Åapk gS vkSj Åij ls [kqyk gS
#i;s çfr oxZ ehVj dh nj ls ml Vadh ds vanj
36 lseh] 75 lseh vkSj 80 lseh gSaA ,d ?ku dk fdukjk lhesaV yxkus dh ykxr Kkr dhft,\
Kkr dhft, tks leku {kerk dk gksxk% (a) Rs.6582 (b) Rs.8256
(a) 60 cm (b) 52 cm (c) Rs.7752 (d) Rs.8752
(c) 46 cm (d) None of these
10. Three cubes each of edge 3 cm long are ,d dejs ds iQ'kZ ij j[ks tk ldus okys lcls yacs
placed together as shown in the adjoining [kaHks dh yackbZ 12 ehVj gS vkSj dejs esa
figure. Find the surface area of the cuboid
so formed : ldus okys lcls yacs [kaHks dh yackbZ 15 ehV
3 lseh yacs fdukjs okys rhu ?kuksa dks ,d lkFk j[kk dejs dh Å¡pkbZ gS
x;k gS tSlk fd layXu fp=k esa fn[kk;k x;k gSA bl (a) 3 m (b) 6 m
çdkj cus ?kukHk dk i`"Bh; {ks=kiQy Kkr dhft, (c) 9 m (d) None of these
15. A rectangular block 6 cm × 12 cm × 15
cm is cut up into an exact number of equal
cubes. Find the least possible number of
cubes.
6 lseh × 12 lseh × 15 lseh ds ,d vk;rkdkj
3cm
CykWd dks cjkcj ?kuksa dh lVhd la[;k esa
tkrk gSA ?kuksa dh U;wure laHko la[;k Kkr d

m
3c
(a) 30 (b) 20
3cm 3cm 3cm (c) 25 (d) 40
(a) 182 sq. cm (b) 162 sq. cm 16. A hall is 15 m long and 12 m broad. If the
(c) 126 sq. cm (d) none of these sum of the areas of the floor and the

r
11. Three cubes of equal volume are joined end ceiling is equal to the sum of the areas
of four walls, the volume of the hall is:

si
to end. Find the surface area of the
resulting cuboid if the diagonal of the cube ,d gkWy 15 ehVj yack vkSj 12 ehVj pkSM
is 63 cm. ;fn iQ'kZ vkSj Nr ds {ks=kiQy dk ;ksx pkj nh
an by
cjkcj vk;ru okys rhu ?kuksa dks fljs ls fljs lVkdj
vkil esa tksM+k tkrk gSA ;fn ?ku dk fod.kZ
63 lseh
ds {ks=kiQy ds ;ksx ds cjkcj gS] rks gkWy dk
gS

n
gS] rks ifj.kkeh ?kukHk dk i`"Bh; {ks=kiQy Kkr dhft,A(a) 720 (b) 900

ja
SSC CGL 05/12/2022 (Shift- 04) (c) 1200 (d) 1800
R s
(a) 509 cm² (b) 504 cm² 17. How many bricks each measuring 25cm
(c) 516 cm² (d) 512 cm² × 11.25cm × 6cm, will be needed to build
a th

12. Four solid cubes, each of volume 1728 cm³, a wall 8m × 6m × 22.5m
are kept in two rows having two cubes in ,d 8 ehVj× 6 ehVj× 22.5 ehVj dk nhokj cukus
each row. They form a rectangular solid with ds fy, 25 lseh× 11.25 lseh× 6 lseh eki okyh
square base. The total surface area (in cm²) fdruh bZaVksa dh vko';drk gksxh\
ty a

of the resulting solid is:

(a) 5600 (b) 600
izR;sd 1728 lseh
3
vk;ru okys pkj Bksl ?kuksa dks nks (c) 6400 (d) 7200
iafDr;ksa esa j[kk tkrk gS vkSj izR;sd iafDr esa nks ?ku j[ks
di M

18. A cistern of capacity 8000 litres measures

tkrs gSaA muls oxkZdkj vk/kj okyk vk;rkdkj Bksl externally 3.3 m by 2.6 m by 1.1 m and
curk gSA ifj.kkeh Bksl dk dqy i`"Bh; {ks=kiQy 2
(lsehits walls are 5 cm thick. The thickness of
esa) Kkr djsaA the bottom is:
SSC MTS 18/10/2021 8000 yhVj dh {kerk okyk ,d VSad ckgjh :i ls
(a) 576 (b) 1152 3-3 ehVj× 2-6 ehVj × 1-1 ehVj eki dk gS vkSj
(c) 2304 (d) 1440 bldh nhokjsa 5 lseh eksVh gSaA uhps dh ek
13. The length of the side of a cube is 5.6 cm. djsaA
What is the volume of the largest sphere (a) 90 cm (b) 1 dm
that can be taken out of the cube? (c) 1 m (d) 1.1cm
,d ?ku dh Hkqtk dh yackbZ 5-6 ls-eh- gSA ?ku
19. lss
The dimensions of an open box are 50 cm,
ckgj fudkys tk ldus okys lcls cM+s xksys dk 40 cm and 23 cm. Its thickness is 3 cm.
A

If 1 cubic cm of metal used in the box

vk;ru D;k gS\ weight 0.5 gms, find the weight of the box.
SSC CGL 05/12/2022 (Shift- 01)
,d [kqys fMCcs dh foek,¡ 50 lseh] 40 lseh v
(a) 91.98 cm³ (b) 99.96 cm³
(c) 96.98 cm³ (d) 90.69 cm³
23 lseh gSaA bldh eksVkbZ 3 lseh- gSA ;fn
14. The length of the longest pole that can be ç;qDr /krq ds 1 ?ku lseh dk Hkkj 0-5 xzke gS
placed on the floor of a room is 12 m and fMCcs dk Hkkj Kkr dhft,A
the length of longest pole that can be placed (a) 8.04 kg (b) 8.14 kg
in the room is 15 m. The height of the room (c) 8.24 kg (d) 9.04 kg
is :
20. How many cubes of 10cm edge can be put ,d Vadh 6 ehVjyEch rFkk4 ehVj pkSM+h gS ft
in a cubical box of 1m edge ehVj25 lseh Å¡pkbZ rd ikuh Hkjk gSA xhy
1
1 ehVj fdukjs okys ,d ?kukdkj fMCcs esa 10 lseh dk dqy i`"Bh; {ks=kiQy gksxk&
fdukjs okys fdrus ?ku j[ks tk ldrs gSa\ (a) 55 m² (b) 53.5 m²
(a) 1000 (b) 100 (c) 50 m² (d) 49 m²
(c) 10 (d) 10000 26. A rectangular field is 40m. long and 14m.
broad. In one corner of it, a pit 12m. long,
21. Water flows into a tank 200 m × 150 m 6m. wide and 5m. deep has been dug out
through a rectangular pipe of 1.5m × 1.25 and the earth taken out of it has been
m at 20 kmph. In what time (in minutes) evenly spread over the remaining part of
will the water rise by 2 meters? the field. Find the rise in level of the field.
200 ehVj × 150 ehVj ds VSad esa 1-5 ehVj
× 1-25 ,d vk;rkdkj •sr 40 ehVj yEck o 14 ehVj pkSM+
ehVjvk;rkdkj ikbZi ds ekè;e ls 20 fdeh çfr gSA blds ,d fdukjs ij 12 ehVj yEck] 6 ehVj pkSM
?kaVsnjdsls ikuhcgrk gSAfdrus le; esa (feuVksa o 5 ehVj xgjk xM~<+k •ksnk x;k o mlls fud
esa) ikuh 2 ehVj Åij mB tk,xk\ feêðh dks iwjs •sr esa iQSyk fn;kA ml feêðh d
(a) 92 min (b) 93 min •sr dk Lrj fdruk Åij mBsxk\
(a) 73.77 cm. (b) 72.12 cm.
(c) 95 min (d) 96 min
(c) 70 cm. (d) 75 cm.

r
22. The external dimensions of a wooden box
27. A field in the form of a rectangle having
closed at both ends are 24 cm, 16 cm and

si
length 20 m and breadth 25 m. There is a
10 cm respectively and thickness of the square pit outside the field having
wood is 5 mm. If the empty box weight dimension 15 m × 15 m. This pit is to be

wood:
an by
7.35 kg, find the weight of 1 cubic cm of filled uniformly upto a height of 4 m with
the soil taken out by digging the

n
nksuksa fljksa ij can ,d ydM+h ds cDls dk ckgjh vk;ke rectangular field. Find out the depth upto
which the rectangular field must be dug if
Øe'k% 24 lseh] 16 lseh vkSj 10 lseh gS vkSj ydM+h the soil is to fill the pit?

ja
dh eksVkbZ 5 feeh gSA ;fn •kyh fMCcs dk Hkkj 7-35 ,d vk;rkdkj •sr ftldh yEckbZ 20 ehVj o pkSM+k
R s
fdxzk gS] rks 1 ?ku lseh ydM+h dk Hkkj Kkr dhft,A 25 ehVj gSA blds ckgj 15 × 15 ehVj dk ,d xM~<+k
a th

(a) 10 g (b) 12.5 g gSA bl xM~<+s dks 4 ehVj rd Hkjus esa •sr ls
(c) 27 g (d) 15 g feêðh •ksnh xbZ mldh xgjkbZ crk,aA
23. A cube of 11 cm edge is immersed 9 9
completely in a rectangular vessel (a) m. (b) m.
ty a

containing water. If the dimensions of base 5 2

are 15 cm and 12 cm. Find the rise in 9 9
(c) m. (d) m.
di M

water level of the vessel: 7 4

11 lseh fdukjs dk ,d ?ku ikuh ls Hkjs ,d vk;rkdkj 28. A tank is in the form of a cuboid with
crZu esa iwjh rjg Mwck gqvk gSA ;fn vk/kj dh foek,¡length 12m. If 18 kilolitre of water is
removed from it, the water level goes down
15 lseh vkSj 12 lseh gSaA crZu esa ty Lrj esa o`f¼by 30cm. What is the width (in m) of the
Kkr dhft,% tank?
(a) 6.85 cm (b) 7 cm ?kukHk ds vkdkj okys ,d VSad dh yackbZ 12 eh
(c) 7.31 cm (d) 7.39 cm ;fn blls 18 fdyks yhVj ikuh fudky fy;k tkrk gS
24. A cuboid of size 50cm. × 40cm. × 30cm. is rks bldk ty Lrj 30 lseh uhps pyk tkrk gSA VS
cut into 8 identical parts by 3 cuts. What dh pkSM+kbZ (ehVj esa) fdruh gS\
is the total surface area (in cm²) of all the (a) 4.5 (b) 4
8 parts? (c) 5 (d) 5.5
,d ?kukHk dk vkdkj50 lseh× 40 lseh× 30 lseh 29. Water flows into a tank which is 200m long
A

gS bls 3 dVksa }kjk 8 leku Hkkxksa esa dkVk tkrk gSA and 150m wide, through a pipe of cross-
lHkh 8 Hkkxksa dk dqy i`"Bh; {ks=kiQy 2
esa) (lseh
D;k section 0.3m × 0.2m at 20 km/hour. Then
the time (in hours) required for the water
gS\ level in the tank to reach 8m
(a) 11750 (b) 14100 20 fdeh@?kaVk dh xfr0.3mls × 0.2m ØkWl&lsD'k
(c) 18800 (d) 23500 ds ,d ikbi ds ekè;e ls ikuh 200 ehVj yacs vkSj
25. A cistern 6m. long and 4m. wide, contains 150 ehVj pkSM+s ,d VSad esa cgrk gSA VSad
water up to a height of 1m. 25cm. The Lrj 8 ehVj rd igqapus esa yxus okyk le; (?ka
total area of the wet surface is. esa) gksxk
 (a) 50 (b) 120 35. A conical vessel has a capacity of 15 L of
(c) 150 (d) 200 milk. Its height is 50 cm and base radius
30. The base of a rectangular reservoir is 15 is 25 cm. How much milk can be contained
m long × 12 m wide. In this water flows at in a vessel in cylindrical form having the
the rate of 16 m/s through a pipe whose same dimensions as that of the cone?
cross-section is 5 cm × 3 cm. To what
height will the water rise in the reservoir ,d 'kaDokdkj crZu esa 15 yhVj nwèk dh {kerk
in 25 minutes. gSA bldh ÅapkbZ 50 lseh vkSj vkèkkj f=kT;k
,d vk;rkdkj tyk'k; dk vk/kj 15 ehVj yack × gSA 'kadq ds leku vk;ke okys csyukdkj crZ
12 ehVj pkSM+k gSA blesa ikuh ,d ikbi ds ekè;e ls fdruk nwèk lek ldrk gS\
16 ehVj@lsdaM dh nj ls cgrk gS ftldk ØkWl&lsD'ku (a) 15 L (b) 30 L
5 lseh × 3 lseh gS 25 feuV esa tyk'k; esa ikuh (c) 45 L (d) none of these
fdruh ÅapkbZ rd c<+ tk,xkA 36. A cylindrical rod of iron, whose height is
(a) 0.2 m (b) 2 m equal to its radius, is melted and cast into
(c) 0.5 m (d) 0.02m spherical balls whose radius is half the
31. A rectangular block of length 20 cm, radius of the rod. Find the number of balls.
breadth 15 cm and height 10 cm is cut up
yksgs dh ,d csyukdkj NM+] ftldh Å¡pkbZ m

r
into exact number of equal cubes. The least
possible number of cubes will be f=kT;k ds cjkcj gS] dks fi?kykdj xksykdkj xsa

si
yackbZ 20 lseh] pkSM+kbZ 15 lseh vkSj ÅapkbZ 10 <kyk
lsehtkrk gS] ftldh f=kT;k] NM+ dh f=kT;
ds vk;rkdkj CykWd dks cjkcj ?kuksa dh lVhd la[;k vkèkh gksrh gSA xsanksa dh la[;k Kkr dhft,

la[;k gksxh an by
esa dkV fn;k tkrk gSA ?kuksa dh de ls de laHko (a) 3
(c) 5
(b) 4
(d) 6

n
(a) 12 (b) 16 37. If the radius of the cylinder is increased
(c) 20 (d) 24 by 25%, then by how much percent the

ja
R s
32. The liquid in a container is sufficient to height must be reduced, so that the
paint an area of 11.28 m². How many boxes volume of the cylinder remains same?
a th

of dimension 30 cm × 25 cm × 12 cm can ;fn csyu dh f=kT;k 25» c<+k nh tkrh gS] rks m
be painted with the liquid in this container.
,d crZu esa 11-28 eh
2
{ks=kiQy dks jaxus ds fy, i;kZIr ÅapkbZ fdrus çfr'kr de dh tkuh pkfg,] rkfd
rjy gSA foekvksa 30 lseh
× 25 lseh × 12 lseh ds fdrus csyu dk vk;ru leku jgs\
ty a

cDls bl crZu ds rjy }kjk jaxs tk ldrs gSa\ (a) 36 (b) 56

SSC CPO 16/03/2019 (Shift-01) (c) 64 (d) 46
di M

(a) 40 (b) 24 38.

The radius of base of a solid cylinder is 7
(c) 32 (d) 12 cm and its height is 21 cm. It is melted
and converted into small bullets. Each
33. The ratio between curved surface area and
total surface area of cylinder is 2 : 3. If bullet is of same size. Each bullet consisted
the total surface area be 924 cm2, find the of two parts . A cylinder and a hemisphere
volume of the cylinder : on one of its base. The total height of
csyu ds oØ i`"Bh; {ks=kiQy vkSj dqy i`"Bh; {ks=kiQy
bullet is 3.5 cm and radius of base is 2.1
ds chp dk vuqikr 2 % 3 gSA ;fn dqy i`"Bh; cm. Approximately how many complete
{ks=kiQy 924 2lseh
gS] rks csyu dk vk;ru Kkr bullets can be obtained?
dhft,A ,d Bksl csyu ds vkèkkj dh f=kT;k 7 lseh gS v
(a) 2156 cm3 (b) 1256 cm3 bldh Å¡pkbZ 21 lseh gSA bls fi?kykdj NksV
A

3
(c) 1265 cm (d) none of these
xksfy;ksa esa cnyk tkrk gSA çR;sd xksyh ,d gh
34. The radii of two cylinders are in the ratio
of 3 : 5 and their heights are in the ratio dh gksrh gSA çR;sd xksyh esa nks Hkkx gk
4: 3. The ratio of their volumes is csyu vkSj mlds ,d vkèkkj ij ,d vèkZxksykA xk
nks csyuksa dh f=kT;k,¡ 3 % 5 ds vuqikr esa gSa vkSj
dh dqy ÅapkbZ 3-5 lseh vkSj vkèkkj dh f=kT
mudh Å¡pkbZ 4 % 3 ds vuqikr esa gSA muds vk;ruksa
lseh gSA yxHkx fdruh iw.kZ xksfy;ka çkIr
dk vuqikr gS ldrh gSa\
(a) 12 : 25 (b) 13 : 25
(c) 4 : 5 (d) 5 : 4 (a) 83 (b) 89
(c) 74 (d) 79
39. A cylindrical vessel of radius 6 cm is 14 ehVj vkarfjd O;kl okyk ,d dqvka 15 ehVj xgjk
partially filled with water. By how much •ksnk x;k gSA blesa ls fudkyh xbZ feêðh dk
will the water level rise if a sphere of radius pkjksa vksj leku :i ls 21 ehVj dh pkSM+kbZ e
5 cm is completely immersed in this rVca/ cukus ds fy, iQSyk;k x;k gSA pcwrjk dh Å
water? fdruh gS\
22 (a) 1 m (b) 2 m
(Take  = ) (c) 3 m (d) 4 m
7
f=kT;k 6 lseh dk ,d csyukdkj crZu vkaf'kd :i 43. If the radius of cylinder is doubled, but
height is reduced by 50%. What is the
ls ikuh ls Hkjk gqvk gSA ;fn 5 lseh f=kT;k dk ,d percentage change in volume?
xksyk bl ikuh esa iwjh rjg Mqcks fn;k tk, rks ikuh ;fn csyu dh f=kT;k nksxquh dj nh tkrh gS] ysfdu Å
22 50» de dj nh tkrh gSA vk;ru esa çfr'kr ifjorZu
dk Lrj fdruk c<+ tk,xk\ (  = 7 ) D;k gS\
(a) 6.67 cm (b) 5.56 cm (a) 50% (b) 25%
(c) 200% (d) 100%
(c) 6.94 cm (d) 4.63 cm
44. How many cubic metres of water is filled
40. A solid cylinder has radius of base 14 cm in a pipe which is 3500 m long and 0.08

r
and height 15 cm. 4 identical cylinders m in diameter?
are cut from each base as shown in the ,d ikbi tks 3500 ehVj yack vkSj 0-08 ehVj O;k

si
given figure. Height of small cylinder is
dk gS] mlesa fdrus ?ku ehVj ikuh Hkj ldrk gS\
5 cm. What is the total surface area of the
remaining part?
an by
,d Bksl csyu ds vkèkkj dh f=kT;k 14 lseh vkSj
(a) 17.5 m³
(c) 21 m³
(b) 17.6 m³
(d) 35 m³

n
45. A cube of metal, whose edge is 10 cm, is
ÅapkbZ 15 lseh gSA çR;sd vk/kj ls 4 leku csyu wholly immersed in water contained in
dkVs x, gSa tSlk fd fp=k esa fn[kk;k x;k gSA NksVs
cylindrical tube whose diameter is 20 cm.

ja
R s
By how much will the water level rise in
csyu dh Å¡pkbZ 5 lseh- 'ks"k Hkkx dk dqy i`"Bh;the tube?
{ks=kiQy fdruk gS\
a th

/krq dk ,d ?ku] ftldh Hkqtk 10 lseh gS] 20 lseh

O;kl okyh ,d csyukdkj uyh esa j•s ikuh esa iw
rjg Mwck gqvk gSA uyh esa ikuh dk Lrj fdru
tk,xk\
ty a

3
(a) 3.3 cm (b) 6 cm
di M

11
2
(a) 3740 (b) 3432 (c) 3 cm (d) None of these
11
(c) 3124 (d) 2816 46. The amount of concrete required to build a
41. A hollow cylindrical tube open at both ends cylindrical pillar whose base has a perimeter
is made of iron 2 cm thick. If the external 8.8 m and whose curved surface area is 17.6
diameter be 50 cm and the length of the m²:
tube is 210 cm, find the volume of iron in ,d csyukdkj LraHk ds fuekZ.k ds fy, vko';d daØh
it. dh ek=kk Kkr djsa ftlds vk/kj dh ifjf/ 8-8 ehVj
nksuksa fljksa ij •qyh ,d •ks•yh csyukdkj uyh 2 gS vkSj ftlds oØ i`"B dk {ks=kiQy 17-6 oxZ
lseh eksVh yksgs dh cuh gSA ;fn ckgjh O;kl 50 lsehgS%
(a) 12.32 m³
rFkk uyh dh yEckbZ 210 lseh gks] rks mlesa yksgs dk (b) 12.23 m³
A

(c) 9.235 m³ (d) 8.88 m³

vk;ru Kkr dhft,A 47. The radius of an iron rod decreased to one-
(a) 63360 cm³ (b) 63,000 fourth. If its volume remains constant, the
(c) 63,500 (d) 64,000 length will become:
42. A well with 14 m inside diameter is dugout ,d yksgs dh NM+ dh f=kT;k ?kVdj ,d pkSFkk
15 m deep. The earth taken out of it has tkrh gSA ;fn bldk vk;ru fLFkj jgrk gS] rks yac
been evenly spread all around it to a width gks tk,xh%
of 21 m to form an embankment. What is (a) 2 times (b) 8 times
the height of the embankment?
(c) 4 times (d) 16 times
48. The ratio of heights of two cylinders is 3 : ,d csyukdkj vkÑfr dh iQkmaVsu isu dh cksry
2 and the ratio of their radii is 6 : 7. What lseh yEch gS rFkk O;kl 5 ehyh ehVj gSA L;k
is the ratio of their curved surface areas?
,d cksry vkSlr 330 'kCn fy•us esa •pZ gks tkrh
nks csyuksa dh ÅapkbZ dk vuqikr 3 % 2 gS vkSj mudh 1
gSA ,d cksry ftlesa 1 yhVj dk5 Hkkx dks •pZ
f=kT;kvksa dk vuqikr 6 % 7 gSA muds oØ i`"Bh; {ks=kiQyksa
dk vuqikr D;k gS\ djus ds fy, fdrus 'kCnksa dks fy•uk gksxk\
(a) 9 : 7 (b) 1 : 1 (a) 48000 (b) 42000
(c) 7 : 9 (d) 7 : 4 (a) 56000 (b) 28000
49. The curved surface of a cylinder is 1000
sq cm. A wire of diameter 5mm is wound 54. Two cylindrical vessels with radii 15 cm
around it, so as to cover it completely. and 10 cm and heights 35 cm and 15 cm
What is the length of the wire used ? respectively are filled with water. If this
water is poured into a cylindrical vessel 15
,d csyu dk oØ i`"B dk {ks=kiQy 1000 lseh
2
gSA 5 cm in height, then the radius of the vessel
feyh ehVj O;kl okyk rkj bldks ?ksjs gq, gSa] rkfd is :
iw.kZ :i ls <+d ldsA ml rkj dh yackbZ Kkr djsaA nks csyukdkj Vadh ftudh f=kT;k Øe'k% 15 lseh
(a) 22 m (b) 20 m 10 lseh vkSj Å¡pkbZ Øe'k% 35 lseh vkSj 15 ls
(c) 18 m (d) None of these ikuh ls Hkjh gqbZ gSA ;fn ;g ty ,d nwljh Vad
mM+syk tkrk gS ftldh Å¡pkbZ 15 lseh gS rks V

r
50. The outer and inner diameters of a circular
pipe are 6 cm and 4 cm, respectively. If its f=kT;k D;k gksxh \

si
length is 10 cm, then what is the total
surface area in sq. cm. ? (a) 25 cm (b) 20 cm
(c) 17.5 cm (d) 18 cm
,d csyukdkj ikbi ds ckÞ; vkSj var% O;kl Øe'k%

an by
6 lseh vkSj 4 lseh gSA ;fn mldh yackbZ 10 lseh gSAtotal surface area of a solidsurface
rc mldk dqy i`"B dk {ks=kiQy (lseh
2
esa) Kkr djsaA
55. The sum of the curved area and
cylinder is 2068

n
cm2. If radius of its base is 7 cm, then what
(a) 35  (b) 110  is the volume of this cylinder?
(c) 510  (d) None of these ,d Bksl csyu ds oØ i`"Bh; {ks=kiQy rFkk la

ja i`"Bh; {ks=kiQy dk ;ksx 20682 lseh gSA ;fn blds

R s
51. A cylindrical tank of diameter 35 cm is full
of water. If 11 litres of water is drawn off vk/kj dh f=kT;k 7 lsehgks] rks bl csyu dk vk;ru
the water level in the tank will drop by :
a th

D;k gksxk\
(Take  = 22/7)
(a) 2480 cm³ (b) 2760 cm³
,d 35 lseh O;kl okyk csyukdkj Vadh ikuh ls Hkjh (c) 3080 cm³ (d) 2060 cm³
gSA vxj mlesa ls 11 yhVj ikuh fudky fy;k tk, rks
56. The height of a solid cylinder is 35 cm. The
ty a

ikuh ds Lrj esa D;k fxjkoV vk,xh \ circumference of its base is 37 cm more
1 6 than the radius. What will be the volume of
di M

(a) 10 cm (b) 12 cm this cylinder?

2 7
,d Bksl csyu dh Å¡pkbZ 35 lseh gSA blds vk/kj
(c) 14 cm
3
(d) 11 cm ifjf/ bldh f=kT;k ls 37 lseh vf/d gSA bl csyu dk
7 vk;ru D;k gksxk\
52. A solid cylinder has total surface area of
SSC CGL MAINS (08/08/2022)
1
462 sq. cm. its curved surface area is rd (a) 4420 cm³ (b) 4850 cm³
3
of the total surface area. Then the radius (c) 4740 cm³ (d) 5390 cm³
of the cylinder is 57. A 15 m deep well with radius 2.8 m is dug
and the earth taken out from it is spread
,d csyu dk dqy i`"B {ks=kiQy 462 2lseh gSA bldk evenly to form a platform of breadth 8 m
oØ i`"B dk {ks=kiQy blds lEiw.kZ i`"Bh; {ks=kiQy and dk height 1.5 m. What will be the length
1 of the platform?
xquk gSA csyu dh f=kT;k Kkr djsaA
A

3 2-8 ehVj f=kT;k okyk] ,d 15 ehVj xgjk dqvka [k

(a) 7 cm (b) 3.5 cm tkrk gS blls fudyh feV~Vh dks cjkcj djds 8 ehV
(c) 9 cm (d) 11 cm pkSM+k vkSj 1-5 ehVj Åapk ,d pcwrjk cuk;k t
53. The barrel of a fountain-pen, cylindrical in pcwrjs dh yackbZ D;k gS\
shape, is 7 cm long and 5 mm is diameter.
A full barrel of ink in the pen will be used  22 
up on writing 330 words on an average.   = 
7 
How many words would be written by a
1 (a) 28.8 m (b) 30.8 m
bottle of ink containing of a litre ?
5 (c) 28.4 m (d) 30.2 m
58. A cylinder can whose base is horizontal and 62. A conical tent has 60° angle at the vertex.
is of internal radius 3.5 cm contains The ratio of its radius and slant height is
:
sufficient water, so that when a solid
sphere is placed inside it. Water just covers ,d 'kaDokdkj racw ds 'kh"kZ ij dk dks.k gSA
60°
the sphere. The sphere fits in the can bldh f=kT;k vkSj frjNh ÅapkbZ dk vuqikr gS
exactly. The depth of water in can before (a) 3 : 2
the sphere was put) (b) 1 : 2
(c) 1 : 3
,d csyu ftldk vkrafjd f=kT;k 3-5 lseh gS] bl (d) can't be determined
csyu ds vanj bruk ikuh Hkjk gqvk gS] fd tc blds63. Water flows at the rate of 5 m per min
vanj ,d xksyk Mkyk tkrk gSa rks ;g xksys dks iw.kZr%
from a cylindrical pipe 8 mm in diameter.
How long will it take to fill up a conical
Mqcks nsrk gS] xksys Mwcus ls igys ikuh dh xgjkbZvessel
D;k whose radius is 12 cm and depth
Fkh \ 35 cm?
8 feeh O;kl okys ,d csyukdkj ikbi ls ikuh 5
(a)
25
(b)
17 ehVj çfr feuV dh nj ls cgrk gSA ,d 'kaDokdk
3 3 crZu ftldh f=kT;k 12 lseh vkSj xgjkbZ 35 ls
gS] dks Hkjus esa fdruk le; yxsxk\

r
7 14 (a) 315 s (b) 365 s
(c) (d)

si
3 3 (c) 5 min (d) none of these
59. The radius and height of a right circular 64. A solid cone of height 36 cm and radius

an by
cone are in the ratio of 5 : 12. If its volume of base 9 cm is melted to form a solid
cylinder of radius 9 cm and height 9 cm.
What percent of material is wasted in this

n
2
is 314 m3, its Slant height is : process?
7
36 lseh ÅapkbZ vkSj 9 lseh vkèkkj f=kT;k ok

ja
,d yEc o`Ùkh; 'kadq dh f=kT;k vkSj ÅapkbZ 5 % 12
Bksl 'kadq dks fi?kykdj 9 lseh f=kT;k vkSj 9
R s
2 ÅapkbZ dk ,d Bksl csyu cuk;k tkrk gSA bl çfØ
ds vuqikr esa gSA ;fn bldk vk;ru314 eh3
a th

7 esa fdrus çfr'kr lkexzh cckZn gksrh gS\

gS] rks bldh frjNh ÅapkbZ gS (a) 5 (b) 25
(c) 10 (d) 15
(a) 26 m (b) 19.5 m
65. Right triangle with sides 3 cm, 4 cm and
ty a

(c) 13 m (d) none of them 5 cm is rotated with the side of 3 cm to

60. How many metres of cloth 10 m wide will form a cone. The volume of the cone so
di M

be required to make a conical tent with formed is:

base radius of 14 m and height is 48 m? 3 lseh] 4 lseh vkSj 5 lseh Hkqtkvksa okys le
14 ehVj ds vkèkkj dh f=kT;k vkSj 48 ehVj dh f=kHkqt dks 3 lseh dh Hkqtk ls ?kqekdj ,d
Å¡pkbZ okys ,d 'kaDokdkj rEcw dks cukus ds fy, cuk;k tkrk gSA bl çdkj cus 'kadq dk vk;ru gS
(a) 12  cub.cm (b) 15  cub.cm
10 ehVj pkSM+s vkSj fdrus ehVj yacs diM+s dh
(c) 16  cub.cm (d) 20  cub.cm
vko';drk gksxh\ 66. An inverted conical shaped vessel is filled
(a) 110 m (b) 55 m with water to its brim. The height of the
(c) 77 m (d) 220 m vessel is 8 cm and radius of the open end
is 5 cm. When a few solid spherical metallic
61. From a circular sheet of paper of radius
25 cm, a sector area 4% is removed. If the 1
balls each of radius cm are dropped in
remaining part is used to make a conical 2
A

surface, then the ratio of the radius and the vessel , 25% water is overflowed. The
number of balls is:
height of the cone is :
,d mYVs 'kaDokdkj vkdkj dk crZu ikuh ls H
25 lseh f=kT;k okys dkxt dh ,d o`Ùkkdkj 'khV gqvk gSA crZu dh ÅapkbZ 8 lseh vkSj [kqys
ls] ,d f=kT;[kaM {ks=kiQy 4» gVk fn;k tkrk gSA 1
;fn 'ks"k Hkkx dk mi;ksx 'kaDokdkj lrg cukus ds f=kT;k 5 lseh gSA 2 tc lseh f=kT;k dh dqN Bk
fy, fd;k tkrk gS] rks 'kadq dh f=kT;k vkSj ÅapkbZxksykdkj èkkfRod xsanksa dks crZu esa fxjk;
dk vuqikr gS rks 25» ikuh cg tkrk gSA xsanksa dh la[;k gS
(a) 16 : 25 (b) 9 : 25 (a) 100 (b) 400
(c) 7 : 12 (d) 24 : 7 (c) 200 (d) 150
67. A right circular cone is cut parallel to its vk/kj f=kT;k 3 lseh vkSj ÅapkbZ 5 lseh okys ,d
/krq ds csyu dks fi?kykdj ÅapkbZ 1 lseh vkSj v
1
base at rd of its height from the base f=kT;k 1 feeh okys
n Bksl 'kadq cuk, tkrs gSaA
n dk
3 eku Kkr dhft,A
what is the ratio of the volume of the (a) 15,000 (b) 14,000
smaller cone formed to that of the frustum (c) 13,500 (d) 14,500
formed? 72. The height of a cone is 30 cm. A small cone
,d yac o`Ùkh; 'kadq dks mlds vkèkkj ds lekUrj vkèkkj is cut off at the top by a plane parallel to
1 1
ls mldh oh Å¡pkbZ ij dkVk tkrk gS] cus NksVs 'kadq the base. If its volume be
27
of the
3
volume of the given cone, at what height
ds vk;ru dk fNUud ds vk;ru ls vuqikr D;k gS\ above the base is the section made?
(a) 8 : 27 (b) 19 : 27 ,d 'kadq dh Å¡pkbZ 30 lseh gSA vk/kj ds leku
(c) 11 : 27 (d) 8 : 19 ,d ry }kjk ,d NksVk 'kadq 'kh"kZ ij dkVk tkrk
68. The base radius and height of a cone are 1
6 cm and 36 cm respectively. If a cone ;fn bldk vk;ru fn, x, 'kadq ds vk;ru dk
27
is cut parallel to its base at the height of gS] rks vk/kj ds Åij fdl Å¡pkbZ ij •aM cuk;k x;k

r
the center of the base. If the volume of gS\
this frustum is 264 cm3. Find the radius

si
(a) 20 cm (b) 18 cm
of the smaller cone?
(c) 27 cm (d) 15 cm
'kadq dk vkèkkj f=kT;k vkSj ÅapkbZ Øe'k%73. 6 lseh

an by
vkSj 36 lseh gSA ;fn 'kadq dks vkèkkj h dhls
ÅapkbZ ij mlds vkèkkj ds lekukarj dkVk tkrk gSAbeing
A tent is in the form of right circular cone
10.5 m high, the diameter of the base
13 m. If 8 men are in the tent, find

n
the average number of cubic metres of air
;fn bl fNUud dk vk;ru 264 lseh3 gSA NksVs 'kadq space per man:
dh f=kT;k Kkr dhft;s\ ,d racw 10-5 ehVj Åaps yEc o`Ùkh; 'kadq ds :
ja
R s
(a) (104)1/3 cm (b) (104)1/2 cm gS] ftlds vk/kj dk O;kl 13 ehVj gSA ;fn 8 vkneh
(c) 5 cm (d) (174)1/3 cm racw esa gSa] rks çfr O;fÙkQ vkSlr ?ku ehVj
a th

69. A conical vessel (solid) is made of iron. Kkr dhft,%

Its base radius is 7 cm and height is 15 3
cm. If the weight of the iron per cubic (a) 32 (b) 59.75
58
ty a

centimetre is 15 g, what is the weight of

9 3
the vessel? (c) 36 (d) 58
13 32
di M

,d 'kaDokdkj ik=k (Bksl) yksgs dk cuk gSA 74.

blds A concial vessel whose internal radius is 10
vk/kj dh f=kT;k7 cm vkSj Å¡pkbZ 15 cm gSA cm and height 72 cm is full of water. If this
water is poured into a cylindrical vessel
;fn yksgs dk Hkkj izfr ?ku lsaVhehVj
15 g gS] rks with internal radius 30 cm, the height of
ik=k dk Hkkj Kkr dhft,A the water level rises in it is:
,d 'kaDokdkj crZu ftldh vkarfjd f=kT;k 10 lseh
SSC CGL 09/12/2022 (Shift- 02)
vkSj ÅapkbZ 72 lseh gS] ikuh ls Hkjk gSA ;fn
(a) 13.55 kg (b) 12.55 kg dks 30 lseh vkarfjd f=kT;k okys ,d csyukdkj ik
(c) 14.55 kg (d) 11.55 kg esa Mkyk tk,] rks blesa ty Lrj dh Å¡pkbZ fdruh
70. Find the area of the iron sheet required to tkrh gS\
prepare a cone 20 cm high with base radius 2 2
21 cm. (a) 2 cm (b) 3 cm
3 3
A

21 lseh vk/kj f=kT;k okys 20 lseh Å¡ps ,d 'kadq dks 2

cukus ds fy, vko';d yksgs dh pknj dk {ks=kiQy (c) 5 cm (d) None of these
3
Kkr dhft,A 75. The volume of a right circular cone is 1232
(a) 3300 cm² (b) 3250 cm² cm3 and its vertical height is 24 cm. Its
(c) 3400 cm² (d) 3350 cm² curved surface area is.
71. A solid metallic cylinder of base radius 3 ,d yac o`Ùkh; 'kadq dk vk;ru 1232 lseh
3
gS rFkk
cm and height 5 cm is melted to make n mldh Å¡pkbZ 24 lseh gSA 'kadq dk oØ i`"B {
solid cones of height 1 cm and base radius
1 mm. Find the value of n. Kkr djsaA
 (a) 154 cm² (b) 550 cm² (a) 432  (b) 603 
(c) 604 cm² (d) 704 cm² (c) 433  (d) 539 
81. The radius of the base of a conical tent is
76. A sector is formed by opening out a cone
of base radius 8 cm and height 6 cm. The 3
16 metre. If 427 sq. metre canvas is
radius of the sector is (in cm.) 7
required to construct the tent, the slant
,d 8 cm. vk/kj f=kT;k rFkk 6 cm. Å¡pkbZ okys height of the tent is :
'kadq dks •ksydj ,d f=kT;•aM cuk;k tkrk gSA ml ,d 'kaDokdkj VSaV dh f=kT;k 16 ehVj gSA v
f=kT;•aM dh f=kT;k crk,¡A 3
dks cukus ds fy, 427 ehVj2 dSuokl dh
(a) 4 (b) 8 7
(c) 10 (d) 6
vko';drk gksxhrks VSaV dh fr;Zd Å¡pkbZ Kkr d
(a) 17 metre (b) 15 metre
77. A sector of circle of radius 3 cm has an (c) 19 metre (d) 8.5 metre
angle of 120°. If it is moulded into a cone, 82. A cardboard sheet in the form of a circular
find the volume of the cone. sector of radius 30 cm and central angle
,d o`Ùk•aM ftldh f=kT;k 3 lseh& vkSj dks.k
120° 144° is folded to make a cone. What is the
radius of the cone ?
gS] dks ,d 'kadq ds :i esa cnyk tkrk gS] rks 'kadq ,d 144 º dsaæh; dks.k rFkk 30 lseh f=kT;k okyk o`

r
dk vk;ru Kkr djksA xÙkk gSA mls eksM+dj cuk, x, 'kadq dh f=k
djsaA

si
 2 2
(a) cm³ (b) cm³ (a) 12 cm (b) 18 cm
3 3 (c) 21 cm (d) None of these

(c)
2 3
cm³ an by (d)
3
cm³
83. The height of a solid cone is 20 cm. A small
cone is cut off from the top of it such that

n
  base of the cone cut off and the base of a
given cone are parallel to each other. If the
78. A plane divides a cone into two parts of volume of the cone cut and the volume of

ja
equal volume. If the plane is parallel to the the original cone are in the ratio of 1 : 8,
R s
base, then the ratio in which the height of find the height of the frustum.
the cone is divided, is- ,d Bksl 'kadq dh Å¡pkbZ 20 lseh gSA bls vk/k
a th

,d ry 'kadq dks nks cjkcj vk;ruksa esa ck¡Vrk gS ;fn lekUrj dkV dj ,d NksVk vU; 'kadq curk gSA ;
ry vk/kj ds lekukarj gS] rks Å¡pkbZ dk vuqikr D;k NksVs 'kadq rFkk ml Bksl 'kadq ds vk;ru dk vu
% 8 gks rks fNUud dh Å¡pkbZ Kkr djks&
gksxk tks 'kadq dks nks Hkkxksa esa ck¡Vrh gSA (a) 6 cm (b) 8 cm
ty a

(a) 1 : 2 (b) 1 : 3 2 – 1 (c) 10 cm (d) 12 cm

84. The height of a cone is 40 cm. If a small
di M

(c) 1 : 3 2 (d) 1 : 3 2  1 cone is cut off at the top by a plane parallel

to the base of the cone, the volume of the
79. A solid cone of height 8 cm and base radius 1
6 cm is melted and recast into identical smaller con is the volume of the larger
64
cones, each of height 2 cm and radius 1 con. What is the height of the frustum
cm. What is the number of cones formed ,d 'kadq dh Å¡pkbZ 40 lseh gSA ;fn 'kadq ds v
? ds lekukarj ,d lery }kjk 'kh"kZ ij ,d NksVk 'kad
,d 6 lseh f=kT;k rFkk 8 lseh Å¡pkbZ okys yac o`Ùkh;dkVk tkrk gS] rks NksVs 'kadq dk vk;ru cM+s
'kadq dks fi?kykdj 2 lseh& Å¡pkbZ rFkk 1 lseh f=kT;k
vk;ru dk
1
gSA fNUud dh Å¡pkbZ fdruh gS
okys u, 'kadqvksa esa cny fn;k tkrk gSA u, 'kadqvksa 64
CRPF HCM 23/02/2023 (Shift - 03)
dh la[;k Kkr djsaA
(a) 5 (b) 15
A

(a) 36 (b) 72 (c) 10 (d) 30

(c) 144 (d) 180 85. Volume of a cone whose radius of base and
80. The radii of the circular end of a conical height are r and h respectively, is 400 cm3.
bucket are 14 cm and 6 cm, whose height What will be the volume of a cone whose
radius of base and height are 2r cm and h
is 6 cm, find the total surface area of cm respectively?
bucket.
,d 'kadq] ftlds vk/kj dh f=kT;k rFkk Å¡pkbZ Øe'k
r
,d 'kaDokdkj fNUud ckYVh ftldh Å¡pkbZ 6 lseh gS rFkkh gS] dk vk;ru 400 lseh3 gSA ,d 'kadq] ftlds
vk/kj dh f=kT;k rFkk Å¡pkbZ Øe'k%
vkSj f=kT;k,¡ 14 lseh rFkk 6 lseh gSaA ckYVh dk dqy 2r lseh rFkkh
i`"Bh; {ks=kiQy Kkr djks& lseh gS] dk vk;ru D;k gksxk\
 SSC CGL MAINS (08/08/2022) 185 175
(a) 100 cm³ (b) 1200 cm³ (a) (b)
6 3
(c) 1600 cm³ (d) 800 cm³ 185 175
(c) (d)
86. If the curved surface area of a right circular 3 2
cone is 10010 sq cm and its slant height 90. Find the volume of a spherical shell whose
is 91 cm, find its total surface area. external and internal diameters are 14 cm
and 10 cm respectively.
;fn ,d yEc o`Ùkh; 'kadq dk oØ i`"Bh; {ks=kiQy
,d xksykdkj [kksy dk vk;ru Kkr dhft;s ftldk
10010 oxZ lseh gS vkSj bldh frjNh Å¡pkbZ 91 ckgjh vkSj vkarfjd O;kl Øe'k% 14 lseh vkSj
lseh gS] rks bldk dqy i`"Bh; {ks=kiQy Kkr dhft,A lseh gSA
(a) 27720 sq cm (b) 4620 sq cm 872
(c) 6930 sq cm (d) 13860 sq cm (a) 42  cm3 (b)  cm3
3
87. A spherical ball of lead 6 cm in radius is (c) 118 cm3 (d) 86 cm3
melted and recast into three spherical 91. The volume of a hemisphere is 89.83 cm³.
balls. The radii of two of these balls are Find its diameter (in cm).
3 cm and 4 cm. What is the radius of the ,d vèkZxksys dk vk;ru 89-83 lseh
3
gSA bldk O;kl

r
third sphere? (lseh esa) Kkr dhft,A

si
6 lseh f=kT;k dh lhls dh ,d xksykdkj xsan dks fi?kykdj (a) 3.5 (b) 7
rhu xksykdkj xsanksa esa cnyk tkrk gSA buesa ls nks xsanksa
(c) 14 (d) 10.5

D;k gS\ an by 92. If a solid sphere of radius 10 cm is moulded

dh f=kT;k,¡ 3 lseh vkSj 4 lseh gSaA rhljs xksys dh f=kT;k
into 8 spherical solid balls of equal radius,
then surface area of each ball (in sq.cm)

n
(a) 4.5 cm (b) 5 cm is ?
;fn 10 lsaVhehVj f=kT;k ds ,d Bksl xksys dks
(c) 6 cm

ja (d) 7 cm
f=kT;k dh 8 xksykdkj Bksl xsanksa esa <kyk
R s
88. The hemispherical bowl of internal radius rks çR;sd xsan dk lrg {ks=kiQy (oxZ lseh esa)
a th

6 cm contains alcohol. This alcohol is to

(a) 100  (b) 101/
be filled into cylindrical shaped small
(c) 99 /12 (d) 54/13
bottles of diameter 6 cm and height 1 cm.
93. A hemispherical bowl made of iron has
How many bottles will be needed to empty inner diameter 84 cm. Find the cost of
ty a

the bowl? tin plating it on the inside at the rate of

6 lseh vkarfjd f=kT;k okys vèkZxksyh; dVksjs esa vYdksgy  22 
di M

gSA bl 'kjkc dks 6 lsaVhehVj O;kl vkSj 1 lsaVhehVj Rs.21 per 100 cm²  take  =  correct
 7 
ÅapkbZ okyh csyukdkj vkdkj dh NksVh cksryksatoesa two places of decimal.
Hkjuk gSA bl dVksjs dks [kkyh djus ds fy, fdruh yksgs ls cus ,d v/Zxksyh; dVksjs dk vkarfjd O
cksryksa dh vko';drk gksxh  22 
84 lseh gSA 21 #i;s çfr 100 lseh
2  =
 yas

(a) 36 (b) 27  7 
(c) 16 (d) 4 dh nj ls n'keyo ds nks LFkkuksa rd vanj dh vk
89. A hemispherical tank full of water is fVu p<+kus dh ykxr Kkr dhft,A
emptied by a pipe at the rate of 7.7 m3 per SSC CGL 02/12/2022 (Shift- 01)
second. How much time (in hours) will it (a) Rs.2,328.48 (b) Rs.2,425.48
2 (c) Rs.2,425.60 (d) Rs.2,355.48
A

take to empty part of the tank, if the 94. A solid copper sphere of radius 9 cm is
3 hammered and moulded into a wire of
internal radius of the tank is 10.5 m? radius 2 cm. What is the length of this
wire?
ikuh ls Hkjh v¼Z xksyh; Vadh dks3 7-7 izfreh
lsdaM 9 cm f=kT;k okys rk¡cs ds Bksl xksys dks gF
dh nj ls fdlh ikbi }kjk [kkyh fd;k tkrk gSA Vadh ihV&ihV dj2 cm f=kT;k okyk rkj fu£er fd;k
2 tkrk gSA bl fu£er rkj dh yackbZ fdruh gS\
ds Hkkx dks [kkyh djus esa fdruk le; (?kaVs esa) SSC CGL 03/12/2022 (Shift- 03)
3
yxsxk] ;fn Vadh dh vkarfjd f=kT;k 10-5 ehVj gS\ (a) 224 cm (b) 183 cm
(c) 198 cm (d) 243 cm
95. A solid hemisphere has radius 21 cm. It
is melted to form a cylinder such that the
ratio of its curved surface area to total
surface area is 2 : 5. What is the radius
 22 
(in cm) of its base  take  = ?
 7  10
,d Bksl v/Zxksys dh f=kT;k
21 cm gSA bls fi?kykdj
,d ,slk csyu cuk;k tkrk gS fd blds oØ i`"Bh;
{ks=kiQy dk lEiw.kZ {ks=kiQYk ls vuqikr 2 % 5 gks 8
tkrk gSA blds vk/kj dh f=kT;k (cm esa
) D;k (a) 320  cubic cm (b) 160 cubic cm
 22  (c) 150 cubic cm (d) 300 cubic cm
gS\ = 7 y sa

100. The base of a right prism is a triangle
SSC CGL 06/12/2022 (Shift- 01) whose sides are 8 cm, 15 cm and 17 cm,
(a) 23 (b) 21 and its lateral surface area is 480 cm2.
(c) 17 (d) 19 What is the volume (in cm2) of the prism?
96. A cylindrical vessel of diameter 24 cm filled
up with sufficient quantity of water, a solid ,d le fçTe dk vk/kj ,d f=kHkqt gS ftldh Hkqtk,¡

r
spherical ball of radius 6 cm is completely 8 lseh] 15 lseh vkSj 17 lseh gSa] vkSj bldk ik
immersed. The increase in height of water
i`"Bh; {ks=kiQy 4802 lseh
gSA fçTe dk vk;ru (lseh
3

si
level is :
,d csyukdkj crZu ftldk O;kl 24 lseh gS i;kZIr esa) D;k gS\

an by
ikuh ls Hkjk gqvk gSA ,d Bksl xksykdkj xsan ftldh
f=kT;k 6cm gS dks crZu esa iw.kZr% Mqcks;k tkrk gS
crZu esa ikuh dk Lrj fdruk c<+ tk,xk \
(a)rks
540
SSC CGL MAINS 03/02/2022
(b) 600

n
(a) 1.5 cm (b) 2 cm (c) 720 (d) 640
(c) 3 cm (d) 4.2 cm 101. A right prism has height 18 cm and its

ja
R s
97. Half of a large cylindrical tank open at the base is a triangle with sides 5cm, 8cm and
top is filled with water and identical heavy 12 cm. What is the lateral surface area (in
a th

spherical balls are to be dropped into the cm2) ?

tank without spilling water out. If the
radius and the height of the tank are equal ,d yEc fçTe dh ÅapkbZ 18 lseh gS rFkk bldk vkè
and each is four times the radius of a ball, ,d f=kHkqt gS ftldh Hkqtk,a 5 lseh] 8 lseh vkS
then what is the maximum number of balls lseh dh gSaA ik'oZ i`"B {ks=kiQy (oxZ lseh
ty a

that can be dropped ? djsaA

,d Åij ls •qyk csyukdkj VSad ikuh ls vk/k Hkjk
di M

gSA ;fn mlesa ikuh ugha fxjus rd dqN leku xksykdkj SSC CGL TIER II (13/09/2019)
xsans Mky nh tk,A ;fn VSad dh f=kT;k vkSj Å¡pkbZ (a) 450 (b) 468
cjkcj gks vkSj xksys dh f=kT;k dh pkj xquk gks rc (c) 432 (d) 486
mu xsanksa dh la[;k Kkr dhft,A 102. The base of a right prism is a square having
(a) 12 (b) 24 side of 15 cm. If its height is 8 cm, then
(c) 36 (d) 48 find the total surface area.
98. The base of a solid right prism is a triangle ,d fçTe dk vk/kj ,d oxZ gksrk gS] tks 15 lseh dh
whose sides are 9 cm, 12 cm and 15 cm.
The height of the prism is 5 cm. Then, Hkqtk dk gSA ;fn bldh ÅapkbZ 8 lseh gS] r
the total surface area of the prism is lrg ds {ks=k dks Kkr djsaA
,d Bksl le fçTe dk vkèkkj ,d f=kHkqt gS ftldh (CGL MAINS 18/10/2020)
Hkqtk,¡ 9 lseh] 12 lseh vkSj 15 lseh gSaA fçTe dh 2
(b) 930 cm2
ÅapkbZ 5 lseh gSA fiQj] fçTe dk dqy lrg {ks=k gS (a) 920 cm2
A

(c) 900 cm (d) 940 cm2

(a) 180 cm2 (b) 234 cm2
(c) 288 cm 2
(d) 270 cm2 103. Area of the base of a pyramid is 57 sq. cm.
and height is 10 cm, then its volume in cm3,
99. The base of a right prism is an equilateral
is
triangle of side 8 cm and height of the
prism is 10 cm. Then the volume of the ,d fijkfeM ds vkèkkj dk {ks=kiQy 57 oxZ lseh
prism is vkSj ÅapkbZ 10 lseh gS] rks bldk vk;ru3 esa
lseh
,d le fçTe dk vk/kj 8 lseh Hkqtk dk ,d leckgq gS
f=kHkqt gS vkSj fçTe dh ÅapkbZ 10 fiQj
lsehfçTe
gSAdk (a) 570 (b) 390
vk;ru gS (c) 190 (d) 590
104. There is a pyramid on a base which is a ,d fu;fer fijkfeM dk vkèkkj ,d oxZ gS vkSj vU;
regular hexagon of side 2a cm. If every pkj Hkqtkvksa esa ls çR;sd ,d leckgq f=kHkqt g
5a Hkqtk dh yackbZ 20 lseh gSA lseh esa fijkfeM dh
slant edge of this pyramid is of length
2 ÅapkbZ gS
cm, then the volume of this pyramid is (a) 10 (b) 8
,d vkèkkj ij ,d fijkfeM gS tks 2a lseh Hkqtk
dk (c) 12 (d) 5
,d fu;fer "kV~Hkqt gSA ;fn bl fijkfeM dk çR;sd
107. The base of a right pyramid is an
equilateral triangle with side 8 cm, and the
5a
frjNk fdukjk yackbZ lseh dk gS] rks bl fijkfeM height of pyramid is 24 3 cm. The volume
2
dk vk;ru gS (in cm3) of the pyramid is :
,d y?kq fijkfeM dk vk/kj ,d leckgq f=kHkqt g
(a) 3a3 cm3 (b) 3 2 a3 cm3
ftldh Hkqtk 8 lseh dh gSA fijkfeM dh ÅapkbZ
24 3
(c) 3 3 a3 cm3 (d) 6a3 cm3
lseh gSA bl fijkfeM dk vk;ru (?ku lseh esa) K
105. The base of a right pyramid is a square djsaA
of side 40 cm long. If the volume of the

r
pyramid is 8000 cm3, then its height is: SSC CGL TIER II (12/09/2019)
(a) 1152 (b) 480
,d le fijkfeM dk vkèkkj 40 lseh yack Hkqtk

si
(c) 576 (d) 384
okyk ,d oxZ gSA ;fn fijkfeM dk vk;ru 8000
108. If the length of each side of a regular

(a) 5 cm an by
lseh3 gS] rks bldh Å¡pkbZ gS
(b) 10 cm
tetrahedron is 18 cm, then the volume of
tetrahedron is:

n
(c) 15 cm (d) 20 cm ;fn ,d fu;fer prq"iQyd dh izR;sd Hkqtk dh yackb
106. The base of a regular pyramid is a square 18 lseh gS] rks bl prq"iQyd dk vk;ru Kkr djsaA

ja
and each of the other four sides is an
R s
equilateral triangle, length of each side (a) 486 2 cm³ (b) 324 2 cm³
a th

being 20 cm. The vertical height of the

(c) 324 3 cm³ (d) 284 3 cm³
pyramid, in cm, is
ty a
di M
A
 SOLUTIONS
1. (a)
SOLUTION 6. (b)
Volume of 3 cubes = volume of new cube We know,
 3³ + 4³ + 5³ = 216 = 6³
mass
a=6 Density =
volume
 Total surface area = 6 × (6)² = 216 cm²
2. (b) M1 M2
 =
V1 V2

400 3200
 =
4³ a³
 a³ = 8 × 4³ = 512
a = 8 cm.

r
7. (c)

si
Remaining wood = a³ – r²h a³ = 1
a a=1
h = a, r =
2
an by Volume of 64 small cubes = 64 × 1 = 64 cm³
Volume of cuboid = l × b × h

n
a³
 a³ –  Let, l = b = h = x (say)
4

ja  x³ = 64
R s
 22  x=4
 a³ 1 – 

 28 
a th

 Diagonal = x 3 = 4 3 cm.
6 12 8. (b)
 8× = cm³
28 7 Let, volume of larger cube = 125cm³
ty a

3. (d) Volume of one small cube = 1cm³

Surface area = 6 × 8² = 384 ft²
 side of one small cube = 1cm.
di M

384 New cuboid has length = 125 × 1 = 125cm.

Quantity of paint required = = 24 kg.
16 Bredth = 1cm.
cost = 36.50 × 24 = Rs. 876 Height = 1cm.
4. (c)  S.A of cuboid = 2 (125 + 1 + 125)
Given,
= 502cm²
6a² = 24
S.A of larger original cube = 6a² = 6 × 5²
a = 2 dm
= 150cm²
We know 1 meter = 10 decimeter
No. of small cubes  Percentage increase
502 – 150 2
volume of larger cube =  100 = 234 %
=
A

150 3
volume of small cubes
9. (a)
103 1000
= = 125 Surface area of open cistern
23 8
2h (l + b) + lb
x = 125 cubes
 2 × 4 [12.5 + 8.5] + 12.5 × 8.5
5. (a)
A.T.Q, = 168 + 106.25 = 274.25 cm²
36 × 75 × 80 = a³ Total cost of cementing = 24 × 274.25
 a = 60 cm. = Rs. 6582
10. (c) 14. (c)
Total surface area of cuboid = 2(lb + bh + hl )
l ²  b² = 12
= 2 (27 + 9 + 27) = 126 cm2
11. (b)  l² + b² = 144 and l ²  b²  h² = 15
A.T.Q,  l² + b² + h² = 225
 h² = 225 – 144 = 81
 h = 9 m.
6 3 15. (d)
6 × 12 × 15 = n × a³
Given that,
Side of largest possible cube (a)
3 a = 6 3 (here, a is the side of cube) = HCF (6, 12, 15) = 3
a = 6 cm  6 × 12 × 15 = 3³ × n
Length of cuboid (l ) = 18 cm  40 cubes possible
Height of cuboid (h) = 6 cm
16. (c)
Breadth of cuboid (b) = 6 cm
T.S.A of the cuboid = 2 (l b + bh + hl) A.T.Q,
= 2 (18 × 6 + 6 × 6 + 18 × 6) 2 (l × b) = 2h (l + b)
= 504 cm2 lb = h (l + b)

r
12. (c) 15 × 12 = h (15 + 12)

si
A.T.Q,
180 20
 h= = m
27 3

an by Volume = l × b × h

n
20
= 15 × 12 × = 1200 m³
3

ja 17. (c)
R s
volume of wall
No. of bricks = volume of each bricks
a th

Given that, 8  6  22.5

a3 = 1728 (here, a is the side of cube) =  10, 00, 000 = 6400
25  11.25  6
a = 12
ty a

18. (b)
Length of cuboid (l) = 24 cm
Let thickness = T
Breadth of cuboid(b) = 24 cm
Then,
di M

Height of cuboid (h) = 12 cm

T.S.A of the cuboid = 2 (lb + bh + hl) (330 – 10) × (260 – 10) × (110 – T) = 8000 ×
2 (576 + 288 + 288) 1000
2304 cm3 = 320 × 250 × (110 – T) = 8000 × 1000
13. (a) 8000 1000
A.T.Q,  110 – T =
320  250
 T = 10 cm. = 1 dm.
19. (a)
a Volume of metal = Ext volume – Int volume
 (50 × 40 × 23) – (44 × 34 × 20)
= 16080 cm³
A

16080  0.5
 Weight of metal =
1000
Given that = 8.04 kg.
2r = 5.6 (r is the radius of the sphere)
20. (a)
r = 2.8 cm
Volume of the sphere Number of small box
4 4 22 100 ×100 ×100
r 3 = × × 2.8 × 2.8 × 2.8 = =1000
3 3 7 10 ×10 ×10
3
= 91.98 cm
21. (d) 25. (d)
Volume of tank = 200 × 150 × 2 = 60,000 m. Area of wet surface
Length of water flow in 1 min.
= 2 [6 × 4 + 4 × 1.25 + 6 × 1.25] – (6 × 4)
20  1000 = 1000 = 2 [24 + 5 + 7.5] – 24
=
60 3
= 73 – 24 = 49 m²
Volume of water per min.
26. (a)
1.5  1.25  1000 12
=
3
= 625 m³/min. 6

 Time required = 60000 = 96 min . 14

625
22. (a)
Volume of empty box
= (24 × 16 × 10) – (23 × 15 × 9) 40
A.T.Q,

r
= 3840 – 3105 = 735 cm³
12 × 6 × 5 = (40 × 14 – 12 × 6) × h
735 cm³  7.35 × 1000 g.

si
where, h = rise in level
7.35  360 = 488 h

23.
1 cm3 

(d)
735
an by
× 1000 = 10g
 h=
360
488
m

n
360
 h= × 100 = 73.77 cm.
488

ja
R s
27. (a)
h
A.T.Q,
a th

15 × 15 × 4 = 20 × 25 × d
9
 d= m.
5
ty a

28. (c)
Volume of cube = volume of water rose A.T.Q,
 11³ = 15 × 12 × h
di M

3
 h = 7.39 cm. 12 × × b × 1000 = 18000
10
24. (c)
b=5m
29. (d)
ATQ,
5
0.3 × 0.2 × 20 × × t = 200 × 150 × 8
18
t
= 240000
3
t = 720000
A

720000
t =
3600
By cutting the given cuboid, dimensions t = 200 hrs
of each part are:- 30. (a)
25cm. × 20cm. × 15cm. ATQ,
T.S.A of each part = 2 (25 × 20 + 20 × 15 Volume of cross section = volume of the
tank
+ 15 × 25) = 2350 cm²
0.05 × 0.03 × 16 × 25 × 60 = 15 × 12 × h
T.S.A of 8 parts = 8 × 2350 = 18800 cm² h = 0.2 m
31. (d) 36. (d)
ATQ, A.T.Q,
Side of cube = 5 cm 4
r1²h1 = r23 × n
3
Volume of cuboid
No. of cubes = r1
Volume of cube Given, r1 = h1, r2 =
2
20 ×15 ×10 From (1)
= = 24
5×5×5 3
 r1  4
32. (a)  r13 = 
2    3 n
 
TSA. of the cuboid = 2 (lb + bh + hl)
= 2 (30 × 25 + 25 × 12 + 12 × 30) = 2820 cm2 r13 4
r13 =  n
Area which is painted = Total surface area 8 3
of cuboid × n n=6
37. (a)
112800
= = 40 Let, r1 = 4

r
2820  r2 = 5
33. (a)

si
Old New
2rh 2 Radius 4 5
=
2r (h  r) 3
an by Volume 1
r² 16
1
25

n
h 2
 = 1 1
hr 3 h

ja 16 25
R s
Let, h = 2x, r = x h 25 16
Now,
a th

25 – 16
2r (h + r) = 924  % Change in height =  100 =
25
36%
22
 2× × x (3x) = 924 38. (a)
ty a

7
r = 7, h = 21
 x² = 49 A.T.Q,
di M

x=7

22
Volume =  49  14 = 2156 cm³
7
34. (a)

r1 3 h1 4 3.5
= , = 1.4
r2 5 h2 3

r12 h1 9 4 36 12
 =  = =
r22 h 2 25 3 75 25
2.1
35. (c)
A

 2 2 3
Given, r2 h  r2 
 × 7² × 21 = n ×  
 3 
1 2
r² h = 15 L  7² × 21 = n [(2.1)² × 1.4 + × (2.1)³]
3 3
 r²h = 15 × 3  7² × 21 = n × (2.1)² [1.4 + 1.4]
= 45 L 49  21
 =n
 Cylinder can carry = 45 L milk 2.1  2.1  2.8
 n = 83
39. (d) 42. (a)

21

5 7

6 Volume of earth taken out = r²h

Let, h = water raised
22
Volume of sphere = Volume of water raised. = × 7 × 7 × 15 = 2310m³
7
4 A.T.Q,
 r13 = r22 h
3
22
4 = [28² – 7²] × h = 2310
 × 125 = 36 × h 7
3  h = 1m.
125 43. (d)

r
 h= = 4.63 cm.
27 r h Volume

si
40. (b) Old 1 2 2
New 2 1 4

an by %Change =
4–2
2
 100 = 100%

n
44. (b)
15 Quantity of water

ja
R s
22 0.08  0.08
=  3500 
7 4
a th

= 17.6 m³
45. (c)
14 A.T.Q,
22
ty a

28 7
r1 of smaller cylinder = = , h1 = 5 10 × 10 × 10 = × 10 × 10 × h
8 2 7
Total surface area of remaining part
di M

70 4 2
= TSA. of big solid cylinder + CSA. of all  h= =3 3 cm.
small cylinders 22 22 11
46. (a)
22 22 7 2r = 8.8 and 2rh = 17.6
=2× × 14 (29) + 8 × 2 × × ×5
7 7 2
= (88 × 29 + 40 × 22) 17.6
 h= =2
= 3432 8.8
41. (a) 8.8  7
r= = 1.4
22  2
Amount of concrete required
22
 r²h = × 1.4 × 1.4 × 2 = 12.32m³
7
A

47. (d)

Old New
r 4 1
r² 16 1
V = R²h – r²h
v 1 1
22
= × 210 (25² – 23²)  h 1 16
7
= 660 (96) = 63360 cm³ The length will become 16 times.
48. (a) 52. (a)
2r (h + r) = 462 ..........(1)
h1 3 r 6
= ; 1 = 462
h2 2 r2 7 2rh = ..........(2)
3
2r1 h1 6 3 9 (1) ÷ (2)
=  =
2r2 h2 7 2 7 hr 3
 =
49. (b) h 1
 h = a, r = 2a
2rh = 1000cm² From (2)
Times of wire = n 22 462
for n turns, wire should make upto 5n mm 2 × 2a × a = = 154
7 3
or 0.005n m height.
77
 Height of cylinder = 0.005n m  a² =
22
Let length of wire be l
7
 l = 2 × r × n  a=
2
l 7

r
= 2 × r r=2× = 7cm.
n 2

si
 1000 = 0.1m² = 2 × r × h 53. (a)

l h = 7cm, r = 5 = 1 cm.
0.1 =
n
an by
× 0.005n 20 4
22 1 1
  7 =
11
cm³

n
0.1 volume of ink =
 l= = 20m. 7 4 4 8
0.005
11
50. (b)

ja cm3  330 words

R s
8
We know,
a th

1
ltr. = 200 cm3
5
330
h  × 8 × 200 = 48,000 words
ty a

11
54. (a)
v1 + v2 = v3
di M

(15² × 35 + 10² × 15) = r² × 15

 7875 + 1500 = 15r²
R  r² = 625
r1 = inner radius = 2  r = 25cm.
55. (c)
R1 = outer radius = 3 A.T.Q.
height = 10cm. = 2rh + 2r (h + r) = 2068
T.S.A = 2Rh + 2rh + 2 (R² – r²)  2r (2h + r) = 2068
22
= 2h (R + r) + 2 (R – r) (R + r) 2× × 7 (2h + 7) = 2068
7
= 2 (R + r) [h + R – r]  h = 20
= 2× 5(10+1) = 110 22
51. (d) Volume = r2h = × 7 × 7 × 20
A

7
3
= 3080 cm
35 56.(d)
r=
2 2r = 37 + r
v = 11 ltr. = 11000cm³  44 
  – 1  r = 37
2  7 
22  35 
 11000 = 
  h
 r = 7, h = 35
7  2   Volume = r²h
80 3 22
 h= = 11 cm. = × 7 × 7 × 35 = 5390 cm³
7 7 7
57. (b) 61. (d)
A.T.Q,
r2h = lbh
22
× 2.8 × 2.8 × 15 = 8 × 1.5 × l
7
l = 30.8 m
58. (d)
A.T.Q
96
× r12 = r2 l
100
96
× 25 × 25 = r2 × l
100
600 = r2 × l
Also, we know, l = r1 = 25
= r2 = 24
Hence, h 2 = 25² – 24²
According to figure =7
Radius of cylinder = Radius of sphere r2 24

r
 =
4 3 h2 7
r2h = r

si
3 62. (b)
4 A
h = × 3.5
3
14 an by
n °
cm

60
h=
3 30° l
59. (c)

ja
R s
5

a th

B C
12 Since, AB = AC and BAC = 60°
 ABC is an equilateral 
ty a

13 Hence, l = 2r
r r 1
di M

 = =
l 2r 2
r 5 63. (a)
=
h 12 1
Volume = r²h
 l = 13 [Pythagorean triplet] 3
Also, 1 22
=  × 12 × 12 × 35 = 5280 cm³
1 2 2200 3 7
r² h = 314 =
3 7 7 Rate of water flow = 5m/min.
1 22 2200 5 100
   5  5  12x ³ = = cm/sec.
3 7 7 60
 x³ = 1 25
x=1 = cm/ sec .
3
A

 l = 13 m.
60. (d)
Radius of cylinder pipe = 16  1 = 0.8 cm.
22 2 10
rl = × 14 × l
7
Volume of cone
l = h²  r² = 48²  14² = 50  Time =
ar. of cylinder × Rate of water
Also,
22 5280
 × 14 × 50 = 10 × length of cloth = = 315 sec .
7 22 –1 –1 25
 Length = 220 m.  8  10  8  10 
7 3
64. (b) 67. (d)
h 1 = 36, r1 = r2 = 9, h2 = 9
1 r12 h1 – r22 h2 h
% Waste =  100 r
3 1
r12 h1
3 H
1 
r12  h1 – h 2 
 3   100 = 3  100 = 25% R
r12  1  12
 h1 
 3
 
65. (a) Volume of small cone r²h
=
A.T.Q, Volume of large cone R²H

2 2
h= H  r= R
3 3
5 Hence,
4

r
4 2
R²  H
r²h 9 3

si
=
3 R²H R²H

an by
5
Vol. small
=
Vol. large 27
8

n
4
Vol. small cone 8 8
 Vol. large frustum = 27 – 8 = 19

ja
R s
3 68. (d)
a th

r=3
H–h
h=4 r
36 = H
ty a

1 22
Volume =  × 3 × 3 × 4 = 12
3 7 h
di M

66. (a)
5 6
We know,
6 36
=  36 – h = 6r
r 36 – h

8
 h = 36 – 6r
h = 6 (6 – r)
1
Vol. of frustum =  [R² + r² + Rr]h
3
1
A

25 =  [6² + r² + 6r] [6 – r]6

n × Volume of 1 sphere = × volume of 3
100
cone 22
264 = 2  [6³ – r³]
7
4  1 3 1 1 42 = 6³ – r³
 n×  
   =   5  5  8
3 2  4 3 r³ = 216 – 42
r³ = 174
1 88
 n=  25  = 100 1
4 4 r = (174) 3 cm.
69. (d) 74. (a)
1 22 r1 = 10, h1 = 72cm
Volume =   7  7  15
3 7 r2 = 30, h2 = ?
= 770 cm³ A.T.Q,
Weight of vessel = 770 × 15
1
= 11550 g.  × 10 × 10 × 72 =  × 30 × 30 × h2
= 11.55 kg. 3
70. (a)
2
 h2 = 2 cm
22 3
rl =  21  21²  20²
7 75. (b)
= 66 × 29
22 1
r ² =  21  21 r²h = 1232, h = 24
7 3
T.S.A = 66 × 29 + 66 × 21
1232  3  7
= 66 × 50 = 3300  r² = = 49
22  24

r
71. (c)
r=7
1 1 1

si
 × 3 × 3 × 5 = n ×   1
3 10 10 l = 7²  24² = 25
 n = 13500
72. (a)
an by C.S.A = rl =
22
7
× 7 × 25 = 550cm²

n
h2 = 30 cm.
76. (c)
h1 Radius of sector = slant height of cone

ja
v1
R s
v2  l = h²  r² = 6²  8² = 10 cm.
h2
a th

77. (b)
ty a

We know,
3
di M

v1 (h1 )³ 120°
=
v 2 (h2 )³
1 h3
 = 1
27 (30)³ Length of sector = circumference of base of
h1 1 cone
 =
30 3 120
h1 = 10
 × 2 (3) = 2r
360
 h2 – h1 = 20cm.
 r = 1cm.
73. (d)
l of cone = r of sector
h = 10.5, r = 6.5
A

 l = 3cm.
1 22
v=  × 6.5 × 6.5 × 10.5 = 464.33 m²  h = l ² – r² = 3² – 1² = 2 2
3 7
Avg. no. of cubic meters of air space per 1 1
man v= r²h = ×  × (1)² ×
3 3
Volume
= 2 2
8 2 2 = cm³
3
464.33 3
= = 58.04 = 58 m³
8 32
78. (b) 82. (a)

v1 144°
v1 30cm.
v2

h2–h1

Circumference of base of cone = length of

v1 1
 = arc
v2 2

We know, 2R = × 2r
360
v1 (h1 )³ 144
=  R= × 30 = 12cm.
v 2 (h2 )³ 360
83. (c)

r
1 h
= 1

si
(2)1/3 h2 h1


h1
=
1

an by
h2 – h1 (2)1/3 – 1 h2

n
79. (c)
h = 8cm, r = 6cm.

ja
R s
A.T.Q,
1 1 v1 1
a th

×6×6×8=n× ×2×1×1 h2 = 20 v = 8
3 3 2

 n = 144 We know,
80. (a) v1 h ³
= 1
ty a

r = 6, R = 14, h = 6 v 2 h2 ³
1 h1 ³
di M

l = h²  (R – r)² =
8 h2 ³
= 36  64 = 10 1 h1
 =
T.S.A = l (R + r) +  (r2 + r2) 2 20
 h1 = 10
22 22  height of frustum = 20 – 10 = 10cm.
 × 10 (20) + (196 + 36) 84. (d)
7 7
22
 (200 +232) h1
7
22 h2
 (432) = 432
7
A

81. (d)
r = 16,
3
rl = 427 h2 = 40cm.
7
We know,
22 2992 v1 h3
 × 16 × l = = 13
7 7 v 2 h2
187 1 h3
 l= = 8.5 m  = 13
22 64 40
 h1 1 89. (b)
 =  h1 = 10
40 4 A.T.Q,
height of frustum = 40 – 10 = 30 cm.
2 2
85. (c) 7.7 × t = ×r 3
3 3
1
× r2h = 400
3 2 22 2
7.7 × t = × × 10.5 × 10.5 × 10.5 ×
Consider, 3 7 3
1 1 2
×  × (2r)2 × h = r h × 4 = 400 × 4 175
3 3 t= hrs
3
= 1600
86. (d) 90. (b)
A.T.Q, 4 4
V= R³ – r³
3 3
3 3
91 4 22 
 14  10   

r
=    –
     
3 7  2  2  

si
4 872
= [218] = cm3

Given that,
an by 91. (b)
3 3

n
rl = 10010
A.T.Q,
22

ja
× 91 × r = 10010 2 22
R s
7  × r³ = 89.83
r = 35 3 7
a th

T.S.A of the cone = r(r + l)  r  3.5

22  d = 7 cm.
× 35 (126) = 13860 cm2
7
92. (a)
ty a

87. (b)
A.T.Q, A.T.Q,
di M

4 22 4 22 3 4 4
 ×6×6×6=  [3 + 43 +x3]  10³ = 8  r³
3 7 3 7 3 3
 6³ = 3³ + 4³ + x³
3
 x = 5 cm. 10 
 r³ = 
   = (5)³
88. (c) 2 
 r = 5 cm.
6
S.A = 4r² = 4 ×  × 5 × 5 = 100
93. (a)
1×n
22
L.S.A = 2r² = 2   42  42 = 11088 cm²
7
A

Rate = 21 per 100 cm²

3
A.T.Q, T. Price = 110.88 × 21 = Rs. 2328.48
2 94. (d)
×  × (6)³ =  (3)² × 1 × n
3 ATQ,
2 6 6 4 22 22
n=   6   9 9 9 =  (2)²  l
3 3 3 3 7 7
n = 16
 l = 81 × 3 = 243 cm
95. (b)
1
r = 21 = (9 + 12 + 15) 5 + 2 × × 9 × 12
2
2
 r³ = r12 h = 180 + 108 = 288 cm²
3
2 99. (b)
 (21)³ = r12 h
3
.........(1)
C.S.A 2r1 h h 2
= = =
T.S.A 2r1 (h  r1 ) h  r1 5
h = 2x  r1 = 3x .......(2)
Put (2) in (1)
2 10
× 21 × 21 × 21 = 9x² × 2x
3
 x³ = 7 × 7 × 7
 x=7
r = 3 × 7 = 21 8
96. (b) V = ar. of base × h

r
3

si
= × 8 × 8 × 10
4

an by h = 160 3 cm³
100. (c)

n
L.S.A of a prism
r Perimeter of the base × height

ja 40h = 480
R s
h = 12 cm
R
We know, Volume of prism = area of base × height
a th

volume of sphere = volume of water rose 1

4 = × 15 × 8 × 12 = 720 cm3
 r³ = R²h 2
3 101. (a)
ty a

4 L.S.A of a prism = Perimeter of the base ×

 × 6³ = 12 × 12 × h height
3
di M

 h = 2cm. = (5 + 8 + 12)18 = 450 cm2

97. (b) 102. (b)
Radius of ball = r We know
 Radius of cylinder = 4r, height = 4r T.S.A of a prism = Perimeter of the base ×
Vol of water in cylindrical tank =  (4r)² × 2r height + 2 × area of base
= 32r³ = 60 × 8 + 2 × 225
Let, n = max. no. of balls. = 480 + 450
A.T.Q, = 930 cm2
4 103. (c)
n  r³ = 32r³
3 1
n = 24 Volume = × Ar. of base × h
98. (c) 3
1
A

=  57  10 = 190 cm³
3
104. (c)
5a
5
e=
2
We know,
9 12  5a 2 3
h² = 
  – (2a)²  h =
 a
15
 2  2
T.S.A = (perimeter of base × h) + 2 ×Ar of base
 A.T.Q,

3a
l = slant height =
2

3
=  20 = 10 3
2

Now, in ABC:-
2a

1 BC = 10, AC = 10 3
Volume = × ar. of base × h
3
 h = AB = AC² – BC²
1 3 3a
= 6  4a² 
3 4 2 = 3 3a³ cm³ = 300 – 100 = 200
105. (c)
 h = 10 2

r
107. (a)
We know,

si
1
Volume of a pyramid = area of base ×

an by 40
height of pyramid
3

n
1 3
= × × 8 × 8 × 24 3
1 3 4
V=

ja
× ar. of base × h = 1152 cm3
R s
3
108. (a)
a th

1 We know,
8,000 = × 40 × 40 × h
3 a3
 h = 15 cm. Volume of tetrahedron =
6 2
ty a

106. (a)
18 ×18 ×18
A = = 486 2 cm3
6 2
di M

B
c 20
a a=
A
 (TABULAR DI)
Direction: (1 - 2) Study the table and answer the
questions: 2016 1800 2500 1800 2000
2017 2500 2300 1850 1800
2018 2300 2400 1840 1760
2019 2440 1950 1900 1600
Table shows District-wise data of the 2020 2250 2100 2000 1750

number of primary school teachers posted

3. In which city were maximum trees planted
in schools of a city.
in 2016 and 2019 taken together?

SSC CGL 2020

East 1650 2375 (a) Chandigarh (b) Ahmedabad
North 1075 2651 (c) Pune (d) Kolkata
West 1280 1520 4. What is the total number of trees planted
South 1170 1085
in Chandigarh in 2017 and in Kolkata in
Central 690 859
2020?
1. What is the average number of female
teachers in the five districts?

SSC CGL 2020

SSC CGL 2020 (a) 4750 (b) 4500
(a) 1690 (b) 2871
(c) 4250 (d) 3550
(c) 1698 (d) 1173
5. From 2016 to 2020, how many more trees
2. What is the ratio of the number of male
teachers to the number of female teachers were planted in Ahmedabad as compared
in the city? to trees planted in Pune?

SSC CGL 2020

SSC CGL 2020
(a) 195 : 283 (b) 78 : 113
(a) 2340 (b) 2000
(c) 586 : 849 (d) 391 : 566
Direction: (3 - 6) Study the table and answer the (c) 1850 (d) 1860
questions: 6. In which year were the maximum number
of trees planted?

Table shows the number of trees planted

in 4 cities from 2016 to 2020. SSC CGL 2020
(a) 2018 (b) 2020
(c) 2017 (d) 2016
 Table shows income (in Rs.) received by 4
employees of a company during the month
of December 2020 and all their income
SSC CGL 2020
sources.
(a) 1 : 1 (b) 58 : 57
(c) 55 : 57 (d) 58 : 53
Direction: (10 - 11) Study the table and answer
the questions:
Salary 35000 38500 29000 42000
Arrears 6000 6300 5000 7500
Bonus 1000 1100 1000 1240
Study the following table and answer the
question:
7. by what percent are the Arrears of Amit
and Suresh taken together less the Arrears
of Nitin and Varun taken together?
Number of students enrolled for Vocational
Courses (VC) in five institutes - A, B, C, D
& E.
SSC CGL 2020 A, B, C, D E
(a) 1.2 (b) 1.5 (VC)
(c) 1.6 (d) 1.4
8. By what percent is the bonus of Varun less
A 120 135 130 135 128 140
than the bonus of Amit and Nitin taken B 125 132 138 132 135 142
together? C 125 120 125 138 140 135
D 100 125 122 140 128 138
E 105 110 115 147 130 145

10. The total number of students enrolled for

SSC CGL 2020
VC in institute C in 2013, 2014 and 2017
(a) 38 (b) 40.9 is what percent of the total number of stu-
(c) 45 (d) 48 dents enrolled in the five institutes in
2018?
9. The data given in the table shows the
number of boys and girls enrolled in three C VC
different streams in a school over 5 years.

SSC CGL 2020

(a) 62 (b) 55
(c) 53 (d) 58

2012 48 36 40 35 35 45 11. The ratio of the total number of students

2014 42 43 42 32 32 42 enrolled for VC in institutes A, C and E in
2016 45 42 38 30 36 38 2016 to the total number of students
2018 39 46 41 23 28 34
enrolled in institutes B and D in 2018, is
2020 36 43 39 30 39 41
 SSC CGL 2020 SSC CGL 2020
(a) 14 : 9 (b) 3 : 2 (a) 10.3 (b) 10.8

(c) 21 : 19 (d) 8 : 7 (c) 11.8 (d) 11.1

Direction: (12 - 13) Study the table and answer Direction: (14 - 16) Study the table and answer
the questions: the questions:

Study the following table and answer the The data given in the table shows the
question: number of students studying in four
different disciplines in 5 institutes.

Number of students enrolled for Vocational

Courses (VC) in institutes A, B, C, D, E & F.
A, B, C, D, E F
Students the table and answer the
(VC)
question:

A 110 150 165 180 205

B 120 180 176 200 220
C 140 220 180 175 225 A 36 48 59 57
B 45 54 55 48
D 125 210 175 180 230
C 55 36 56 51
E 150 200 160 200 240 D 45 48 55 53
F 165 230 200 220 210 E 48 44 52 55

14. By what percent is the number of students

12. The ratio of the total number of students
studying Computer Science in institutes A
enrolled for VC in institutes A, C and E in
2015 to the total number of students and B more than the number of students
enrolled in institutes B and D in 2017, is: studying Arts in institutes B and C?

A, C E VC A B
B C
B D

SSC CGL 2020 SSC CGL 2020

(a) 9 : 10 (b) 3 : 4 (a) 2 (b) 24
(c) 3 : 2 (d) 10 : 11 (c) 14 (d) 5
13. The total of students enrolled for VC in 15. Number of students studying Computer
institutes B, C and E in 2015 is x% more Science in the institutes A and C taken
than the total number of students enrolled together is what percent of the number of
in institutes A, D and F in 2016. The value
students studying Arts in the institutes B
of x is closest to:
and D taken together?
 SSC CGL 2020 SSC CPO 16 March 2019 (Afternoon)
(a) 200 (b) 83.3 (a) 3 (b) 9
(c)108 (d) 120 (c) 8 (d) 5
Direction: (20 - 22) Study the table and answer
16. What is the ratio of number of students
studying Science in institutes C and D the questions:
taken together to the number of students
studying Computer Science in institutes A
and E taken together? The given table shows the number (in
C D thousands) of cars of five different models
A A, B, C, D and E produced during Years
E 2012-2017. Study the table and answer the
questions that follows:
SSC CGL 2020
(a) 43 : 56 (b) 42 : 55 A, B, C, D E
(c) 41 : 56 (d) 3 : 4
Direction: (17 - 19) Study the table and answer
the questions: A B C D E Total
2012 18 26 22 23 31 120
2013 22 18 32 40 18 130
2014 32 43 26 35 34 170
The table below shows the admission and 2015 18 22 26 14 20 100
2016 36 12 44 38 50 180
transfer in standards 1-3 of a school. 2017 12 48 40 22 28 150
20. In the year 2015, which type of car
constitutes exactly 20% o f the total
2015 2016
number of cars produced that year?
Std. Existing Admission Transfer Admission Transfer
1 232 12 8
2 241 6 11
3 248 16 13

17. In the given table, in Standard 1, how many SSC CGL 3 March 2020 (Morning)
students were there at the end of year
(a) D (b) B
2016?
(c) E (d) A
21. If 2013 and 2014 are put together, which
SSC CPO 16 March 2019 (Afternoon) type of cars constitute exactly 25% of the
(a) 223 (b) 228 total number of cars produced in those two
(c) 236 (d) 232 years?
18. In the given table, what was the total
strength in Standards 1-3 at the end of
2015?

SSC CGL 3 March 2020 (Morning)

SSC CPO 16 March 2019 (Afternoon) (a) E (b) C
(a) 723 (b) 721
(c) B (d) D
(c) 710 (d) 704
 SSC CGL 3 March 2020 (Morning)
(a) 62.33% (b) 45% A 70 90 65 64 88
B 84 92 75 68 49
(c) 33.33% (d) 50% C 66 80 85 80 84
D 62 74 75 88 60
Direction: (23 - 24) Study the table and answer E 54 64 55 72 85
F 72 84 65 60 65
the questions:
What are the average marks of students B,
C, D and F in Math?
Study the following table and answer the B, C, D F
question:

SSC CGL 2020

Percentage of marks obtained by six (a) 120.75 (b) 125.5

students in five subjects A, B, C, D & E. (c) 82.5 (d) 123.75
A, B, C, D E Direction: (26 - 27) Study the table and answer
the questions:

Manju 68 85 86 72 92
Amit 64 65 80 96 80
Rekha 88 75 65 74 90 Study the following table and answer the
Anuj 80 55 68 66 84
Abhi 72 65 72 54 74 questions:
Vikram 60 70 73 84 86

23. The total marks obtained by Anuj in all the

five subjects are?
Number of cars sold by dealers A, B, C, D
& E during first six months of 2018.

SSC CGL 2020 2018 A, B, C, D

E
(a) 328 (b) 303
(c) 324 (d) 331
24. The total marks obtained by Amit in A 620 640 628 635 430 625
B 600 642 635 580 450 620
subjects A, B and C is what percent less C 640 635 640 540 625 740
D 520 645 722 740 600 780
than the total marks obtained by Vikram
E 548 638 720 740 650 800
in subjects B, C, D and E?
26. The ratio of the total number of cars sold
A, B C by dealer B in January, April and June to
B, C, D E the total number of cars sold by dealers A
and D in March is:
SSC CGL 2020 B
(a) 42 (b) 35 A D
(c) 38 (d) 40
25. Study the following table and answer the SSC CGL 2020
question: (a) 4 : 3 (b) 10 : 9
(c) 8 : 9 (d) 7 : 5
 in July 2018? vacant in AC classes on Monday. Tuesday
D and Friday?
AC

D AC
SSC CGL 2020
(a) 1020 (b) 959
SSC CGL 2020
(c) 1014 (d) 975
(a) 35 : 62 (b) 62 : 35
28. Study the table and answer the question.
(c) 39 : 62 (d) 62 : 39
30. This table shows the number of students
In the table, production and sale (in 1000 studying in various streams in different
tonnes) of a certain product of a company colleges.
over 5 years is given.

College
Streams
A B C D E
Art 580 460 320 470 370
Science 620 680 540 360 400
Commerce 480 520 350 520 330
2015 1250 1000
2016 1400 1290 If the data about students of the commerce
2017 1450 1100 stream in all colleges is represented by a
2018 1500 1450 pie-chart, what is the central angle of the
2019 1600 1390 sector representing college D, to the near-
In which year(s) sale is more than 90% of est degree?
the production?

D
SSC CGL 2020
(a) 2016, 2018 (b) 2017,2018 (a) 80° (b) 82°
(c) 2015, 2017, 2019 (d) 2016, 2017 (c) 88° (d) 85°
29. The following table shows the daily seats 31. This table shows the percentage of students
occupancy in different classes of a train. passing out of five different colleges over
Numbers in bracket represent the total three years. It is given that from each
seats available for a particular class. college, 200 students appeared every year.

Colleges
Years
A B C D E
Monday 850 460 480 240 145
Tuesday 840 400 450 230 120
2015 68 65 80 92 72
Wednesday 830 390 480 220 130 2016 72 68 88 95 75
Thursday
Friday
790
840
480
470
490
500
250
210
125
130
2017 74 77 92 98 73
 passed out students of college C?

SSC CPO 15 March 2019 (Morning)

C
(a) 7.46 (b) 6.80
(c) 4.96 (d) 5.52
34. In the given table, what is the percentage
(a) 69° (b) 79° of expenditure on income in the year 2002?
(c) 77° (d) 67° (round off)
32. The table shows the production of different
types of cars (in thousands).
SSC CPO 15 March 2019 (Morning)
(a) 85% (b) 78%
(c) 82% (d) 81%
35. In the given table, if a person reduced his
Years
2012 2013 2014 2015 2016 expenditure by 10% by how much would
Cars his total savings increased?
A 30 35 48 45 56
B 42 48 40 38 56
C 48 36 38 35 44
D 51 21 30 46 54
E 20 42 40 35 43
SSC CPO 15 March 2019 (Morning)
If the data related to the production of cars (a) 69.8 (b) 83
of type E is represented by a pie chart, then (c) 74.6 (d) 78.2
the central angle of the sector represent- Direction: (36 - 38) Study the table and answer
ing the data of production of cars in 2013 the questions:
will be:
E

The table shows the number of candidates

appearing in the interview for a post in six
banks (H, I, J, K, L, M) and the percentage
(a) 102° (b) 84° of candidates found eligible.
(c) 70° (d) 80° (H, I, J, K, L, M)
Direction: (33 - 35) Study the table and answer
the questions:

Banks Candidates % of Candidates

The table shows Income and expenditure Appearing Qualifying
of a person for 3 years (in thousands) H 1500 14
I 2200 26
J 3000 17
K 980 20
L 1200 28
M 2500 21
Statement of Income and expenditure
36. The number of candidates found ‘not
Year Income Expense Savings
eligible’ in Bank K is approximately what
2000 110 103 +7 percent of the number of candidates found
2001 223 214 +9 ‘not eligible’ in Bank I? (to the nearest
2002 243 197 +46 integer)
2003 189 232 –43
 (a) 42 (b) 48 SSC CGL 6 March 2020 (Afternoon)
(c) 44 (d) 51
(a) 4 (b) 1
37. What was the average number of candidates
appeared in the interview for Banks H, J (c) 2 (d) 3
and L taken together? Direction: (41 - 44) Study the table and answer
H, J L the questions:

SSC MTS - 13 August 2019 (Morning) The given table represents the exports (in
(a) 1900 (b) 1500 crores) of four items A, B, C and D over a
(c) 1800 (d) 2000 period of six years. Study the table carefully
38. What is t he rati o of the num be r of and answer the questions that follows:
candidates found eligible in Bank H to the
number of candidates found eligible in Bank
L? A, B, C D
H
L

SSC MTS- 13 August 2019 (Morning) ITEMS A B C D

(a) 2 : 5 (b) 3 : 7 YEAR
(c) 5 : 8 (d) 5 : 6 2010 240 128 180 214
Direction: (39 - 40) Study the table and answer 2011 250 134 244 282
2012 225 138 230 247
the questions: 2013 370 169 340 224
2014 425 182 300 309
2015 400 209 306 275

The table below shows income (in rupees) 41. In which year, the exports of item D were
for a particular month, together with their 1.4 times the average exports of item B
sources in respect of 5 employees A, B, C, during the six years?
D and E.
D
A, B, C, D
B
E
SSC C GL 5 March 2020 (Morning)
(a) 2014 (b) 2013
Employee A B C D E
Salary 52000 48,500 42,000 31,000 25,000 (c) 2011 (d) 2012
Overtime 0 0 1500 2500 3,200
Arrears 5500 4500 4,000 3000 1,500 42. What is the ratio of the total exports of item
Bonus 3500 3000 2,500 2000 2,000 A in 2014 and 2015 to the total exports of
Miscellaneous 5000 3000 2,000 1500 0 item C in 2011 and 2015?
income
Total 66000 59000 52,000 40,000 31,700
A
39. How many employees got more arrears than B
the average arrears received by all the
employees?
SSC CGL 5 March 2020 (Morning)
(a) 4:3 (b) 7:5
(c) 5:4 (d) 3:2
SSC C GL 5 March 2020 (Evening)
43. The total exports of item D in 2010, 2012 and
(a) 3 (b) 2 2014 is what percentage of the total exports
(c) 1 (d) 4 of all the four items in 2011 and 2012?
 (a) 44.8% (b) 45% during six years?
(c) 46.2% (d) 44%
D
44. The total exports of item A from 2012 to
B
2014 is what percentage less than the total
exports of all the four items in 2015? (a) 2011
(correct to one decimal place) (b) 2012
2012 2014 A 2015 (c) 2014
(d) 2013
48. The total exports of item A from 2012 to
SSC CGL 5 March 2020 (Morning) 2014 is what percentage less than the total
exports of all the four items in 2015? (Cor-
(a) 15.2% (b) 13.8%
rect to one decimal place)
(c) 16.7% (d) 14.3%
A
Direction: (45 - 48) The given table represents
the exports (in crores) of four items A, B, C and
D over a period of six years. Study the table and
answer the question that follows.
(a) 15.2%
A,
(b) 16.7%
B, C D
(c) 14.3%
(d) 13.8%
Direction: (49-50) Study the given table and an-
swer the question that follows.

The table shows the income and expendi-

45. What is the ratio of the total exports of item ture of companies A and B from 2015 to
A in 2014 and 2015 to the total exports of 2019 (in Rs. crore)
item C in 2011 and 2015?
A B
A
C

Year Company A Company B

(a) 3 : 2 Income Expenditure Income Expenditure
(b) 4 : 3 2015 110 90 160 120

(c) 5 : 4 2016 150 125 180 125

2017 200 140 225 140
(d) 7 : 5
2018 240 175 320 275
46. The total exports of item D in 2010, 2012 2019 300 220 360 300
and 2014 is what percentage of the total
exports of all the four items in 2011 and 49. 40% of the total expenditure of company B
2012? from 2015 to 2019 is what per cent less
D than 65% of the total income of company A
from 2015 to 2018? (correct to one deci-
mal place)
 Selection Post - Phase IX (14 March 2022) 2015 2017 B
(a) 14.7% 2016 2018 A
(b) 13.5% x x
(c) 15.6% Selection Post - Phase IX (15 March 2022)
(d) 18.5% (a) 90 and 100 (b) 60 and 70
(c) 70 and 80 (d) 80 and 90

Answer Key
1.(c) 2.(d) 3.(b) 4.(c) 5.(d) 6.(c) 7.(c) 8.(a) 9.(b) 10.(b)

11.(b) 12.(c) 13.(d) 14.(d) 15.(d) 16.(d) 17.(a) 18.(a) 19.(a) 20.(c)

21.(d) 22.(d) 23.(d) 24.(d) 25.(d) 26.(a) 27.(c) 28.(d) 29.(d) 30.(d)

31.(b) 32.(d) 33.(b) 34.(d) 35.(c) 36.(b) 37.(a) 38.(c) 39.(a) 40.(c)

41.(b) 42.(d) 43. (d) 44. (d) 45.(a) 46.(c) 47.(d) 48.(c) 49.(c) 50.(d)
 Data Interpretation/vkdM+kass dk fuo
( Practice Sheet With Solution)
Direction (01): Study the given table and answer the 25» dk ykHk vftZr djus ds fy,] 2019& 2020 esa ldy jk
question that follows. djksM+ esa) fdruh gksuh pkfg,] ;fn oqQy ykxr leku
fuEukafdr rkfydk dk vè;;u djsa vkSj uhps fn, x, ç'u dk SSC CGL 14/07/2023 (Shift-01
mÙkj nsaA (a) 7800 (b) 8000
The table shows the classification of 100 students based (c) 8250 (d) 8125
on the marks obtained by them in History and Direction (03): Study the given pie chart and answe
Geography in an examination. the question that follows.

r
fn,vadksa
;g rkfydk ,d ijh{kk esa bfrgkl vkSj Hkwxksy esa Nk=kksa }kjk çkIr x, ikbZdspkVZ dk vè;;u dhft, vkSj uhps fn, x, iz
vk/kj ij 100 Nk=kksa dk oxhZdj.k n'kkZrh gSA mÙkj nhft,A

si
Population of six villages in 2020
Marks out of 50/50 esal sv ad o"kZ 2020 esa 6 xk¡oksa dh tula[;k

an by
40 and 30 and 20 and 10 and 0 and
Subject above above above above above 12%

n
fo"k; 40 vkSj 30 vkSj 20 vkSj 10 vkSj 0 vkSj 18% A
mlls vf/d mlls vf/d mlls vf/d F
mlls vf/d mlls vf/d 13%
B

ja
History
9 32 80 92 100
R s 17%
bfrgkl
Geography E
4 66 81 100
Hkwxksy 21
a th
8%
Average D 32%
(Aggregate) 7 27 73 87 100 C
vkSlr(dqy)
1. Based on the table, what is the number of students scoring 3. The population of village D in 2020 was 10,500. Wha
ty a

less than 20% marks in aggregate? was the population of village A in 2020? Tota
population of these six villages is 100%.
rkfydk ds vkèkkj ij] oqQy 20» ls de vad izkIr djus okys Nk=kksa
di M

2020 esa xk¡oD dh tula[;k 10,500 FkhA

2020 esa xk¡oAd
dh oqQy la[;k fdruh gS\ tula[;k fdruh Fkh\ bu Ng xk¡oksa dh oqQy100%
tula[;kgS
SSC CGL 14/07/2023 (Shift-01) SSC CGL 14/07/2023 (Shift-02
(a) 13 (b) 11 (a) 15,570 (b) 15,750
(c) 10 (d) 12 (c) 17,550 (d) 7,875
Direction (02): The following bar chart represents the 4. What is the central angle corresponding to the secto
gross amount (in Rs.lakhs) and total cost (in Rs. lakhs) of a firm. indicating the expenses incurred on Health?
LokLF; ij gksus okys O;; dks n'kkZus okys {ks=k ds vuq
fuEufyf[kr ckj pkVZ ,d iQeZ dh ldy jkf'k (#i;s yk[k esa) vkSj
oqQy ykxr (#i;s yk[k esa) n'kkZrk gSA Amount spent of various expenses
Y fofHkUu O;;ksa ij [kpZ dh xbZ jkf'k
Total/dqy
7500
Gross Amount/ldy jkf'k
A

50 lect

6500
E
00 ric

6000
5500 200 20000
En 0 Rent
ity

5000
ter
Rupees in crores

4500 tain
.
4000
3500 n
io
000 cat
8 du
E 15000
X Health
2016-17 2017-18 2018-19 2019-20
Years
SSC CGL 14/07/2023 (Shift-03
2. In order to make a profit of 25%, what should the gross
amount have been (in Rs. crores) in 2019-2020, if the (a) 72° (b) 54°
total cost remained the same? (c) 110° (d) 108°

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 answer the following question. of sweets manufactured by six factories.
nh xbZ rkfydk dk è;kuiwoZd vè;;u dhft, vkSj fuEufyf•r ç'u fuEufyf[kr lkj.kh N% dkj[kkuksa }kjk fufeZr feBkb;
dk mÙkj nhft,A djrh gSA
The table shows the percentage of students of four departments
- Mechanical, Civil, Computer Science and Applied - with each
student being in only one department. The table also shows
the number of students of these four departments in five
different colleges, with the total number of students being 2080.
rkfydk pkj foHkkxksa & eSdsfudy] flfoy] daI;wVj lkbal vkSj ,IykbM & ds
fo|kfFkZ;ksa ds çfr'kr dks n'kkZrh gS] ftlesa çR;sd fo|kFkhZ dsoy ,d foHkkx esa
gSA rkfydk] fo|kfFkZ;ksa dh dqy la[;k 2080 gksus ds lkFk ikap vyx&vyx
dkWystksa esa bu pkj foHkkxksa ds fo|kfFkZ;ksa dh la[;k Hkh n'kkZrh gSA
7. What is the respective ratio of sweets manufactured b
Number of Computer
College Mechanical Civil Applied factories P, Q and R together in the year 1998, to th
students Science
sweets manufactured by factories S, T and U togethe
IIT Delhi 430 – 20% – 10%

r
IIT Kanpur 350 20% – 25% –
in the same year?
o"kZ 1998 esa dkj[kkuksa
P, Q vkSjR esa fufeZr feBkb;ksa v

si
IIT Bombay – 20% 18% – 32%
IIT Madras – – 25% 18% 35%
dkj[kkuksa
S, T vkSjU esa fufeZr feBkb;ksa dk vuqikr D;k
IIT Guwahati 400 20% 22% – 20%

an by
SSC CGL 14/07/2023 (Shift-04
5. If the number of students in IIT Bombay is 20% less
than the number of students in IIT Madras, then the (a) 568 : 499 (b) 61 : 72

n
number of students in IIT Bombay is: (c) 61 : 42 (d) 40 : 51
;fn IIT ckWEcs esa fo|kfFkZ;ksa
IIT dh
enzkl
la[;k]
esa fo|kfFkZ;ksa dh la[;kDirection (08): Select the correct statement wit

ja
ls 20» de gS] rks
IIT ckWEcs esa fo|kfFkZ;ksa dh la[;k fdruh gS\ to the below bar graph.
respect
R s
SSC CGL 14/07/2023 (Shift-03) uhps fn, x, naM vkjs[k ds laca/ esa lgh dFku dk p;
a th
(a) 300 (b) 500 Production of Rice (in lakh tonnes)
by three states A, B and C
(c) 200 (d) 400
Direction (06): Study the given bar-graph and answer 70
the question that follows. 60
ty a

fn, x, naM&vkjs[k dk vè;;u dhft, vkSj uhps fn, x, iz'u dk 50

mÙkj nhft,A 40
di M

The bar graph shows the sales of cycles (in 1000 numbers) 30

from four different companies during 2018. 20

naM vkys[k 2018 ds nkSjku pkj fofHkUu daifu;ksa ls lkbfdyksa dh fcØh10(1000

la[;k esa) n'kkZrk gSA 0
2016 2017 2018 2019 2020 2021
A B C
Sales of cycles (in 1000 numbers)
from four different companies in 2018
80
8. Production of Rice (in lakh tonnes) by three state A, B and C
70
rhu jkT;ksa
A, B vkSjC }kjk pkoy dk mRiknu (yk[k Vu esa
Sales (in 1000 numbers)

70
60 55
50 SSC CGL 17/07/2023 (Shift-01
40
30 25
30 (a) Rice production by A in 2016 is less than the ric
20 production by C in 2020.
A

10
2016 esaA }kjk pkoy dk mRiknu
2020 esaC }kjk fd, x
0
A B C D pkoy mRiknu ls de gSA
Different Companies
(b) B produced more rice than A and C in 2018.
6. What is the average sales of all the companies (in 1000 2018 esaB us A vkSjC ls vf/d pkoy dk mRiknu fd;k
numbers) for the year 2018? (c) The highest production of rice was in the year 2017
o"kZ 2018 ds fy, lHkh daifu;ksa dh vkSlr fcØh (1000 la[;k esa) pkoy dk lokZf/d mriknu o"kZ
2017 esa gqvk FkkA
fdruh gS\ (d) Rice production by B in 2017 is equal to the ric
SSC CGL 14/07/2023 (Shift-04) production by A in 2020.
(a) 45,000 (b) 34,000 2017 esaB }kjk pkoy dk mRiknu
2020 esaA ds pkoy mR
(c) 54,000 (d) 24,000 ds cjkcj gSA

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 monthly sale of laptops of five companies A, B, C, D and E Year Company
in a store. A G Y T F
fuEufyf[kr ikbZ pkVZ esa ,d LVksj esaA,ik¡p daifu;ksa
B, C, D vkSjE 2010 9.6 10.4 9.3 9.8 8.7
ds ySiVkWi dh ekfld fcØh dks n'kkZ;k x;k gSA 2011 10.4 12.6 7.2 13.8 6.2
2012 12.6 9.8 10.4 14.9 9.8
E 2013 16.8 15.4 11.4 16.3 11.3
12%
2014 19.3 13.4 13.4 11.8 7.8
A 2015 18.7 16.7 12.7 15.7 13.7
D 40%
25%
11. What is the difference (in thousands) between th
students placed in all companies in the year 2014 an
C B 2012?
8% 15%
o"kZ 2014 vkSj 2012 esa lHkh daifu;ksa esa ukSdj
la[;k ds chp varj fdruk gS (gtkj esa)\

r
9. If 2,500 laptops were sold by the store in a month,
then what is the difference between the number of SSC CGL 17/07/2023 (Shift-03

si
laptops sold of Company A and that of Company C? (a) 9000 (b) 7800
;fn ,d eghus esa LVksj2}kjk
]500 ySiVkWi csps x,] rksAdaiuh
ds

an by
(c) 8200 (d) 7500
csps x, ySiVkWi dh la[;k vkSjCdaiuh
ds csps x, ySiVWki dh la[;k
12. Study the given pie-chart carefully and answer th
ds chp dk varj D;k gS\

n
following question.
SSC CGL 17/07/2023 (Shift-02)
fn, x, ikbZ&pkVZ dk è;kuiwoZd vè;;u dhft,A

ja
(a) 750 (b) 650
R s If scholarship has to be paid out of the donation fund
(c) 800 (d) 700 then what is the percentage of donation fund used fo
a th
Direction (10): Study the given table and answer the this purpose (rounded off to two decimal places)?
question that follows.
;fn Nk=ko`fr dk Hkqxrku nku iQaM ls fd;k tkuk gS
nh xbZ rkfydk dk vè;;u djsa vkSj uhps fn, x, iz'u dk mÙkj nsaA
fy, mi;ksx fd, x, nku iaQM dk izfr'kr (n'keyo dks nks
The table shows the classification of 100 students based on the
iw.k±afdr) fdruk gSa\
ty a

marks obtained by them in Statistics and Mathematics in an

examination out of 50. The entire fund that school gets from different source
di M

is equal to Rs.10 lakh

rkfydk ,d ijh{kk esa lka[;dh vkSj xf.kr esa Nk=kksa }kjk 50 esa ls izkIr vadksa ds
vk/kj ij 100 Nk=kksa dk oxhZdj.k n'kkZrh gSA Ldwy dks fofHkUu lzksrksa ls feyus okyk iwjk iQaM 1

40 and 30 and 20 and 10 and 0 and

Pa yment
Subject
11%
above above above above above Government
agencies Sc
Math 8 33 90 82 100 12% ho
la
School
NGO's rs maintains
26
Statistics 5 22 60 87 100 Internal
sources 38% % h ip 30%
15%
10. If at least 60% marks in Mathematics are required for
Donation Reserved
pursuing higher studies in Mathematics, then how 35% 33%
many students will be eligible to pursue higher studies
in Mathematics? Source of funds in school Uses of funds by school
;fn xf.kr esa mPp f'k{kk tkjh j[kus ds fy, xf.kr esa de ls de 60»
A

vad vk'o;d gSa] rks fdrus Nk=k xf.kr esa mPPk f'k{kk tkjh j[kus ds SSC CGL 17/07/2023 (Shift-04
fy, ik=k gksaxs\ (a) 74.29% (b) 72.15 %
SSC CGL 17/07/2023 (Shift-03) (c) 80.25 % (d) 75.25%
(a) 33 (b) 27 13. The following table shows the scores of three attempt
(c) 90 (d) 10 of Archery Players in a tourament. the player scorin
the highest average score was decleared the best playe
Direction (11): The following table gives the information of
the number of students (in thousands) placed in five different Who the best player ?
companies during six different years. fuEu rkfydk ,d Vquk±esaV esa rhjankth ds f[kykfM
fuEufyf•r rkfydk Ng vyx&vyx o"kks± ds nkSjku ikap vyx&vyx daifu;ksa
Ldksj dks n'kkZrh gSA tks f[kykM+h mPpÙke vkSlr
esa j•s x, Nk=kksa dh la[;k (gtkjksa esa) dh tkudkjh nsrh gSA f[kYkkM+h gksxkA mPPkre vkSlr Ldksj djus okys

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 Player Score-1 Score-2 Score-3
answer the question.
Angela 62.50 65.00 64.50
fuEufyf[kr n.M vkjs[k dk vè;;u dhft, vkSj iz'u dk mÙ
Maria 69.05 70.00 67.52
Sales of XYZ Phones over the years
Sareena 73.81 72.50 74.20
60

Sales of XYZ phones

Preeti 74.30 75.00 77.50 48
50
40 40
Deepika 64.29 67.50 63.28 40
30
30
SSC CGL 17/07/2023 (Shift-04) 20 18
20
(a) Angela (b) Maria 10
(c) Deepika (d) Preeti 0
2015 2016 2017 2018 2019 2020
Direction (14): Study the following table and answer Years
the question below.
fuEufyf[kr rkfydk dk vè;;u djsa vkSj uhps fn, x, iz'u dk mÙkj nsaA
Sales of XYZ Phone over the years = fiNys oqQN o"kks
XY
Percentage of enrolled Ratio of male to
iQksu dh fcØh

r
School Total number of students who opted female students
Name students enrolled for Biology who opted for Biology Sales of XYZ Phone = XYZ iQksu dh fcØh]
Years = o"kZ

si
(LoQwy dk (thofoKku dk pÕku djus(thofoKku dk pÕku djus 16.
(ukekafdr fo|kfFkZÕkksa Find the percentage increase of sales of XYZ phone
uke) dh oqQy l[aÕkk) okys ukekfadr fo|kfFkZÕkkas
okys Nk=kkas vkSj Nk=kkvkas from 2019 to 2020? (Rounded up to 2 decimal place
dk i fzr'kr) dk izfr'kr)

an by
A 900 30% 7:8 2019 ls 2020 rd XYZ iQksu dh fcØh esa izfr'kr o`f¼ K
B 400 38% 9 :10 (nks n'keyo LFkkuksa rd)

n
C 1000 24% 5 :19
D 800 18% 5:7 SSC CGL 18/07/2023 (Shift-02
(a) 122.22 % (b) 110.11%

ja
14. What is the ratio of the total number of male students to
R s
that of female students who opted for Biology in schools A (c) 119.19 % (d) 121.89 %
and D together? Direction (17): A toffee company prepares toffee of tw
a th
LowQy A vkSjD dks lkFk feykdj muesa thofoKku dk p;u djus
different flavours X and Y. The production of two flavours ove
a period of 4 years is expressed in the bar graph given below
okys oqQy Nk=kksa vkSj oqQy Nk=kkvksa dk vuqikr D;k gS\
,d VkWiQh daiuh nks vyx&vyxXÝysoj
vkSjY dh VkWiQ
SSC CGL 18/07/2023 (Shift-01)
djrh gSA 4 o"kZ dh vofèk esa nks Ýysoj dk mRiknu uhps
ty a

(a) 38 : 31 (b) 21 : 38
esa n'kkZ;k x;k gSA
(c) 31 : 28 (d) 31 : 38
di M

15. In a factory, utensils are manufactured in three plants, Production of two different toffee flavours by a
company over different years (in 1000 packs)
plant A, B and C. How many plates are manufactured
35
by plant B if total plates are 3260? 30
32
TOFFEE 1000 PACKS

30
,d iQSDVªh esa crZuksa dk fuekZ.k rhu la;a=kksa
(plants) A, B vkSjC 25
25

esa fd;k tkrk gSA ;fn fufeZr IysVksa dh dqy la[;k 3260 gS] rks
20
20 20 20 20 20

la;a=k
(Plant) B }kjk fdruh IysVksa dk fuekZ.k fd;k tkrk gS\ 15
10
Share in manufacturing 5
0
100% 2017 2018 2019 2020

X Y
80%
Plant-A
60% 17. What is the difference between the average productio
Plant-B of flavour X in 2017 and 2018 and the averag
A

40% Plant-C production of flavour Y in 2019 and 2020.

20% 2017 vkSj 2018 esa Ýysoj
X ds vkSlr mRiknu vkSj 20
0% 2020 esa Ýysoj
Y ds vkSlr mRiknu ds chp fdruk varj gS
Bowl Spoons Plates Glass
SSC CGL 18/07/2023 (Shift-03
Shares in manufacturing – fuekZ.k dk fgLlk (a) 6000 packs (b) 6400 packs
Bowl – dVksjk Spoons – IysV (c) 7500 packs (d) 7000 packs
Glass – fxykl Direction (18): The given table shows the percentag
of marks obtained by five students in five different
SSC CGL 18/07/2023 (Shift-02)
subject in a school.
(a) 1467 (b) 1304 nh xbZ rkfydk ,d LowQy esa ik¡p vyx&vyx fo"k;k
(c) 1254 (d) 1141 }kjk izkIr vadksa ds izfr'kr dks n'kkZrh gSA

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs4

 vx
azst h foKku xf. kr fgUnh d IaÕkV
wj of a year are shown in the following bar graph.
(100) (100) (100) (100) (100) fuEukafdr ckj xzkiQ esa ,d Jfed }kjk ,d o"kZ ds ck
Hitansh
67 70 64 55 75
vftZr dh xbZ etnwjh
` esa)
( dks n'kkZ;k x;k gSA
fgrk'ak
Akshara 12000
75 88 95 60 75
v{kjk 10000 9700
10000 9500 9400 9600 9250
Pihu 8500 8600
8250
69 72 89 64 77 8000
figq 8000 7200

Shreya 5600
71 66 80 68 72 6000
JÕskk
Rahul 4000
59 64 59 67 90
jkgqy 2000

18. What is Hitansh's overall percentage (rounded up to 2 0

decimal places) in the examination?
ijh{kk esa fgrka'k dk oqQy izfr'kr (2 n'keyo LFkkuksa rd) fdruk gS\
SSC CGL 18/07/2023 (Shift-03)

r
What is the average wage (in `) received by th
(a) 66.22% (b) 64.89% labourer in the first five months of the year?

si
(c) 65.31% (d) 75.11% o"k
Z ds igys ikap eghuksa esa Jfed }kjk izkIr vkSlr
` esa)
etnwjh
D;k g
Direction (19): Quantity of various food items used by a
restaurant during 4 months of a year (in kg). SSC CGL 19/07/2023 (Shift-02

an by
,d jsLrkjka }kjk o"kZ ds 4 eghuksa ds nkSjku mi;ksx dh tkus okyh
(a) 9,300 (b) 9,000
fofHkUu [kk| inkFkk±s dh ek=kk (fdyks esa)A (c) 9,100 (d) 9,200

n
22. The following pie chart shows the annual expenditur
on different items.
fuEufyf•r ikbZ pkVZ fofHkUu
ij okf"kZd
enksaO;; dks n'kkZrk g

ja
R s
What is the minimum angle in the given pie chart?
fn, x, ikbZ pkVZ esa U;wure dks.k fdruk gS\
a th

10 usi ena
Ho aint
% ng nc
m
30%
Meal
ty a

19. What is the average quantity of food item C used

25%
during all the 4 months together?

e
Children’s
,d lkFk lHkh 4 eghuksa ds nkSjku mi;ksx fd, x, [kk|
C dhinkFkZ
di M

education
vkSlr ek=kk fdruh gSA
in.
12.5%
Medicines
Enterta
Cl .5%
s
he

SSC CGL 18/07/2023 (Shift-04)

12
ot

10%

(a) 303.7 kg (b) 292.5 kg

(c) 211.8 kg (d) 253.6 kg
SSC CGL 19/07/2023 (Shift-03
20. The following figure shows the scores of students in
three tests organised by a coaching institution. Who (a) 36° (b) 45°
scored the best on average? (c) 32° (d) 30°
fuEukafdr vkjs• ,d dksfpax laLFkku }kjk vk;ksftr rhu ijh{kkvksa
23. esaThe
fo|kfFkZ;ksa
number of mobile sim-chards in 4 states/UT ar
given in the bar diagram.
ds vadksa dks fu:fir djrk gSA fdlds vkSlr vad lcls vf/d gSa\
Scores in different tests
4 jkT;ksa@dsUnz 'kkflr izns'kksa esa eksckby fle
120 vkjs[k esa nh xbZ gSA
101 99
A

100 95 93
90
94
88
93
90
Study the diagram and answer the question.
84 79 80
80
71
76
71 vkjs[k dk vè;;u djsa vkSj iz'u dk mÙkj nsaA
Scores

60
30
No. of mobile Sim Card owncrs

40
25
20

0 20
A B C D E
(in lacs)

Student Name 15 Jio

Test 1 Test 2 Test 3 BSNL

10
Airtel

5
SSC CGL 19/07/2023 (Shift-01)
0
(a) D (b) C Bihar U.P Delhi Punjab
States of India
(c) E (d) B

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs5

 cards sold is: that follows.
fnYyh esa ch,l,u,y fle&dkMZ vkSj ,;jVsy fle&dkMZ dh fcØh
ckj xzkiQ dk è;kuiwoZd vè;;u djsa vkSj fuEufyf•r ç'u d
dk vuqikr &&&& gSA The following bar graph shows the sale of Bluetoot
SSC CGL 19/07/2023 (Shift-03) earphones on online market places over the years.
(a) 1 : 2 (b) 2 : 1 fuEufyf•r ckj xzkiQ fiNys dqN o"kks± esa vkWuyk
(c) 3 : 2 (d) 2 : 3 CywVwFk b;jiQksu dh fcØh dks n'kkZrk gSA
24. The following figure shows the different expenses that are
Sale of Bluetooth earphones (in lakhs)
incurred in manufacturing toys. If the total expenditure 94.5
100
is `3,00,000, then how much expenditure was incurred 73.5
84
80
on Material Cost and Selling Expenses (in `)? 64.75
60 49.7
fuEufyf•r vkÑfr f•ykSuksa ds fuekZ.k esa gksus okys fofHkUu40O;; 42dks
n'kkZrh gSA ;fn dqy `3,00,000
O;; gS] rks lkexzh ykxr vkSj 20

foØ; O;; ( ` esa) ij fdruk O;; fd;k x;k\ 0

2016 2017 2018 2019 2020 2021

r
% Share in total expenditure
Sale of Bluetooth earphones (in lakhs)

si
Material Find the approx, percentage increase in the sale o
Selling Cost earphones from 2017 to 2020.

an by
15%
Expenses os t
28%
b
rc
ou % 2017 ls 2020 rd b;jiQksu dh fcØh esa yxHkx çfr'kr o`f¼
La 1 2

n
SSC CGL 20/07/2023 (Shift-01
Assembling
Packaging cost (a) 69% (b) 57%

ja
25% 20%
(c) 78% (d) 71%
R s
27. Study the given pie-chart carefully and answer th
a th
following question. What is the difference between th
SSC CGL 19/07/2023 (Shift-04) funds (in Rs.) acquired by the school from donatio
(a) 1,20,000 (b) 1,15,000 and those from government agencies?
(c) 1,29,000 (d) 84000 fn, x, ikbZ&pkVZ dk è;kuiwoZd vè;;u dhft, vkSj
ty a

25. Study the given bar graph carefully and answer the ç'u dk mÙkj nhft,A Ldwy }kjk nku ls vftZr iQaM
following question. ,tsafl;ksa ls çkIr iQaM
Rs. esa)
( ds chp fdruk varj gS\
di M

fn, x, naM vkjs• dk è;kuiwoZd vè;;u dhft, vkSj fuEufyf•r The entire fund the school gets from different source
ç'u dk mÙkj nhft,A is equal to Rs.10 lakh
What is the ratio of the number of companies with LowQy dks fofHkUu lzksrksa ls feyus Rs.10
okyk iwjk
yk[k
more demand than production to the number of cjkcj gS
companies with more production than demand? Pa yment
mRiknu ls vf/d ek¡x okyh daifu;ksa dh la[;k dk] ek¡x ls vf/d
11%

Government
agencies Sc
mRiknu okyh daifu;ksa dh la[;k ls vuqikr D;k gS\ 12% ho
la
School
maintains
Internal NGO's 26 rs h
38% % ip 30%
Demand and Production of Washing Machines
sources
of Five Companies for December 2022 15%
Number of washing Machines

Donation Reserved
6000 35% 33%
5000
A

4600 4800

3600 4000 Source of funds in school Uses of funds by school

3000 2800 2900
SSC CGL 20/07/2023 (Shift-02
2300
(a) 2,30,000 (b) 2,50,000
(c) 2,80,000 (d) 2,40,000
P Q R S T 28. Study the given bar graph carefully and answer th
Companies
following question. The demand of Company P is wha
Demand Production
percentage of the demand of Company Q?
SSC CGL 19/07/2023 (Shift-04) fn, x, naM vkjs• dk è;kuiwoZd vè;;u dhft, vkSj fuE
(a) 3 : 2 (b) 4 : 3 ç'u dk mÙkj nhft,A daiuhP dh ek¡x daiuhQ dh ek¡x
(c) 3 : 5 (d) 2 : 3 fdruk çfr'kr gS\

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs6

 Demand and Production of Washing Machines
of Two Companies for January 2022
Cotton and Khadi ?
Demand Production dikl vkSj [kknh ds oqQy mRiknu ds eè; fdruk varj g
Number of washing machines Note: Ignore negative sign, if any
4550

3276
fVIi.kh% Í.kkRed fpUg ij è;ku u nsa] ;fn dksbZ gS]
2630

1520 Production (in mn units)

130
125
125

P Q 120 121 119 120

120
118 118
Companies

Production
115 114
115
112
110 111 110 109
110
SSC CGL 20/07/2023 (Shift-02) 105
105
(a) 68% (b) 65%
100
(c) 70% (d) 72%
95
29. The following pie charts show the data of the number Jute Cotton Silk Polyster Khadi

r
of appeared and passed students of class 12 in Material

sections A, B, C, D and E. Company A Company B Company C

si
fuEu ikbZ pkVZ d{kk 12 dsA,lsD'ku
B, C, D vkSjE esa mifLFkr
vkSj mÙkh.kZ Nk=kksa dh la[;k dk MsVk n'kkZrs gSaA SSC CGL 21/07/2023 (Shift-01

an by
(a) 10 (b) 11
A, 42°
(c) 8 (d) 7

n
10%
E, 58° 20%
B, 50° 15% 32. Study the given table and answer the question tha
follows.

ja
D, 110°
nh xbZ rkfydk dk vè;;u dhft, vkSj uhps fn, x, ç'u dk mÙkj
R s 25% 30%
C, 100°

The table gives the number of graduate student

a th
Appeared students = 1800 Passed students = 800
enrolled in 4 different colleges A, B, C, and D in a cit
over the years 2010 to 2014 and also the number o
What is the percentage of students who appeared for
students who passed in the final examination durin
the exam in section E (correct to one decimal place)?
these years.
lsD'kuE esa ijh{kk esa mifLFkr Nk=kksa dk izfr'kr D;k gS (n'keyo
ty a

ds ckn ,d vad rd)\ ;g rkfydk o"kZ 2010 ls 2014 rd fdlh 'kgj ds fofHkUu
A, B, C vkSjD esa ukekafdr Lukrd Nk=kksa dh la
di M

SSC CGL 20/07/2023 (Shift-04)

lkFk gh bu o"kks± ds nkSjku vafre ijh{kk esa mÙk
(a) 29.1% (b) 16.8%
(c) 18.2% (d) 16.1%
la[;k Hkh n'kkZrh gSA
30. The following figure shows the percentage growth in Year 2010 2010 2011 2011 2012 2012 2013 2013 2014 2014
College Enrolled Pass Enrolled Pass Enrolled Pass Enrolled Pass Enrolled Pass
units of fans and coolers sold in some years. What is A 680 620 600 560 720 700 800 760 750 700
the different between the averages of the growth rate B 550 530 450 420 600 550 650 620 700 680
in coolers and fans? C 480 450 520 500 580 550 620 600 720 700

fuEukafdr vkjs• dqN o"kks± esa csps x, ia•ksa vkSj dwyjksa dh bdkb;ksa
D 710 650 750 710 680 640 720 690 740 710

esa çfr'kr o`f¼ n'kkZrk gS dwyjksa vkSj ia•ksa dh o`f¼ nj ds Find

vkSlrthe
esaratio of the average of students enrolled from
fdruk varj gS\ college D to the average of students who passed from
college D over all the years.
Percentage growth rate in sales
lHkh o"kks± ds nkSjku
D ls
dkWyst
ukekafdr Nk=kksa ds vkSl
60 D ls mÙkh.kZ Nk=kksa ds vkSlr ls vuqikr Kkr dhft,A
A

2021
27

2020
72 SSC CGL 21/07/2023 (Shift-03
18
Year

(a) 19 : 17 (b) 18 : 17
54
2019
30
(c) 21 : 17 (d) 20 : 17
2018
20
45 33. Study the given data and answer the question tha
0 10 20 30 40 50 60 70 80
follows.
Sales (in %) fn, x, vkadM+ksa dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙ
Coolers Sold Fans sold
Data regarding population of different states in the yea
2015 is shown in the pie-chart and table.
SSC CGL 21/07/2023 (Shift-01)
(a) 34 (b) 30 o"kZ 2015 esa fofHkUu jkT;ksa dh tula[;k ls lacaf
(c) 32 (d) 42 vkSj rkfydk esa fn•k;k x;k gSA

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs7

 T.N question that follows.
U.P
9%
15% fn, x, xzkiQ dk è;kuiwoZd vè;;u dhft, vkSj uhps fn,
Bihar A.P ds mÙkj nhft,A
11% 25%
Wheat imports (in thousand tonnes)
Goa
7000
12% 6000
M.P 6000
5200
Delhi 20%
8% 5000 4400

4000 3465 3600

3000 2700
2413
Sex and Literacy-wise
population Ratio 2000 1600

States Sex Literacy 1000

M F Literate Illiterate 0
1970 1971 1972 1973 1974 1975 1976 1977
A.P 5 3 2 7

r
M.P. 3 1 1 4
Delhi 2 3 2 1 The decrease in imports in 1971 was what percentag

si
Goa 3 5 3 2 of the imports in 1970 (rounded off to the neares
Bihar 3 4 4 1 integer) ?
U.P. 3 2 7 2

an by
T.N. 3 4 9 4 1971 esa vk;kr esa deh] 1970 esa vk;kr dk fdruk çfr'kr F
iw.kkZad rd)\
If the total population of the given states is 31,5000,

n
then what was the total number of illiterate people in SSC CGL 21/07/2023 (Shift-04
Goa and M.P.? (a) 54% (b) 56%

ja
;fn fn, x, jkT;ksa dh oqQy tula[;k 31]5000 gS] rks xksok vkSj
R s
(c) 52% (d) 53%
eè; izns'k esa fuj{kj yksxksa dh oqQy la[;k fdruh Fkh\ 36. Study the given data and answer the question tha
a th
SSC CGL 21/07/2023 (Shift-03) follows.
(a) 65520 (b) 81900 fn, x, MsVk dk vè;;u dhft, vkSj uhps fn, x, ç'u
(c) 120500 (d) 90870 mÙkj nhft,A
ty a

34. The given pie-chart shows the marks scored by a The data shows the number of candidates (i
student in different skills in an examination thousand) appearing for Civil Service (CS) an
di M

mathematical ability, verbal ability, reasoning, coding Engineering Service (ES) Examinations in the year
and puzzle solving. The values are given in degrees. 2020 2021, 2022 in USA.
fn;k x;k ikbZ pkVZ ,d ijh{kk esa fofHkUu dkS'kyksa xf.krh; {kerk];w,l,
MsVk] (USA) esa o"kZ 2020 2021 2022 esa f
ekSf•d {kerk] rdZ'kfÙkQ] dwVys•u vkSj igsyh gy djuk esa(CS) ,d vkSj bathfu;fjax lsok
(ES) ijh{kkvksa ds fy, mifL
Nk=k }kjk çkIr & vadksa dks n'kkZrk gS eku fMxzh esa fn, x,okys
gSaAmEehnokjksa dh la[;k (gtkj esa) n'kkZrk gSA
Answer the following question. If total marks were 3000, Civil Service Engineering Service
Graduates out of Graduates out o
then what would be the marks in resoning? Year
Total No. of
the total
Total No. of
the total
Candidates Candidates
candidates candidates
fuEufyf•r ç'u dk mÙkj nsaA ;fn dy vad 3000 Fks] rks rdZ'kfÙkQ Appeared
appeared (in %)
Appeared
appeared (in %)
2020 75 60 89 57
esa vad D;k gksaxs\ 2021 110 64 118 65
2022 120 80 135 78

What is total number of graduates who appeared fo

A

Puzzle both CS and ES together in the year 2022?

75° Mathematics
o"kZ 2022 CSesa vkSjES nksuksa ds fy, mifLFkr gq
100°
Coding
dh dqy la[;k fdruh gS\
45° Verbal SSC CPO 03/10/2023 (Shift-01
Reasoning 65° (a) 2,01,000 (b) 2,01,200
75° (c) 2,01,100 (d) 2,01,300
37. Study the given data and answer the question tha
SSC CGL 21/07/2023 (Shift-04) follows.

(a) 375 (b) 833 fn, x, MsVk dk vè;;u dhft, vkSj uhps fn, x, ç'u
(c) 625 (d) 541 mÙkj nhft,A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs8

 80
appearing in an entrance exam from four different

Number of Students
70
cities and the ratio of candidates passing and failing 60
in the same.
50
rkfydk pkj vyx&vyx 'kgjksa ls ,d ços'k ijh{kk esa mifLFkr 40

gksus okys mEehnokjksa dh la[;k (yk• esa) vkSj mlesa mÙkh.kZ

30 vkSj

vuqÙkh.kZ gq, mEehnokjksa dk vuqikr n'kkZrh gSA 20

10
City A B C D 0
1st division 2nd division 3rd division Fail
Number of candidates 2.15 1.14 1.56 2.35 2008 20 60 60 30
2009 30 60 50 50
Ratio of candidates passing and failing within the city. 2010 40 60 70 30

'kgj ds Hkhrj mÙkh.kZ vkSj vuqrh.kZ gq, mEehnokjksa dk The

vuqikr
percentage of students who passed in the secon
division in the year 2010 was:
City Passing Failing
o"kZ 2010 esa f}rh; Js.kh esa mÙkh.kZ gksus oky

r
A 4 2
B 5 3 fdruk Fkk\

si
C 6 4 SSC CPO 03/10/2023 (Shift-01
D 4 6 (a) 28% (b) 32%

an by
What is the number of failed candidates (in lakh) in (c) 25% (d) 30%
city D?

n
40. What is the central corresponding to the expenditur
D 'kgj esa vuqÙkh.kZ gq, mEehnokjksa dh la[;k (yk[k esa) fdruh gS\on Education in the given figure?
spent
fn, x, ikbZ pkVZ esa f'k{kk ij [kpZ fd, x, O;;

ja
R s
SSC CPO 03/10/2023 (Shift-01)
dsaæh; dks.k D;k gS\
(a) 1.70 (b) 1.85
a th
(c) 1.25 (d) 1.41
38. The following table represents the population of four Education Health
5425 5250
different areas and the percentage of males, females,
ty a

and children P among them. Rent

4200 Clothing
fuEu rkfydk pkj vyx&vyx {ks=kksa dh tula[;k vkSj muesa ls 6125
di M

iq#"k] efgykvksa ,oa cPpksa dk çfr'kr n'kkZrh gSA Food

6475 Travelling
7525
Area Population Males Females Children
P 10,000 40% 40% 20%
Q 20,000 30% 40% 30%
R 16,000 50% 30% 20% SSC CPO 03/10/2023 (Shift-01
S 18,000 45% 35% 20% (a) 42° (b) 52°
(c) 55.2° (d) 55.8°
What is the total number of children in areas P and
Q together? 41. The given pie-chart shows the distribution of th
marks scored by Ram in five different subjects (i
{ks=k
P vkSjQ feykdj cPpksa dh dqy la[;k fdruh gS\ percentage).
fn;k x;k ikbZ&pkVZ ikap vyx&vyx fo"k;ksa es
A

SSC CPO 03/10/2023 (Shift-01)

(a) 7000 (b) 6000 vadksa dk forj.k (çfr'kr esa) n'kkZrk gSA
(c) 8000 (d) 9000
Social
39. Study the given bar-graph and answer the question 12% Hindi
that follows. 15.50%

fn, x, naM vkjs• dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA English
Science 22.50%
The bar-graph shows the higher secondary students 30%
of a school for the years 2008, 2009 and 2010.
Maths
uhps fn, x, naM vkjs• esa ,d Ldwy ds o"kZ 2008] 2009 vkSj 20%

2010 ds mPp ekè;fed fo|kfFkZ;ksa dks fn•k;k x;k gSA

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs9

 central angle of Science? During March and April.
xf.kr ds dsUæh; dks.k dk foKku ds dsUæh; dks.k ls vuqikr
ekpZ vkSj vçSy ekg ds nkSjku
A lsns'k
vkus okys i;ZVd
fdruk gS\ la[;k dk vuqikr Kkr dhft,A
SSC CPO 03/10/2023 (Shift-01) SSC CPO 03/10/2023 (Shift-01
(a) 5 : 2 (b) 2 : 3 (a) 7 : 10 (b) 9 : 10
(c) 4 : 3 (d) 3 : 5 (c) 3 : 7 (d) 4 : 7
42. Study the given graph carefully and answer the
44. The following data gives a year-wise outlay (in lakh
question that follows.
of rupees) in a certain 4-year plan (2018-2022
fn, x, xzkiQ dk è;kuiwoZd vè;;u dhft,] vkSj fuEufyf[kr of a state
iz'u dk mÙkj nhft,A fuEu vk¡dM+s ,d jkT; dh ,d fuf'pr 4 o"khZ; ;kstuk (20
esa o"kZ& okj ifjO;; (yk• #i;s esa) çnku djrs gSaA
Exports of Apples (in Lakhs of Rupees)

12 11.4 Year Health Housing Transport Education

10.8
9.5 2018-2019 6200 6500 7100 8000

r
10 9.9 2019-2020 7000 6600 7200 8100
8 2020-2021 7300 6800 7400 8200

si
6.5 7.8
2021-2022 7400 7000 7500 8400
6 5.2

an by
What was the percentage increase during 2020-202
4 over 2018-2019 in Education outlay?
f'k{kk ifjO;; esa 2018&2019 dh rqyuk esa 2020&

n
2

0
çfr'kr o`f¼ fdruh Fkh\

ja
02 03 00 4 005 006 00 7 008 SSC CPO 03/10/2023 (Shift-01
20 20
R s
2 2 2 2 2
(Years) (a) 2.8% (b) 2.5%
a th
In which year was there maximum percentage increase (c) 3.2% (d) 2.0%
in the export of apples to that of the previous years? 45. The table given below shows the number of product
fdl o"kZ lsc ds fu;kZr esa fiNYks
lsc ds
o"kZ
fu;kZr dh rqyuk sold by 4 shopkeepers A, B, C and D on fou
different days.
esa vf/dre izfr'kr o`f¼ gqbZ\
ty a

SSC CPO 03/10/2023 (Shift-01) fuEu rkfydk 4 vyx&vyx laLFkkvksa

A, B, C vkSjD esa fof
foHkkxksa esa deZpkfj;ksa dh la[;k dks n'kkZrh g
di M

(a) 2006 (b) 2005

(c) 2003 (d) 2008 Days/Person A B C D
43. Darjeeling gets tourists from two countries A and B. Monday 1000 800 – 750
Percentage wise distribution of the influx of tourists Tuesday 500 – 900 650
is given below for the first six months of the year. Wednesday 800 1000 700 –
nkftZfyax esa nks
A ns'kksa
vkSjB ls i;ZVd vkrs gSaA o"kZ ds igys 900 Thursday – 1200 800
Ng eghuksa esa i;ZVdksa ds vkxeu dk çfr'kr okj forj.k uhps
If the total products sold by A and B, including a
fn;k x;k gSA
four days, is 3000 and 3500, respectively, th
Total number of tourists = 60000 products sold by B on Tuesday is how many mor
Month Ratio (A : B) or less than the products sold by A on Thursday?
Jan 5:7
Feb 6:7 ;fn lHkh pkj fnuksa dks feykdj
A vkSjB }kjk csps x,
A

Mar
April
3:2
8:7
mRikn Øe'k% 3000 vkSj 3500 BgSa]
}kjkrks
eaxyokj dk
May 4:5 x, mRiknA }kjk xq#okj dks csps x, mRiknksa ls
June 3:4 ;k de gSa\
Jan
June 5%
SSC CPO 03/10/2023 (Shift-02
10% Feb
10% (a) 200 less (b) 300 more
(c) 300 more (d) 300 less
May March
30% 20% 46. Study The the given pie-chart and answer the questio
that follows.
April
25% fn, x, ikbZ pkVZ dk vè;;u dhft, vkSj fuEufyf•r ç
mÙkj nhft,A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 the year 2019. follows.
nh xbZ rkfydk dk vè;;u dhft, vkSj mlds ckn fn, x
ikbZ pkVZ] o"kZ 2019 esa jk"Vªh; ctV O;; dks n'kkZrk gSA
dk mÙkj nhft,A
The table shows the number of students doing variou
courses in various cities.
rkfydk fofHkUu 'kgjksa esa fofHkUu dkslZ djus
Others
20% la[;k n'kkZrh gSA
City MBA MCA M.Sc. M.Com. M.A. Total
International Ahmedabad 1234 1384 1440 1289 1332 6679
12% n Bengaluru 1156 1783 1874 1003 1340 7156
o
st Military
ere % Bhopal 1187 1347 1532 1321 1486 6873
Int ebt 6 62%
Chennai 1342 1473 1129 1765 1666 7375
d
New Delhi 1230 1098 1128 1865 1777 7098
Hyderabad 1456 1234 1556 1653 1789 7688
Kolkata 1239 1785 1865 1504 1762 8155
Percentage Distribution In Chennai, the number of students in MBA is near

r
If Rs. 850 billion were spent in year 2019 for interest what percentage of students in M.A? (rounded off t

si
on debt and military, then what would have been the 2 digits after decimal)
total expenditure for that year (in Rs.billion)? psUubZ MBAesa] esa Nk=kksa dhM,A
la[;kesa Nk=kksa dh
yxHkx fdruk çfr'kr gS\ (n'keyo ds ckn 2 vadksa rd i

an by
;fn o"kZ 2019 esa ½.k ij C;kt vkSj lsuk ds fy, 850#- fcfy;u
SSC CPO 03/10/2023 (Shift-02
•pZ fd, x, gksrs] rks ml o"kZ ds fy, dqy O;; (#- fcfy;u esa)

n
(a) 92.12% (b) 80.55%
fdruk gksrk
\
(c) 76.25% (d) 99.23%
SSC CPO 03/10/2023 (Shift-02) 49. The given table shows marks obtained by fou

ja
R s students in four subjects.
(a) 1,450 (b) 1,300
nh xbZ rkfydk pkj Nk=kksa }kjk fo"k;ksa esa izkIr
(c) 1,700 (d) 1,250
a th
47. Study the given pie-chart and answer the question Subjects Maths English Sciece Hindi
Student (100) (100) (100) (100)
that follows.
A 75 80 75 85
fn, x, ikbZ&pkVZ dk vè;;u djsa] vkSj fuEufyf•r ç'u dk mÙkj B 80 70 70 80
ty a

nsaA C 70 85 85 75
D 85 75 80 70
The pie-chart shows the percentage of 3600 students
di M

enrolled in different hobby classes in a school The average percentage of marks obtained by a
fuEu ikbZ&pkVZ ,d Ldwy esa fofHkUu gkWch d{kkvksastudents
esa ukekafdr
in Hindi is:
3600 fo|kfFkZ;ksa ds çfr'kr dks n'kkZrk gSA lHkh Nk=kksa
}kjk fganh esa izkIr vadksa dk vkSlr izfr'k
SSC CPO 03/10/2023 (Shift-02
Si (a) 77.6% (b) 77.5%
n
18 gin Cooking (c) 77.0% (d) 77.8%
% g
Da 22% 50. Study the given pie-chart and answer the questio
nci
21 n g that follows.
%
Drama
fn, x, ikbZ&pkVZ dk vè;;u dhft, vkSj fuEufyf[kr
Painting 13% mÙkj nhft,A
15% The pie-chart shows total expenditure of `3,00,00,00
on different items for constructing a flat in a town.
A

Stitching
11% ikbZ&pkVZ] ,d dLcs esa ,d ÝysV ds fuekZ.k d
enksa ij`3,00,00,000 ds dqy O;; dks n'kkZrk gSA
The number of students enrolled in Painting classes is
Tiles
approximately what percentage of those enrolled in 35° Steel
Singing classes? 55°
Timber
isafVax d{kkvksa esa ukekafdr fo|kfFkZ;ksa dh la[;k] flafxax d{kkvksa esa
45°
Labour
ukekafdr fo|kfFkZ;ksa dh la[;k ds yxHkx fdrus çfr'kr ds cjkcj gS\ Cement 65°
70°
SSC CPO 03/10/2023 (Shift-02)
Bricks
(a) 108% (b) 83% 90°

(c) 98% (d) 78%

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 and Timber is: dh dqy la[;k vkSj uoacj esa csdjh
P vkSj R }kjk ,dlk
Je vkSj bekjrh ydM+h ij dqy O;; dk izfr'kr Kkr dhft,A feykdj csps x, dsd dh dqy la[;k ds chp fdruk varj g
SSC CPO 03/10/2023 (Shift-02)
SSC CPO 03/10/2023 (Shift-02
5 1 (a) 9 (b) 10
(a) 30 % (b) 30 %
9 9 (c) 7 (d) 8
1 7 54. The following table shows the percentage of student
(c) 30 % (d) 30 % passing out of four colleges (A, B, C and D) over fou
3 9
years. Study the table and answer the question tha
Direction: (16-17): Study the given table and answer follows.
the question that follows.
nh xbZ rkfydk dk vè;;u dhft,] vkSj uhps fn, x, iz'u dk fuEufyf•r rkfydk pkj o"kks± esa pkj dkWystksa
A, B, C vkSj(D
mÙkj nhft,A ls mÙkh.kZ gksusNk=kksa
okys dk çfr'kr n'kkZrh gSA rkfy
The table shows the production of five types of cars P, Q,
djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA
R, S and T by a company from the year 1989 to 1994.
College College College College
nh xbZ rkfydk ,d daiuh }kjk o"kZ 1989 ls 1994 rd ik¡p çdkj dh

r
Years
A B C D
dkjksa
P, Q, R, S vkSjT ds mRiknu dks n'kkZrh gSA 2019 68 78 56 78

si
2020 72 86 78 82
Type of Car 1989 1990 1991 1992 1993 1994 Total
P 8 20 16 17 21 6 88 2021 78 91 82 85

an by
Q 16 10 14 12 12 14 78 2022 80 87 84 87
R 21 17 16 15 13 8 90
S 4 6 10 16 20 31 87

n
T 25 18 19 30 14 27 133 In which colleges is the average percentage of passin
Total 74 71 75 90 80 86 476 students over the given four years the least?
fn, x, pkj o"kks± esa fdl dkWyst esa mÙkh.kZ Nk=

ja
51. During the period 1989-94, which type of car featured
R s
a continuous increase in production? lcls de gS\
1989&94 dh vof/ ds nkSjku] fdl izdkj dh dkj ds mRiknu
a th
SSC CPO 03.10.2023 (Shift-3
esa yxkrkj c<+ksÙkjh gqbZ gS\
(a) College D (b) College A
SSC CPO 03/10/2023 (Shift-02)
(c) College B (d) College C
(a) S (b) P
ty a

(c) Q (d) R 55. The following table shows the earning of different firm
(in Rs. crore) in different years.
52. During the period 1989-94, which type of car featured
fuEu rkfydk fofHkUu o"kks± esa fofHkUu iQeks±
di M

a continuous decrease in production?

1989&94 dh vof/ ds nkSjku] fdl izdkj dh dkj ds mRiknu esa) n'kkZrh gSA
esa yxkrkj deh gqbZ gS\
Year 2012 2013 2014 2015 2016
SSC CPO 03/10/2023 (Shift-02) Firm
(a) R (b) T P 50 45 49 56 57
(c) Q (d) S Q 46 48 50 55 62
53. Study the given table and answer the question that R 70 65 63 68 73
follows. S 38 47 56 60 69
nh xbZ rkfydk dk vè;;u dhft, vkSj uhps fn, x, iz'u dk
The earning (in Rs. crore) of which firm was highes
mÙkj nhft,A in the year 2015?
The table shows the number of cakes sold by four
different bakeries months. 4 different o"kZ 2015 esa fdl iQeZ dh dekbZ (djksM+ #i;s esa) l
A

rkfydk 4 vyx&vyx eghuksa esa pkj vyx&vyx csdfj;ksa }kjk SSC CPO 03.10.2023 (Shift-3
csps x, dsd dh la[;k n'kkZrh gSA (a) S (b) Q
Month August September October November (c) R (d) P
Bakery
P 250 241 213 168
56. Study the given table and answer the question tha
Q 175 189 201 122
follows.
R 164 145 129 168 nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk m
S 224 154 184 209
The table shows the consumption of food grains pe
What is the difference between the total number of day (in grams) in the given set of following years.
cakes sold by bakeries Q and S together in September
and the total number of cakes sold by bakeries P and rkfydk fuEufyf•r o"kks± ds fn, x, lsV esa çfr fnu
R together in November? •ir (xzke esa) n'kkZrh gSA

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 The number of boy students in section B i
Coarse
Year Rice Wheat Pulses approximately what percentage of the number of gi
grains
1960  62 161 62 131 61
students in section E?
1970  72 197 81 116 64 lsD'kuB esa yM+dksa dh la[;k] E
lsD'ku
esa yM+fd;ksa dh
1980  82 187 116 103 46 yxHkx fdruk çfr'kr gS\
1990  92 189 124 101 44
2000  02 201 137 86 64 SSC CPO 03.10.2023 (Shift-3
(a) 25% (b) 45%
For which food grain consumption was there
successive increase over the given set of years? (c) 55% (d) 15%
fn, x, o"kks± esa fdl •k|kUu dh •ir esa yxkrkj o`f¼ gqbZ\59. Study the given table and answer the question tha
SSC CPO 03.10.2023 (Shift-3) follows.

(a) Wheat (b) Pulsesa nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk m
(c) Rice (d) Coarse grains The table shows the number of pages printed by fou
57. Study the given pie-chart and answer the question printers in three days.

r
that follows.
rkfydk rhu fnuksa esa pkj fçaVjksa }kjk eqfær i`"Bk
fn, x, ikbZ&pkVZ dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA

si
The pie-chart shows the expenditure of a family on Days Printers
different items and their savings throughout the year A B C D

an by
2021. Friday 350 451 412 206
ikbZ&pkVZ o"kZ 2021 ds nkSjku fofHkUu enksa ij ,d ifjokj Saturday
ds •pZ 420 325 365 251

n
vkSj mudh cpr dks n'kkZrk gSA Sunday 238 198 258 326

rt What is the respective ratio between total number o

ja
po
Savings ns
Tra 5%
R s pages printed by printer D on Friday and Sunda
Other 15%
20% together and total number of pages printed by printe
a th
ation A on Friday and Saturday together?
Ed uc ren
h ild
'kqØokj vkSj jfookj dks DfçaVj
}kjk eqfær i`"Bksa dh d
o f C
Food 1 2%
23%
vkSj 'kqØokj vkSj 'kfuokj dks A }kjk
fçaVjeqfær i`"Bksa
Clothing

Housing
la[;k ds chp lacaf/r vuqikr D;k gS\
ty a

15%
5%

SSC CPO 03.10.2023 (Shift-3

di M

If the total income of the family was Rs.1,00,000, how

much money (in Rs.) was spent on the education of (a) 51 : 34 (b) 34 : 51
the children? (c) 55 : 38 (d) 38 : 55
;fn ifjokj dh dqy vk; 1]00]000 #i;s Fkh] rks cPpksa dh f'k{kk
60. Study the given pie-chart and answer the questio
ij fdruk iSlk (#i;s esa) •pZ fd;k x;k\ that follows.
SSC CPO 03.10.2023 (Shift-3)
fn, x, ikbZ&pkVZ dk vè;;u djsa vkSj fuEufyf•r ç'u dk
(a) 13,500 (b) 13,000
(c) 12,000 (d) 12,500 The pie-chart is about Country-wise Global Export
Presentation.
58. Study the given graph and answer the following
question. ikbZ&pkVZ ns'k&okj oSf'od fu;kZr çLrqfr ds ckjs
fn, x, xzkiQ dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA Spain, 9°
Number of boys and girls students in each section.
çR;sd vuqHkkx esa yM+ds vkSj yM+fd;ksa dh la[;kA
A

100
90 USA, 96°
80 Australia,
STUDENTS

70 75°
60
50 India,
40 UK, 125° 55°
30
20
10
0
A C B
D E Total Export = Rs.14400 billion
SECTION By how much does the value of the exports of U
Number of boys Number of Girls
exceed that of Australia (in Rs. billion)?

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 esa) fdruk vf/d gS\ students (boys and girls) in school Y and Z togethe
and the total number of students (boys and girls) i
SSC CPO 03.10.2023 (Shift-3) school X and Y together?
(a) 2,000 (b) 1,440 LdwyY vkSjZ esa ,d lkFk feykdj Nk=kksa (yM+dksa
(c) 2,550 (d) 1,500 dh dqy la[;k vkSj LdwyX vkSjY esa ,d lkFk feykdj N
61. Study the given table and answer the question that (yM+dksa vkSj yM+fd;ksa) dh dqy la[;k ds chp fd
follows.
SSC CPO 04/10/2023 (Shift-01
nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA (a) 720 (b) 780
The table shows the number of employees in five (c) 820 (d) 840
different organisations A, B, C, D and E in different
departments. 63. The following table shows the mobile-addicte
population of four different cities Hyderabad, Mumba
rkfydk ikap vyx&vyx laxBuksa
A, B, C, D vkSjE esa fofHkUu Chennai, and Pune, and the ratio of males to female
foHkkxksa esa deZpkfj;ksa dh la[;k n'kkZrh gSA among them.
fuEu rkfydk pkj vyx&vyx 'kgjksa gSnjkckn] eqa

r
Dept 
A B C D E iq.ks dh eksckby&O;luh tula[;k vkSj muds chp iq#"

si
Org 
dk vuqikr n'kkZrh gSA
HR 1050 1015 976 888 1004

an by
City Mobile-Addicted population Male : Female
Finance 1017 960 786 1025 963 Hyderabad 2,20,000 7 : 8
Marketing 1382 1384 1275 1300 1290 Mumbai 3,50,000

n
Chennai 2,00,000 11 : 9
Production 1542 1545 4550 1570 1580 Pune 1,50,000 9 : 11
Accounts 786 745 801 800 735 Find the total number of addicted males in Chenna

ja
Legal 48 54 36 30 53
psUubZ esa O;luh iq#"kksa dh dqy la[;k Kkr djsa
R s
Total 5825 5703 5424 5613 5625
SSC CPO 04/10/2023 (Shift-01
a th
Which organisation has minimum number of (a) 1,20,000 (b) 1,00,000
employees?
(c) 80,0000 (d) 1,10,000
fdl laxBu esa deZpkfj;ksa dh la[;k U;wure gS\ 64. Study the given table and answer the question tha
ty a

follows.
SSC CPO 03.10.2023 (Shift-3)
nh xbZ rkfydk dk vè;;u dhft,] vkSj uhps fn, x, iz
(a) D (b) B
di M

mÙkj nhft,A
(c) C (d) E
The data given in the table is for the month o
62. Study the given pie-chart and answer the question December 2022.
that follows. fuEu rkfydk esa fn;k x;k MsVk fnlEcj 2022 eghus
fn, x, ikbZ&pkVZ dk vè;;u dhft, vkSj fuEufyf[kr iz'u dk Company Total no. of employees No. of female employees
mÙkj nhft,A P
Q
4560
4258
2210
1650
The pie-chart shows the number of students (boys R 3052 1280
S 4350 1920
and girls) studying in standard 10 of various schools
(X, Y, Z in the year 2022).
What is the difference between the number of ma
ikbZ&pkVZ] o"kZ 2022 esa fofHkUu
(X, Y, Z) Ldwyksa
ds d{kk 10 esa employees in company P and that in company R?
i<+us okys Nk=kksa (yM+ds vkSj yfM+fd;k¡) dh la[;k dks n'kkZrk gSAiq#"k deZpkfj;ksa dh la[;k] vkSj
daiuhP esa R esa
daiui
A

Total Number of Students (Boys and Girls) = 3600 deZpkfj;ksa dh la[;k ds chp dk varj fdruk gS\
Nk=kksa dh dqy la[;k (yM+ds vkSj =yM+fd;ka)
3600 SSC CPO 04/10/2023 (Shift-01
(a) 605 (b) 580
Z (c) 578 (d) 592
25%
65. Study the given graph and answer the question that follow
X
45% fn, x, xzkiQ dk vè;;u djsa vkSj fuEufyf•r ç'u dk m
The given graph represents the percentage of book
Y on different subjects that a person has in his library
30%
fn;k x;k xzkiQ ,d O;fÙkQ ds iqLrdky; esa fofH
iqLrdksa dk çfr'kr n'kkZrk gSA

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 fofHkUu fo"k;ksa ij iqLrdksa dk çfr'kr forj.k (yk[k esa) nh xbZ rkfydk esa n'kkZbZ xbZ gSA
Types of Batteries
Year
C1 4AH 7AH 32AH 35AH 55AH 55AH Total
10% 1992 75 144 114 102 108 108 543
1993 90 126 102 84 126 126 528
C4 C2 1994 96 114 75 105 135 135 525
40% 20% 1995 105 90 150 90 75 75 510
1996 90 75 135 75 90 90 465
1997 105 60 165 45 120 120 495
1998 115 85 160 100 145 145 605
C3
30%
Which type of battery had the highest total sales ove
the last seven years?
If the total number of books is 60, then what is the
ratio of the number of books on subject C1 to the fiNys lkr o"kZ esa fdl izdkj dh cSVjh dh dqy f
average number of books per subject? vf/d Fkh\
;fn iqLrdksa dh dqy la[;k 60 gS] rksC1fo"k;
ij iqLrdksa dh

r
SSC CPO 04/10/2023 (Shift-01
la[;k dk çfr fo"k; iqLrdksa dh vkSlr la[;k ls vuqikr D;k gS\ (a) 4 AH (b) 55 AH

si
SSC CPO 04/10/2023 (Shift-01) (c) 32 AH (d) 35 AH

an by
(a) 3 : 5 (b) 2 : 5 69. Study the given table and answer the question tha
(c) 2 : 3 (d) 3 : 7 follows.
nh xbZ rkfydk dk vè;;u djsa] vkSj uhps fn, x, ç'u dk

n
66. The table below shows the number of students
commuting to school by different modes of transport: The table shows the number of students studying i

ja
uhps nh xbZ rkfydk ifjogu ds fofHkUu lk/uksa }kjk Ldwy6 vkus
different classes of 6 different schools.
R s
okys fo|k£Fk;ksa dh la[;k dks n'kkZrh gS% fuEu rkfydk 6 vyx&vyx Ldwyksa dh 6 vyx&vyx
a th
Modes of transport School bus Bicycle Walking Other vehicles esa i<+us okys fo|kfFkZ;ksa dh la[;k dks n'kkZrh
No. of boys 80 280 130 200
No. of girls 140 190 90 250
Schools V VI VII VIII IX X
Which mode of transport is used by maximum number P 152 160 145 156 147 144
Q 148 166 150 155 157 143
ty a

of students? R 161 152 140 145 143 165

fo|k£Fk;ksa dh vf/dre la[;k }kjk ifjogu ds fdl lk/u dk S
T
159
147
142
144
149
158
140
163
142
154
168
150
mi;ksx fd;k tkrk gS\
di M

U 150 160 162 160 160 140

Total 917 924 904 919 903 910
SSC CPO 04/10/2023 (Shift-01)
Which class has the maximum number of student
(a) School bus (b) Other vehicles from all schools together?
(c) Bicycle (d) Walking lHkh Ldwyksa esa feykdj fdl d{kk esa fo|k£Fk;ksa dh
67. The given table shows the number of students from
4 schools during 2022-2023. SSC CPO 04/10/2023 (Shift-01
(a) V (b) VI
nh xbZ rkfydk 2022&2023 ds nkSjku 4 Ldwyksa ds fo|k£Fk;ksa dh
la[;k dks n'kkZrh gSA (c) VII (d) X
70. The pie chart I given below shows the runs score
Schools Total no. of students Boys Girls
by a player against 6 different countries durin
A 2060 1339 721
2019-2020.
A

B 1880 1034 846

C 2200 990 1210 uhps fn, x, ikbZ pkVZ 1 esa fdlh f•ykM+h }kjk o
D 1680 924 756 ds nkSjku 6 vyx&vyx ns'kksa ds fo#¼ cuk, x
What is the percentage of girls in school B? n'kkZ;k x;k gSA
LdwyB esa yM+fd;ksa dk çfr'kr fdruk gS\ Pie chart 2 shows the runs scored by the same playe
against the same 6 countries during 2020-2021
SSC CPO 04/10/2023 (Shift-01)
(a) 60% (b) 55% ikbZ pkVZ 2 esa mlh f•ykM+h }kjk o"kZ 2020&
(c) 45% (d) 65%
6 ns'kksa ds fo#¼ cuk, x, juksa dks n'kkZ;k x;k gS
68. The sales of batteries (in lakhs) manufactured by a Total runs scored during 2019-2020 = 4000.
company over the years are shown in the given table. 2019&2020 ds nkSjku cuk, x, dqy ju ¾ 4000

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 New that follows.
zealand
14%
Australia
20%
fn, x, ikbZ pkVZ dk vè;;u dhft, vkSj fuEufyf•r ç
mÙkj nhft,A
Sri Lanka
18% England The pie-chart shows the number of boys studying i
17%

South Africa
play group of various schools (A, B, C).

% n
16 ista
ikbZ pkVZ] fofHkUu(A,Ldwyksa
B, C) ds Iys xzqi esa i<+

15%
k
Pa yM+dksa dh la[;k dks n'kkZrk gSA
Total runs scored during 2020-2021 = 5000 Total number of boys = 2000
2020-2021 ds nkSjku cuk, x, dqy ju
= 5000 yM+dksa dh dqy la[;k
= 2000

New Australia
zealand 12% A
15% 13%

r
England
Sri Lanka 22% C B

si
20% 55% 32%
South Africa
% n
1 5 ista

an by
16%
k
Pa

n
Find the difference between the total runs scored
against New Zealand and Pakistan during 2019-2020 What is the average number of boys studying i

ja
and the total runs scored against Sri Lanka and schools A and C?
R s
Pakistan during 2020-2021.
LdwyA vkSjC esa i<+us okys yM+dksa dh vkSlr la[
o"kZ 2019&2020 ds nkSjku U;wthySaM vkSj ikfdLrku ds fo#¼ cuk,
a th
x, dqy juksa vkSj o"kZ 2020&2021 ds nkSjku Jhyadk vkSj ikfdLrku SSC CPO 04/10/2023 (Shift-02
ds fo#¼ cuk, x, juksa ds chp esa varj Kkr dhft,A (a) 580 (b) 680
SSC CPO 04/10/2023 (Shift-01) (c) 920 (d) 740
ty a

(a) 650 (b) 600 73. The table below shows the production of jute fabr
(c) 550 (d) 500 (in lakh metre) by two Companies P and Q durin
71. Study the given pie-chart and answer the question 2001 to 2005.
di M

that follows.
uhps nh xbZ rkfydk 2001 ls 2005 ds nkSjku nks
P vk
d
fn, x, ikbZ&pkVZ dk vè;;u djsa] vkSj fuEufyf•r ç'u dk mÙkj nsaA
Q }kjk fd, x, twV ds diM+s ds mRiknu (yk• ehV
The pie-chart shows the sale of different fruits in a
shop in a day. n'kkZrh gSA
fuEu ikbZ&pkVZ ,d nqdku esa ,d fnu esa fofHkUu iQyksa dh Year fcØh Company P Company Q
dks n'kkZrk gSA 2001 100 96
2002 120 130
Others 2003 108 100
20% Banana 2004 190 180
30% 2005 240 275
Orange In which year/years, did Company Q produce mor
A

15% Grapes fabric than Company P?

10%
Apple fdl o"kZ@lky esa daiuh
Q us daiuhP dh rqyuk esa v
25%
diM+s dk mRiknu fd;k\
If a total of 1200 kg of fruits were sold in a day, SSC CPO 04/10/2023 (Shift-02
calculate the amount of bananas sold (in kg). (a) 2002, 2005 (b) 2003, 2005
;fn ,d fnu esa dqy 1200kg iQy csps x,] rks csps x, dsyksa (c) 2005 (d) 2002
dh ek=kkkg( esa) dh x.kuk dhft,A 74. The following table shows the number of mobile phone
SSC CPO 04/10/2023 (Shift-02) sold by four different stores in four different years.
(a) 260 (b) 330 fuEu rkfydk pkj vyx&vyx o"kks± esa pkj vyx&
(c) 360 (d) 290 }kjk csps x, eksckby iQksu dh la[;k dks n'kkZrh g

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 Stores 2012 2013 2014 2015 xf.kr esa vuqÙkh.kZ Nk=kksa dh la[;k] vU; lHkh
P 178 183 205 225 Nk=kksa dh la[;k ls fdruh de gS\
Q 133 198 220 235 SSC CPO 04/10/2023 (Shift-02
R 284 250 220 198
(a) 95 (b) 256
S 225 243 199 205
(c) 190 (d) 161
Find the difference between the number of mobile 77. Study the given table and answer the question tha
phones sold by store Q in 2013 and store S in 2015. follows.
2013 esa LVksj
Q }kjk csps x, eksckby iQksu dh la[;k] vkSjnh xbZ rkfydk dk vè;;u dhft, vkSj uhps fn, x, ç'u
S }kjk csps x, eksckby iQksu dh la[;k ds chp mÙkj nhft,A
2015 esa LVksj
dk varj Kkr djsaA The table shows the number of people who responde
to a survey about their favourite style of music.
SSC CPO 04/10/2023 (Shift-02)
rkfydk mu yksxksa dh la[;k n'kkZrh gS ftUgksau
(a) 7 (b) 8 ilanhnk 'kSyh ds ckjs esa ,d loZs{k.k esa çfrfØ;k
(c) 12 (d) 10
75. The number of hours spent by a student on different Age 18 -20 21 -30 31+
Classical 5 5 15

r
activities as shown in the following pie chart:
Pop 5 11 10
fuEu ikbZ pkVZ esa ,d Nk=k }kjk fofHkUu xfrfof/;ksa esa fcrk, x,
Hip -Hop 6 9 8

si
?kaVksa dh la[;k dks fn•k;k x;k gS% Rock 9 7 13
Total 28 32 46

an by
Number of hours
What percentage of the total sample indicated tha
Rock is their favourite style of music (rounded off t
Others

n
2 2 decimal places)?
School
6
loZs{k.k fd, x, dqy yksxksa ds fdrus çfr'kr us la
fd mudh ilanhnk 'kSyh jkWd laxhr gS (nks n'ke

ja
Sleep k
R s
or
8 e
W iw.kkZafdr)\
m
Ho 5
a th
Play SSC CPO 04/10/2023 (Shift-02
3 (a) 29.65% (b) 8.25%
(c) 27.36% (d) 22.35%
The central angle corresponding to the sleeping time
78. Study the pie-chart given below, and answer th
ty a

is what percentage of the whole central angle?

question that follows.
lksus ds le; dk laxr dsaæh; dks.k] iwjs dsaæh; dks.k dk fdruk
uhps fn, x, ikbZ pkVZ dk vè;;u dhft,] vkSj fu
çfr'kr gS\ ç'u dk mÙkj nhft,A
di M

SSC CPO 04/10/2023 (Shift-02) Percentage-wise break up of students in terms o

specialisation in B Tech
(a) 33.33% (b) 50%
(c) 55% (d) 66.66%
chVsd esa fo'ks"kKrk ds lanHkZ esa Nk=kksa dk
76. The following pie chart shows the number of students Total Numbers of Students = 6700
who failed in different subjects in an examination. Nk=kksa dh dqy =la[;k
6700
fuEukafdr ikbZ pkVZ ,d ijh{kk esa fofHkUu fo"k;ksa esa vuqÙkh.kZ
Nk=kksa dh la[;k dks n'kkZrk gSA
IT
Examine the chart and answer the following question. 21% CSE
The total number of students who have failed is 350. 22%
pkVZ dk voyksdu dhft,] vkSj fuEufyf•r ç'u dk mÙkj nhft,A EEE
12%
vuqÙkh.kZ gksus okys Nk=kksa dh dqy la[;k 350 gSA ECE
A

10%
CIVIL
8% MECH
English 27%
Social
22%
21%
What is the total number of students havin
Chemistry Hindi 10% specialisation in CSE, ECE and CIVIL?
12%
s CSE, ECE vkSjCIVIL esa fo'ks"kKrk j•us okys Nk=
ic
ys
h
P 8 %
Mathematics
27% la[;k fdruh gS\
SSC CPO 04/10/2023 (Shift-02

The number of students failed in mathematics is less than (a) 2480 (b) 2860
the number of students failed in all other subjects by: (c) 2840 (d) 2680

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 of four towns below poverty line and the proportion percentage (correct to 2 decimal places) of th
of males and females: expenditure invested in Wheat and Bajra?
fuEu rkfydk xjhch js•k ls uhps ds pkj 'kgjksa dh çfr'kr tula[;k pkoy vkSj tkS esa fuos'k fd;k x;k O;; xsgwa vkSj
vkSj iq#"kksa ,oa efgykvksa dk vuqikr n'kkZrh gS% fd, x, O;; dk fdruk çfr'kr (2 n'keyo LFkkuksa rd lgh)
Town Percentage of population Below poverty line Above poverty line SSC CPO 04.10.2023 (Shift-3
below poverty line Male : Female Male : Female
A 10 1:3 3:2 (a) 206.66% (b) 260.66%
B 15 3:4 1:2 (c) 260.55% (d) 206.55%
C 20 2:3 5:3 s
D 25 4:5 5:4 82. The following bar graph shows the milk production (i
million tons) details for 4 years for three brands (P, Q, R
If the total population of town C is 8000 then what is the
number of females above the poverty line in state C? fuEufyf•r ckj xzkiQ rhu czkaMksa
(P, Q, R) ds fy, 4 o"kks± d
;fn 'kgj C dh dqy tula[;k 8000 gS] rks jkT;
C esa xjhch mRiknu (fefy;u Vu esa) dk fooj.k fn•krk gS%
js•k ls Åij okyh efgykvksa dh la[;k fdruh gS\ 60

50
SSC CPO 04/10/2023 (Shift-02) 50

Milk production (in tons)

r
(a) 2600 (b) 2400 40
40 40

(c) 200 (d) 2000 30 30 30 30

si
30
80. The table below shows the number of trees planted by 25 25
20 20 20
the government in 4 different years. 20

an by
uhps nh xbZ rkfydk 4 vyx&vyx o"kks± esa ljdkj }kjk yxk, x,
10
isM+ksa dh la[;k n'kkZrh gSA

n
0
2016 2017 2018 2019
Year Neem Oak Birch
Brand P Brand Q Brand R
2019 2000 3000 4000

ja
R s
2020 3000 4000 5000 What was the average production of milk (in millio
2021 5000 6000 7000 tons) per year over the period of 4 years?
a th
2022 7000 8000 9000 4 o"kks± dh vof/ esa çfr o"kZ nw/ dk vkSlr mRiknu (
Find the ratio between the number of neem trees D;k Fkk\
planted in the year 2021 and the number of birch SSC CPO 04.10.2023 (Shift-3
ty a

trees planted in the year 2022. (a) 50 (b) 30

o"kZ 2021 esa yxk, x, uhe ds isM+ksa dh la[;k vkSj o"kZ 2022 esa
(c) 70 (d) 90
yxk, x, cpZ isM+ksa dh la[;k ds chp vuqikr Kkr dhft,A 83. Study the given bar-graph and answer the questio
di M

SSC CPO 04.10.2023 (Shift-3) that follows.

(a) 3 : 7 (b) 3 : 8
fn, x, ckj&xzkiQ dk vè;;u djsa vkSj fuEufyf•r ç'u dk
The bar-graph displays the number of bicycle
(c) 5 : 8 (d) 5 : 9
produced at a factory from 1998 to 2002.
81. The given pie chart shows the distribution of land
under various food crops.
ckj&xzkiQ 1998 ls 2002 rd ,d dkj•kus esa mRikfnr
la[;k çnf'kZr djrk gSA
fn;k x;k ikbZ pkVZ fofHkUu •k| iQlyksa ds varxZr Hkwfe ds forj.k
dks n'kkZrk gSA 1400

1200
1200
1100
The expenditure invested on these food crops is 1000 900
800
Rs.25,000 per annum. Study the pie chart carefully 800
600
and answer the question that follows. 600
A

bu •k| iQlyksa ij fuos'k fd;k x;k O;; çfr o"kZ 25]000 #i;s gSA ikbZ 400

pkVZ dk è;kuiwoZd vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA

200

0
1998 1999 2000 2001 2002

What was the average number of bicycles produce

Barley
12% at the factory per year from 1998-2002?
Jwar
8%
Rice 1998&2002 rd çfr o"kZ dkj•kus esa mRikfnr lkbfd
Bajra
50%
la[;k D;k Fkh\
10%
SSC CPO 04.10.2023 (Shift-3
Wheat
20% (a) 940 (b) 920
(c) 880 (d) 960

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 purchased during the years. foHkkxksa esa deZpkfj;ksa dh la[;k n'kkZrh gSA
,d esxkLVksj esa] fuEufyf•r MsVk ls o"kks± ds nkSjku •jhnh xbZ bdkb;ksa
Dept 
dk irk pyrk gSA A B C D E
Org 
What is the percentage growth in total units purchased HR 1050 1015 976 888 1004
(of all the three products) from 2015 to 2020?
Finance 1017 960 786 1025 963
2015 ls 2020 rd •jhnh xbZ dqy bdkb;ksa (rhuksa mRiknksa dh) esa
Marketing 1382 1384 1275 1300 1290
çfr'kr o`f¼ D;k gS\ Production 1542 1545 4550 1570 1580
Accounts 786 745 801 800 735
Year Tea Coffee Cold drink Legal 48 54 36 30 53
2015 650 450 1008 Total 5825 5703 5424 5613 5625
2016 730 402 1092
Which department has a minimum number o
2017 710 354 770
employees in each organisation?
2018 570 438 812
çR;sd laxBu esa fdl foHkkx esa deZpkfj;ksa dh U;w

r
2019 820 390 1043
SSC CPO 04.10.2023 (Shift-3
2020 830 501 1330

si
(a) Production (b) Legal
SSC CPO 04.10.2023 (Shift-3) (c) Accounts (d) Marketing

an by
(a) 26.23 % (b) 21.68 % 87. Study the given graph and answer the question tha
follows.

n
(c) 22.75 % (d) 25.76 %
fn, x, xzkiQ dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙ
85. Study the given table and answer the question that The graph shows data regarding number of boys an

ja
follows. girls studying in class 5 in four different schools.
R s
nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA xzkiQ pkj vyx&vyx Ldwyksa esa d{kk 5 esa i<+u
yM+fd;ksa dh la[;k ls lacaf/r MsVk fn•krk gSA
a th
The table shows the percentage population of four
states below poverty line and proportion of male to 365
400 350
female. 350 280

rkfydk pkj jkT;ksa dh xjhch js•k ls uhps dh çfr'kr tula[;k vkSj 300 260 250
ty a

250 195

iq#"k ls efgyk dk vuqikr n'kkZrh gSA 200

150
153 165

100
di M

State Percentage of Proportion of male to female 50

population below Below poverty line Above poverty line 0
poverty line P Q R S
Schools
Male : Female Male : Female
Girls Boys
P 28 3:2 7:3
Q 15 4:7 6:5
What is the difference between total number of boy
R 24 5:3 3:3
studying in school P and Q together and the tota
S 13 2:5 1:3 number of girls in the same schools together?
LdwyP vkSjQ esa i<+us okys yM+dksa dh dqy la[;k v
If the total population of state R is 26,000, then what is esa i<+us okyh yM+fd;ksala[;k
dh ds
dqychp fdruk varj gS\
the number of females above poverty line in state R?
SSC CPO 04.10.2023 (Shift-3
;fn jkT; R dh dqy tula[;k 26]000 gS] rks jkT;
R esa xjhch js•k
(a) 40 (b) 35
A

ls Åij efgykvksa dh la[;k fdruh gS\

(c) 45 (d) 38
SSC CPO 04.10.2023 (Shift-3) 88. The monthly budget of an average household is give
(a) 9,560 (b) 9,950 in the following table.

(c) 9,880 (d) 9,780 ,d vkSlr ifjokj dk ekfld ctV fuEufyf•r rkfydk esa fn;k x;
86. Study the given table and answer the question that Category Food Clothing Rent Savings
follows. Amount 8000 2000 6000 4000
nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA The percentage of the monthly budget spent on clothing is
The table shows the number of employees in five
different organisations A, B, C, D and E in different
diM+ksa ij ekfld ctV •pZ dk çfr'kr gS%
departments. SSC CPO 04.10.2023 (Shift-3

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1

 (c) 8 % (d) 10% dh vk; dks n'kkZrh gS%
89. Study the given graph and answer the question that Sectors Fina Comm Produc Sales Transpo
nce unicati tion -rtation
follows. on
fn, x, xzkiQ dk vè;;u dhft, vkSj mlds ckn iwNs x, ç'u dk Amount
(in Lakhs
85 60 67 115 75

mÙkj nhft,A Rs)

The following graph shows the distribution of the cost What is the difference (in lakh Rs) between the incom
percentage of different items. from the maximum earning sector and the minimum
earning sector?
fuEukafdr xzkiQ fofHkUu oLrqvksa ds ykxr çfr'kr ds forj.k dks
vf/dre dekbZ okys {ks=k vkSj U;wure dekbZ ok
n'kkZrk gSA okyh vk; ds chp dk varj (yk[k :- esa) fdruk gS\
Cost percentage distribution of different items
SSC CPO 05/10/2023 (Shift-01
(a) 30 (b) 55
Mobile
Phones (c) 40 (d) 48
Laptop
30% 20% 92. The table below shows the number of laptops sol

r
by the 4 sellers and the ratio of Brand1 laptops sol
Camera
to Brand 2 laptops sold.

si
15%
Tablet uhps nh xbZ rkfydk 4 foØsrkvksa }kjk csps x, yS
TV'S
20% rFkk csps x, czkaM 1 ySiVkWi vkSj csps x, cz

an by
15%
vuqikr n'kkZrh gSA

n
Total laptops Brand 1 laptops sold:
If a person purchased all the items for Rs 3,00,000, Seller
then the cost of a TV is less than that of a laptop by: sold Brand 2 laptops sold
A 4000 3:1
;fn fdlh O;fDr us lHkh oLrq,¡ :i;s 300]000 esa [kjhnha] rks ,d

ja
R s B 6000 2:1
Vhoh dh ykxr] ySiVkWi ds ykxr dh rqyuk esa ------- de gSA C 8000 5:3
a th
SSC CPO 05/10/2023 (Shift-01) D 5000 1:4

(a) Rs 55,000 (b) Rs 45,000 The number of Brand1 laptops sold by A seller is ho
many more than the number of Brand 2 laptops sol
(c) Rs 50,000 (d) Rs 40,000
by B seller?
90. Study the given bar graph carefully to answer the
ty a

A foØsrk }kjk csps x, czkaM 1 ySiVkWi B dh

foØsrk
la[;k
question that follows.
csps x, czkaM 2 ySiVkWi dh la[;k ls fdruh vf/d gS
fn, x, ckj xzkiQ dk è;kuiwoZd vè;;u dhft, vkSj fuEufyf•r
di M

SSC CPO 05/10/2023 (Shift-01

ç'u dk mÙkj nhft,A
(a) 1200 (b) 1000
Marks obtained by four students in Mid-I and Mid-II.
(c) 600 (d) 800
feM&
I vkSj feM&
II esa pkj Nk=kksa }kjk çkIr vad 93. The pie chart below shows the spendings of a stat
40 on various sports during a particular year:
36 35 35
35 33
30 30
uhps fn, x, ikbZ pkVZ esa ,d fo'ks"k o"kZ ds nkS
30
25
ij jkT; }kjk fd, x, O;; dks n'kkZ;k x;k gS%
Marks

20 21
20
15
10 Cricket
Other
5 20%
30%
0
Sunil Hemanth Rehman Mohan
Students Tennis
A

Mid-I Mid-II 15%

Hockey
Marks obtain by Hemanth in Mid-I are what percentage 13%
Basket
of the total marks obtained by all the students in Mid-I? ball
12%
feM&I esa gsear }kjk çkIr vad] I feM&
esa lHkh Nk=kksa }kjk çkIr
dqy vadksa ds fdrus çfr'kr ds cjkcj gSa\ If the total amount spent on sport during the yea
was Rs 75,00,000 then how much (in Rs) was spen
SSC CPO 05/10/2023 (Shift-01) on cricket and hockey together?
(a) 15% (b) 30% ;fn o"kZ ds nkSjku [ksyksa ij O;; dh xbZ dqy jk
(c) 20% (d) 25% :i;s Fkh] rks fØdsV vkSj gkWdh ij dqy feykdj
(:- esa) O;; dh xbZ Fkh\
91. The following table shows the income of a company
in a year from various sectors: SSC CPO 05/10/2023 (Shift-01

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 fn, x, vkys• dk vè;;u dhft, vkSj uhps fn, x, ç'u dk
(c) Rs 24,75,000 (d) Rs 21,55,000 mÙkj nhft,A
94. Study the given table and answer the question that The graph shows data related to number of student
follows.
enrolled for a vocational course in two institute
nh xbZ rkfydk dk vè;;u dhft,] vkSj uhps fn, x, ç'u dk (X and Y) during three years.
mÙkj nhft, A vkys• rhu o"kks± ds nkSjku nksX laLFkkuksa
vkSjY) esa O;kolk
(
The table shows data of population and literacy/illiteracy ikBÔØe ds fy, ukekafdr Nk=kksa dh la[;k ls lacaf/r M
rate among the population in three different villages.
320
nh xbZ rkfydk rhu vyx&vyx xkaoksa dh tula[;k esa] tula[;k 350 320

Number of students
270
vkSj lk{kjrk @ fuj{kjrk nj ds vkadM+s n'kkZrh gSA 300
250
240
180 200
Respective Percentage Number of 200
ratio between of literates illiterates 150
Village number of (Males and (Males and 100
males and females) females) 50
females out of total out of total 0

r
2010 2011 2012
population population
Institute X
P 2:5 45 380

si
Institute Y
Q 3:7 65 270
R 1:3 80 150 What is the difference between the average numbe

an by
of students enrolled in institute X in 2010 and 201
The number of illiterates in village R is what percentage and that in institute Y in 2011 and 2012?
less than that in village P?

n
2010 vkSj 2011 esa laLFkku
X esa ukekafdr Nk=kksa dh
xkaoR esa fuj{kjksa dh la[;k]P xkao
esa fuj{kjksa dh la[;k ls
rFkk 2011 vkSj 2012 esa laLFkku
Y esa ukekafdr Nk=kks
fdrus çfr'kr de gS\

ja
la[;k ds chp fdruk varj gS\
R s
SSC CPO 05/10/2023 (Shift-01)
SSC CPO 05/10/2023 (Shift-01
a th
11 8 (a) 15 (b) 20
(a) 60 % (b) 60 %
19 19
(c) 18 (d) 22
9 10 97. The given pie-chart represents the survey report o
(c) 60 % (d) 60 %
19 19
ty a

favourite games of a group of young people.

95. Study the given table and answer the question that fn;k x;k ikbZ&pkVZ ;qok yksxkas ds ,d lewg ds
follows.
di M

losZ{k.k fjiksVZ dks n'kkZrk gSA

nh xbZ rkfydk dk vè;;u djsa] vkSj uhps fn, x, ç'u dk mÙkj nsaA
The table shows the number of students studying in Other
6 different classes of 6 different schools. 5%
Vo
ll e
fuEu rkfydk 6 vyx&vyx Ldwyksa dh 6 vyx&vyx d{kkvksa 17
yb a
ll
Football
% 30%
esa i<+us okys fo|kfFkZ;ksa dh la[;k dks n'kkZrh gSA Hockey
12%
Schools V VI VII VIII IX X
P 152 160 145 156 147 144 Cricket
Q 148 166 150 155 157 143 36%
R 161 152 140 145 143 165
S 159 142 149 140 142 168
T 147 144 158 163 154 150
A

If a total of 4980 people were surveyed, then wha

U 150 160 162 160 160 140 is the central angle produced by the sector indicatin
Total 917 924 904 919 903 910
football?
Which class has the minimum number of students ;fn dqy 4980 yksxksa dk loZs{k.k fd;k x;k] rks
from all schools together?
n'kkZus okys {ks=k }kjk fufeZr dsaæh; dks.k D;k
lHkh Ldwyksa esa feykdj fdl d{kk esa fo|kfFkZ;ksa dh la[;k U;wure gS\
SSC CPO 05/10/2023 (Shift-01
SSC CPO 05/10/2023 (Shift-01)
(a) 100º (b) 108º
(a) VII (b) IX (c) 105º (d) 90º
(c) X (d) VIII 98. In a survey, 200 respondents were asked whethe
96. Study the given graph and answer the question they owned a vehicle or not. Their responses ar
that follows. tabulated below.

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 okgu gS ;k ughaA mudh çfrfØ;k,¡ uhps lkj.khc¼ gSaA difference between sales of grapes and oranges (in kg)
What percentage of respondents do NOT own a car? ;fn ,d fnu esa dqy 1200 fdyksxzke iQy csps x,] rks
fdrus çfr'kr mÙkjnkrkvksa ds ikl dkj ugha gS\ larjs dh fcØh ds chp varj Kkr djsa (fdyks esa)
Men Women SSC CPO 05.10.2023 (Shift-2
(a) 60 (b) 45
Own Scooter 30 20
Vehicle (c) 65 (d) 55
Car 25 10
101. The following table shows the marks obtained out o
Both 15 5 100, by 5 students in 5 different subjects.
Do not own vehicle 50 45
fuEufyf•r rkfydk 5 vyx&vyx fo"k;ksa esa 5 Nk=kk
SSC CPO 05.10.2023 (Shift-2) ls çkIr vadksa dks n'kkZrh gSA
(a) 72.5% (b) 65% Students Hindi English Maths Science Social
(c) 68.5% (d) 75% Ravi 78 84 88 75 79
99. Study the given table chart carefully and answer the Abhishek 84 86 78 89 80

r
question that follows.
Ayush 88 80 90 82 85
fn, x, rkfydk pkVZ dk è;kuiwoZd vè;;u djsa vkSj fuEufyf•r

si
Rohit 72 85 88 89 78
ç'u dk mÙkj nsaA Aryan 86 88 77 79 89
The given table represents the number of employees

an by
in five different organisations, i.e., A, B, C, D and E What are the total marks scored by Ayush in all th
in 2019, 2020, 2021 and 2022. subjects?

n
nh xbZ rkfydk 2019] 2020] 2021 vkSj 2022 esa ikap vyx&vyx vk;q"k }kjk lHkh fo"k;ksa esa dqy fdrus vad çkIr fd
laxBuksa] A,
;kuh
B, C, D vkSjE esa deZpkfj;ksa dh la[;k n'kkZrh gSA
SSC CPO 05.10.2023 (Shift-2

ja
Organisation
R s Years (a) 445 (b) 435
2019 2020 2021 2022
(c) 465 (d) 425
a th
A 4000 7800 2600 5200
B 3800 9200 9100 7300
102. Study the given table and answer the question tha
follows.
C 4500 5500 7800 3250
D 7000 3800 8000 4650
nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk m
ty a

E 6600 6200 7600 2880 The table shows the number of students studying i
six different classes of six different schools.
The ratio of the number of employees in 2019 in rkfydk Ng vyx&vyx Ldwyksa dh Ng vyx&vyx d
di M

organisations D and E taken together to that in 2021

i<+us okys Nk=kksa dh la[;k n'kkZrh gSA
in the same organisations is:
2019 esa laxBu
D vkSjE esa deZpkfj;ksa dh la[;k dk 2021 esa
School Class V Class VI Class VII Class VIII Class IX Class X

P 152 160 145 156 147 144

leku laxBuksa esa deZpkfj;ksa dh la[;k ls vuqikr gS% Q 148 166 150 155 157 143

SSC CPO 05.10.2023 (Shift-2) R 161 152 140 145 143 165

S 159 142 149 140 142 168

(a) 34 : 39 (b) 39 : 34
T 147 144 158 163 154 150
(c) 14 : 9 (d) 9 : 14
U 150 160 162 160 161 140
100. Study the given pie-chart and answer the question Total 917 924 904 919 904 910
that follows.
fn, x, ikbZ&pkVZ dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA
The number of students studying in class VII from
The pie-char shows the sale of different fruits in a shop school U is what percentage of the total number o
A

students from all the classes together from that a

in a day.
school (rounded off to 2 digits after decimal)?
ikbZ&pkVZ
,d fnu esa ,d nqdku esa fofHkUu iQyksa dh fcØh dks n'kkZrk gSA
LdwyU esa d{kkVII esa i<+us okys Nk=kksa dh la[;k
d{kkvksa esa i<+us okys Nk=kksa dh dqy la[;k d
Others (n'keyo ds ckn 2 vadksa rd iw.kkZafdr)\
20%
Banana
30% SSC CPO 05.10.2023 (Shift-2
Orange
15%
(a) 16.48% (b) 15.63%
Gr a
pe
10% s (c) 17.36% (d) 18.25%
Apple
25% 103. The following table shows the data of 10th grad
students of Portland High school.

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 fn•krh gSA (c) B (d) D
Boys Girls 106. Study the following pie chart and answer the belo
Enrolled in Science 25 20 question:
Not enrolled in Science 20 30 fuEufyf•r ikbZ pkVZ dk vè;;u djsa vkSj uhps fn,
What is the approximate percentage of the 10th grade mÙkj nsa%
boys of Portland High school who are NOT enrolled in Total estimated yearly expenses = 3,00,000
Science?
iksVZySaM gkbZ Ldwy ds 10oha d{kk ds yM+dksa dk vuqekfur çfr'kr Others
D;k gS tks foKku esa ukekafdr ugha gSa\ 10%
Food items
SSC CPO 05.10.2023 (Shift-2) 25%
Savings
(a) 33.33% (b) 55.55% 20%
(c) 22.22% (d) 44.44% Travel
8%
104. Study the following table carefully and answer the Rent
questions based on it. 20% Education

r
17%
fuEufyf•r rkfydk dk è;kuiwoZd vè;;u djsa vkSj ml ij vk/kfjr
ç'uksa ds mÙkj nsaA

si
The following table shows the domestic sales of vehicles
of four manufactures from 2010 to 2015. How much per cent is more expense on rent than th

an by
fuEufyf•r rkfydk 2010 ls 2015 rd pkj fuekZrkvksa ds okguksaexpense
dh on others?
?kjsyw fcØh n'kkZrh gSA fdjk, ij gksus okyk •pZ vU; ij gksus okys •pZ ls fdr

n
Manufacturer 2010 2011 2012 2013 2014 2015 vf/d gS\
A 560000 580000 600000 620000 650000 680000
SSC CPO 05.10.2023 (Shift-2

ja
B 540000 590000 570000 630000 710000 550000
R s
C 610000 580000 620000 680000 690000 630000 (a) 150% (b) 10%
D 630000 570000 700000 640000
a th
(c) 50% (d) 100%
With respect to which of the following combinations,
107. The following table gives the number of different type
is the sales of vehicles lowest over the given period?
of batteries sold by a company over the year
fuEufyf•r esa ls fdl la;kstu ds laca/ esa] nh xbZ vof/ esa okguksa
(numbers in hundreds).
dh fcØh lcls de gS\
ty a

fuEu rkfydk fiNys dqN o"kks± esa fdlh daiuh }kjk

SSC CPO 05.10.2023 (Shift-2) çdkj dh cSVfj;ksa dh la[;k (lSdM+ksa esa la[;k) nsrh
di M

(a) 2010 (b) 2015

Type of battery 7AH 32AH 35AH 55AH
(c) 2011 (d) 2010
Years
105. Study the given table and answer the question that
follows. 2018 20 30 40 50
nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA 2019 40 35 45 45
2020 50 40 40 40
The table shows the number of employees in five
different organisations A, B, C, D and E in different 2021 40 50 50 45
departments. 2022 45 45 35 35
rkfydk ikap vyx&vyx laxBuksa
A, B, C, D vkSjE esa fofHkUu What is the difference in the number of 7AH batterie
foHkkxksa esa deZpkfj;ksa dh la[;k n'kkZrh gSA sold in 2021 and 2022?
Dept 
A B C D E 2021 vkSj 2022 esa csph
7AHxbZ
cSVfj;ksa dh la[;k esa D;k
Org 
SSC CPO 05.10.2023 (Shift-3
A

HR 1050 1015 976 888 1004

Finance 1017 960 786 1025 963 (a) 300 (b) 200
Marketing 1382 1384 1275 1300 1290 (c) 500 (d) 700
Production 1542 1545 4550 1570 1580 108. Study the given pie-chart and answer the questio
Accounts 786 745 801 800 735 that follows.
Legal 48 54 36 30 53 fn, x, ikbZ&pkVZ dk vè;;u djsa vkSj fuEufyf•r ç'u dk
Total 5825 5703 5424 5613 5625 The pie-chart shows the expenditure of a family o
Which organisation has maximum number of different items and their savings throughout th
employees? year 2021.
fdl laxBu esa deZpkfj;ksa dh la[;k lcls vf/d gS\ ikbZ&pkVZ o"kZ 2021 ds nkSjku fofHkUu enksa
SSC CPO 05.10.2023 (Shift-2) vkSj mudh cpr dks n'kkZrk gSA

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 Saving sold by stores A and B together in January and th
Other Transport
20%
15%
5% total number of scooters sold by stores D and
together in March?
Education
of Children tuojh esa LVksj
A vkSjB }kjk csps x, LdwVjksa dh dqy
Food
23%
12%
ekpZ esa LVksj
D vkSjE }kjk csps x, LdwVjksa dh dqy la
Housing
vuqikr Kkr dhft,\

10 h in g
15%

%
t
Clo SSC CPO 05.10.2023 (Shift-3
If the total income for the year was Rs. 1,00,000, then (a) 18 : 23 (b) 19 : 25
the difference in the expenses (in Rs.)between housing (c) 17 : 25 (d) 20 : 23
and transport was: 111. The given graph shows the income and expenditur
;fn o"kZ ds fy, dqy vk; 1]00]000 #i;s Fkh] rks vkokl vkSj of a family. Study the graph and answer the questio
ifjogu ds chp •pks± esa varj (#i;s esa) Fkk% that follows.

SSC CPO 05.10.2023 (Shift-3) fn;k x;k xzkiQ ,d ifjokj dh vk; vkSj O;; dks n'kkZ
dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA

r
(a) 9,500 (b) 10,500

si
(c) 9,000 (d) 10,000 140
120
120 110
109. Study the given bar graph carefully and answer the

Rs. in Thousands
100
100

an by
90 90
question that follows. Increment of four employees (in 80
80 70 70
Rs.) in the months of May, June and July.
60
fn, x, ckj xzkiQ dk è;kuiwoZd vè;;u djsa vkSj fuEufyf•r ç'u
50

n
40
40
dk mÙkj nsaA ebZ] twu vkSj tqykbZ ekg esa pkj deZpkfj;ksa dh20 osru

o`f¼ (#i;s esa)A

ja
0
R s 2015 2016 2017 2018 2019
90 Years
80
80 73 75
a th
70 70 Income Expenditure
70 65
INCREMENT

60
60 53 55 55
50 45
The expenditure of the family increased by _____ from
40
40 2017 to 2018.
30
2017 ls 2018 rd ifjokj dk •pZ_____ c<+ x;kA
ty a

20
10
0
SSC CPO 05.10.2023 (Shift-3
di M

Pretam Kamal Ajit Ramesh (a) 25% (b) 22%

EMPOLYEES
May June July (c) 20% (d) 11%
112. Study the given table and answer the question tha
What is Kamal's overall average increment in three follows.
months? nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk m
rhu eghuksa esa dey dh dqy vkSlr osru o`f¼ D;k gS\ The table shows the number of employees in thre
SSC CPO 05.10.2023 (Shift-3) organisations during three years.
(a) Rs. 60 (b) Rs. 75 rkfydk rhu o"kks± ds nkSjku rhu laxBuksa esa deZpkfj;
(c) Rs. 70 (d) Rs. 65 Year 2018 2019 2020
110. The following table represents the number of scooters
sold by 5 stores in 4 different months in the year 2022. Organization
A

fuEufyf•r rkfydk o"kZ 2022 esa 4 vyx&vyx eghuksa esa 5 nqdkuksa P 75 69 85

Q 96 81 74
}kjk csps x, LdwVjksa dh la[;k n'kkZrh gSA
R 85 63 90
Stores  What is the respective ratio between total number o
A B C D E
Months  employees in organisations P and R together in 201
and total number of employees in organisations Q an
January 212 168 173 182 190 R together in 2020?
February 312 250 212 175 182 2019 esa laxBu
P vkSjR esa deZpkfj;ksa dh dqy la[;k v
March 280 275 265 255 245 esa laxBu
Q vkSjR esa deZpkfj;ksa dh dqy la[;k ds ch
vuqikr D;k gS\
April 270 280 290 305 275
SSC CPO 05.10.2023 (Shift-3

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 The average number of children per family is:
(c) 23 : 41 (d) 33 : 41 çfr ifjokj cPpksa dh vkSlr la[;k gS%
113. Study the given table and answer the question that
SSC CPO 05.10.2023 (Shift-3
follows.
nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsaA (a) 1.72 (b) 1.49

The table shows the number of students in various (c) 1.23 (d) 1.59
courses in various cities A, B, C, D and E in India 115. Study the given pie-chart carefully and answer th
during the academic year 2022-2023. question that follows.
rkfydk 'kS{kf.kd o"kZ 2022&2023 ds nkSjku Hkkjr ds
A, fofHkUufn, 'kgjksa
x, ikbZ&pkVZ dk è;kuiwoZd vè;;u djsa vkSj fu
B, C, D vkSjE esa fofHkUu ikBÔØeksa esa Nk=kksa dh la[;k n'kkZrh gSA nsaA
dk mÙkj
City MBA Science Engineering Medicine MCA Total
Degree wise Break-up of Expenditure of a Family in
A 1200 2560 4520 1100 960 10340
B 1350 2650 5100 1050 850 11000
Month.
C 1250 2640 4580 1120 750 10340 ,d eghus esa ,d ifjokj ds O;; dk fMxzhokj C;kSjkA

r
D 1960 1250 3500 1850 1600 10160
E 1200 2100 3500 960 540 8300 Total amount spent in a month = Rs. 56800.

si
During the academic year 2022-2023, which city had ,d eghus dk dqy O;;= 56800 #.
the largest number of students?

an by
'kS{kf.kd o"kZ 2022&2023 ds nkSjku fdl 'kgj esa Nk=kksa dh la[;k Others Education
lcls vf/d Fkh\ 35.8°

n
of children
SSC CPO 05.10.2023 (Shift-3) Savings 95°

ja
(a) B (b) D 65.3°
R s
(c) A (d) C Housing
Transport
a th
114. Study the given table and answer the question that 79.2°
follows. 84.7°
nh xbZ rkfydk dk vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsa
ty a

Number of children Number of families What is the amount spent by the family on housing?
0 4 ifjokj }kjk vkokl ij fdruh jkf'k •pZ dh xbZ gS\
di M

1 22
SSC CPO 05.10.2023 (Shift-3
2 15
3 6 (a) Rs. 11,814 (b) Rs. 12,496
4 2 (c) Rs. 10,615 (d) Rs. 6,248
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 Answer Key
1.(a) 2.(b) 3.(b) 4.(d) 5.(d) 6.(a) 7.(a) 8.(d) 9.(c) 10.(a)

11.(c) 12.(a) 13.(d) 14.(d) 15.(b) 16.(a) 17.(a) 18.(a) 19.(b) 20.(c)

21.(b) 22.(a) 23.(a) 24.(c) 25.(a) 26.(a) 27.(a) 28.(d) 29.(d) 30.(a)

31.(c) 32.(b) 33.(a) 34.(c) 35.(a) 36.(d) 37.(d) 38.(c) 39.(d) 40.(d)

41.(b) 42.(b) 43.(b) 44.(b) 45.(a) 46.(d) 47.(b) 48.(b) 49.(b) 50.(a)

r
51.(a) 52.(a) 53.(c) 54.(b) 55.(c) 56.(a) 57.(c) 58.(c) 59.(d) 60.(a)

si
61.(c) 62.(a) 63.(d) 64.(c) 65.(b) 66.(c) 67.(c) 68.(c) 69.(b) 70.(c)

an by
71.(c) 72.(b) 73.(a) 74.(a) 75.(a) 76.(d) 77.(c) 78.(d) 79.(b) 80.(d)

n
ja
81.(a) 82.(d) 83.(b) 84.(a) 85.(c) 86.(b) 87.(b) 88.(d) 89.(b) 90.(d)
R s
a th
91.(b) 92.(b) 93.(c) 94.(d) 95.(b) 96.(b) 97.(b) 98.(a) 99.(a) 100.(a)

101.(d) 102.(c) 103.(d) 104.(d) 105.(a) 106.(d) 107.(c) 108.(d) 109.(d) 110.(b)
ty a

111.(a) 112.(d) 113.(a) 114.(d) 115.(b)

di M
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 SOLUTIONS
1. (a) 7. (a)
Sweets manufactured by P,Q,R togather in 1990
50  20
20% of 50   10 = 178 + 240 + 150 = 568
100
Sweets manufactured by S,T,U in 1998
Number of student who got less than 10 marks = 130 + 177 + 192 = 499
= 100 – 87 = 13 Ratio = 568 : 499
2. (d) 8. (d)
In years 2019 – 20 Statement (d) is correct.
Total cost = Rs.6500 Rice production by B in 2017 is equal to the ric
production by A in 2020.
1
Assumed profit = 25% = 9. (c)
4
Difference between of laptops sold of company A and C
ATQ,

r
32
5  (40 – 8)% of 2500 =  2500 = 800
 6500  = Rs.8125 100

si
4 10. (a)
3. (b) Total marks = 50

an by
Population of Village D in 2020 is 10,500 60% marks of total marks
ATQ, 60

n
8%  10500 = 50  = 30 marks
100
10500 Total number of student who got more then 30 marks

ja
1% 
8 = 33
R s
Hence, the number of student eligible to purasue highe
10500 studies in mathematics is 33.
a th
Now the Population of Village A =  12
8 11. (c)
= 15750 Students placed in all companies in 2014 = 65.7 × 100
Hence the population of Village A in 2020 is = 15750 Students placed in all companies in 2012 = 57.5 × 100
Difference = (65.7 – 57.5) × 1000 = 8200
ty a

4. (d)
Exp. Incurred on health = 15000 12. (a)
Total Exp. = Electiricity + Rant + Entertain + Education Donation = 35%
di M

+ Health % of Fund used from donation as a scholarship

= 5000 + 20,000 + 2000 + 8000 + 15000 = 50, 000 26
=  100 = 74.29%
15000 35
Now, the angle of Exp. on health  50,000  360 = 108° 13. (d)
5. (d) Sum of Angela = 192
Sum of Maria = 206.57
IIT Bombay + IIT Madras = 2080 – 1180 = 900
Sum of Deepika = 195.07
1 4  IIT Bombay Sum of Preeti = 226.80
20%  – 
5 5  IIT Madras Here sum of Preeti is highest, so Preeti was the best playe
14. (d)
IIT Bombay IIT Madras In school (A) = 900 × 30% = 270
:
4 + 5

A

270
9 Male = × 7 = 126
15
9 units  900
270
1 units  100 Female = × 8 = 144
15
No of students in IIT Bombay = 4 × 100 = 400
6. (a) In School (D) = 800 × 18% = 144
Average sales of all companies 144
Male = × 5 = 60
 25  30  55  70   1000 12
=
4 144
Female = × 7 = 84
 180  12
   1000 = 45000 Total male = (126 + 60) = 186
 4 

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 Male : Female Avg quantity of item C during is month together
186 228
(280295280315)
31 : 38 =
4
15. (b)
1170
Total number of plate = 3260  = 292.5 kg
4
% of plates manufactured by B = 60% – 20% = 40%
20. (c)
According to question,
Average marks of A
100%  3260
847195 250
= = = 83.33
3260 3 3
1% 
100 Average marks of B

3260 93  76  101 270

40%  × 40% = 1304 = = = 90

r
100% 3 3
Hence the number of plates are manufactured by B is Average marks of C

si
1304.
99  79  71 249
= = = 83

an by
16. (a) 3 3
In 2019, sales of xyz phones = 18 Average marks of D

n
In 2020, sales of xyz phones = 40
80  90  94 264
Diff. = 40 – 18 = 22 = = = 88
3 3

ja
R s
22 Average marks of E
Required% =  100 = 122.22%
18
a th
88  93  90 271
= = = 90.33
17. (a) 3 3
In 2017, Production of x = 20 Option (c) is correct.
In 2018, Production of x = 20 21. (b)
ty a

Average production of x in 2017 & 2018 First five months average wage
di M

9500  9400  9600  8000  8500

= 20  20  20 (In 1000) =
5
2
45000
Similarly, = = 9000
5
In 2019, Production of y = 32
22. (a)
In 2020, Production of y = 20
 100% = 360°
Average production y in 2019 & 2020
360
Minimum (10%) =  10 = 36°
32  20 100
= = 26 (In 1000)
2 23. (a)
Diff. = 26 – 20 = 6 × 1000 = 6000 (In Delhi )
Hence, the diff. between Average production of x & y BSNL : Airtel
A

in 6000 = 10 : 20 = 1 : 2
18. (a) 24. (c)
Total marks = 500 Total expenditure = 3,00,000
Marks obtain by Hitansh's % of material cost = 15%
= 67 + 70 + 64 + 55 + 75 = 331 % of selling Exp. = 28

Overall% Total = 15 + 28 + 43%

ATQ,
331
  100 = 66.22% 100%  30,0000
500
300000
Hence, the percentage marks obtain by Hitansh's is 43%   43
66.22%. 100 = 1,29,000

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 The number of companies with more demand than MP = 20%
production = P, Q and R
the number of companies with more production than Literate Illeterate
demand = S and T . 
1 : 4 = × 20% = 16%
5
Hence, the Ratio = 3 : 2 Goa = 12%
Literate Illeterate 20.8%
26. (a)

Bluetooth sale in 2017 = 49.7 Lakhs 3 : 2 = × 12% = 4.8%
5
Bluetooth sale in 2020 = 84 Lakhs
20.8
Increment = 84 – 49.7 = 34.3 Lakh = 315000 × 20.8% = 315000 × = 65520
100
34.3
% increment =  100% 34. (c)
49.7
Total marks = 3000
49
=  100% = 69% Sector of Reasoning = 75°
71

r
27. (a) ATQ,
360°  3000

si
Difference = Donation – Government agencies
= 35% – 12% = 23% 3000
1 

an by
 23% × 10,00000  Rs.230000 360

28. (d) 300

n
75   75  625
P : Q 360
Hence, the marks of reasoning 625.

ja
 3276 : 4550
R s
1638 : 2275 35. (a)
Wheat imports in 1970  3465
a th
1638
%=  100% = 72% Wheat imports in 1971  1600
2275
Diff. = 3465 – 1600
29. (d) = 1865
ty a

Degree of the sector of student who appered in section

E = 58° 1865
%=  100 = 53.82  54%
3465
di M

58
Required% of appered student = ×100 = 16.1% 36. (d)
360
Required number of students =
30. (a)
(60  72  54  45) 231   80   78 
Coolers sold =   100  120   100  135    1000
4 4  
(27  18  30  205) 95  (96 + 105.3) × 1000
Fans sold = 
4 4  201300
231 95 136 37. (d)
Difference = –  = 34
4 4 4 6
31. (c) Failed candidates in city D = × 2.35 = 1.41 lakhs
10
A

Difference b/w the total production of cotton and khadi 38. (c)
 (118 – 114) + (120 – 120) + (125 – 121) Children in P and Q together:-
=4+0+4=8 20 30
  10,000   20,000
32. (b) 100 100
3600 = 2000 + 6000 = 8000
Avg student enrolled from college D =
5 39. (d)
3400 Total students in 2010 = 200
Avg student who passed from college D =
5 Students with 2nd division = 60
3600 3400 60
Ratio  : = 18 : 17 Required% =  100 = 30%
5 5 200

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2

 Total Expenditure = Products sold by A on Thursday = 3000 – (2300)
5425 + 5250 + 6125 + 7525 + 6475 + 4200 = 35000 = 700
Expenditure on education = 5425 Products sold by B on Tuesday = 3500 – (3000)
360 = 500
Required angle = × 5425
35000 Required sale = 700 – 500 = 200 less
90  4  25  217 9  217 46. (d)

35000 35 ATQ,
9  31 2 6 + 62 = 68% 850
=
5 2 850
Total expenditure = 100%  × 100 = 1250
558 68
= = 55.8°
10 47. (b)
41. (b)
15 5
 100   100  83%

r
Maths  100%  360° Required% =
18 6
360

si
20%  × 20% 48. (b)
100
1342

an by
Science 100  360° Required% = × 100
1666
360

n
30%  × 30% 671
100  100 = 80.55%
=
833
360 360

ja
Now, required ratio = × 20% : × 30% 49. (b)
R s
100 100
20 85  80  75  70
a th
= =2:3 Required average =
30 4
42. (b)
310
6.5 – 5.2 = = 77.5
 100  25% 4
In 2003 =
ty a

5.2
77.5
7.8 – 6.5 Now, required percentage = × 100 = 77.5%
 100 = 20% 100
di M

In 2004 =
6.5
50. (a)
9.9 – 7.8
In 2005 =  100  27% 45  65
9.8 Required percentage = × 100
360
10.8 – 9.9
In 2006 = × 100 = 9.09% 1100 275 5
9.9   30 %
=
36 9 9
11.4 – 9.5
In 2008 = × 100 = 20% 51. (a)
9.5
Clearly from given table, Car S Shows continuou
 Max increase in 2005
increase
43. (b)
52. (a)
3
A

A during March  20 12  3 Clearly from given table, Car R shows continuou

 5  decrease.
A during April 8 40
 25 53. (c)
15
Total cakes by Q and S in September = 189 + 154
9
=  9 : 10 = 343
10
Total cakes by P and R in November = 168 + 168 = 33
44. (b)
 difference = 343 – 336 = 7
8200 – 8000
Required% =  100 54. (b)
8000
From given table average percentage of passin
20000 20 5 students is least in college A.
    2.5%
8000 8 2 = 68 + 72 + 78 + 80 = 298

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 In 2015, highest earning is of R, ie, 68. The highest total sale is battery 32 AH
56. (a) 114 + 102 + 75 + 150 + 135 + 165 + 160 = 901
Clearly, from given table wheat has continuous 69. (b)
increase.
Clearly from table maximum number of students ar
57. (c)
in Class VI = 160 + 166 + 152 + 142 + 144 + 16
12% of total expenditure spent on education = 924
12 70. (c)
Now, the required sum = × 100000 = 12000
100
Runs in New Zealand and Pakistan = 14% + 16% = 30%
58. (c)
30
boys in B 30 = × 4000 = 1200
Required percentage = 100 ×   100 100
girl in E 55
Runs agains Srilanka and Pakistan = 20% + 15% = 35%
= 54.54  55%
59. (d) 35
= × 5000 = 1750

r
100
206  326 532 76 38
Required ratio = = = =  Difference = 1750 – 1200 = 550

si
350  420 770 110 55
60. (a) 71. (c)

an by
UK : Australia = 125° – 75° = 50° 30
Bananas sold = × 1200 = 360
50 100

n
 × 14400 = 2000
360 72. (b)
61. (c)  55  13  2000 68000

ja
According to given graph minimum employees in C Avg. of A and C =   ×  = 680
R s 2 100 100
organisation.
73. (a)
a th
62. (a)
Students in Y and Z = 55% P Q
Student in X and Y = 75% 2002  120 130 (more)
 Difference = 75 – 55 = 20% 2005  205 240 (more)
ty a

3600 From table, it is clear that company Q produce mor

 20 = 720
100 than company P in 2002 and 2005.
di M

63. (d) 74. (a)

Addicated males in Channai:- Required difference = S – Q = 205 – 198 = 7
11 75. (a)
= × 2,00,000 = 110000
20
Total = 24 h
64. (c)
Sleep = 8 hour
Male in P = 4560 – 2210 = 2350
Male in R = 3052 – 1280 = 1772 8
Required percentage = × 100 = 33.33%
24
 Required difference = 2350 – 1772 = 578
76. (d)
65. (b)
Required number of students are:-
10
books in C1 100  60 73 – 27
A

4 2
 =  × 350
avg. books 60 10 5 100
4
46
66. (c) = × 350 = 23 × 7 = 161
100
From given table:-
77. (c)
Bicycle has maximum students = 280 + 190 = 470
67. (c) Total people = 28 + 32 + 46 = 106

846 9  7  13
Required% = × 100 Required% = × 100
1880 106
4230 2115 29
=  = 45% = × 100 = 27.36%
94 47 106

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 22  10  8 Difference = 30% – 15% = 15%
Required number of students = × 6700 = 2680
100
15
79. (b) Required amount = × 300000 = Rs.45000
100
3 80 90. (d)
Required females =  × 8000 = 2400
8 100
30
80. (d) Required% = × 100
33  30  36  21
5000 5
Required ratio = 
9000 9 30
  100 = 25%
81. (a) 120

50  12 62 91. (b)
Required% = = × 100 Required difference = 115 – 60 = 55 lakh
20  10 30
92. (b)
620
= = 206.66%

r
3 3 1
Required number = × 4000 – × 6000
82. (d) 4 3

si
= 3000 – 2000 = 1000
40  50  30  20  25  30 

an by
40  30  20  20  30  25 93. (c)
Average production of milk 
4 20  13

n
Required amount = × 7500000
360 100
  90
4 = 33 × 75000 = 24,75,000

ja
83. (b)
R s
94. (d)
800  600  900  1100  1200
a th
Required average = 380 – 150
5 Required% =  100
380
4600
=  920 230 1150 10
5 =  100 =  60 %
ty a

380 19 19
84. (a)
95. (b)
830  501  1330 – 650  450  1008  100
di M

Required% = Clearly, from the graph class IX has manimum no. o

650  450  1008 students.
2661 – 2108 96. (b)
=  100
2108 Required difference:-
55300 320  240 320  200
= = 26.23% –
2108 2 2
85. (c) = 280 – 260 = 20
Required female:-
97. (b)
3 76
=   26000 = 9880 30
6 100 Required angle = × 360° = 108°
100
86. (b)
A

98. (a)
Clearly, from graph leagal department has least
number of employess in each organisation. Total respondents = 200
87. (b) Those who own a car = 25 + 15 + 10 + 5 = 55
Required difference = (280 + 365) – (350 + 260) Those who not own a car = 200 – 55 = 145
= 645 – 610 = 35 145
88. (d) Required% = × 100 = 72.5%
200
Total expenditure = 8000 + 2000 + 6000 + 4000
99. (a)
= 20000
Clothing = 2000 Required ratio:-

2000 7000  6600 13600 34

Required% = × 100 = 10%  
20,000 8000  7600 15600 = 39

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 Required difference:- Overall avg. increament of Kamal:-
(15 – 10)% = 5% of 1200 kg 65  55  75 195
 = Rs.65
3 3
1
= × 1200 = 60 kg 110. (b)
20
101. (d) 212  168 380
Required ratio = =  19 : 25
Required marks = 88 + 80 + 90 + 82 + 85 = 425 255  245 500

102. (c) 111. (a)

2017 expenditure  80
162  100
Required% = = 17.36% 2018  expenditure  100
933
100 – 80
103. (d) Required% = × 100 = 25%
80
20 112. (d)
Required percentage = × 100 = 44.44%

r
45 69  63 132 33
Required ratio = = =
74  90 164

si
104. (d) 41
From table it is clear that, lowest sale combination is:- 113. (a)

an by
B, 2010 Clearly, from table required city is:- 'B'
105. (a) 114. (d)

n
From the table it is clear that A has maximum employees. Required average:-
106. (d)
4  0   22  1  15  2

ja
R s
20 – 10 10  6  3  4  2
Required% =   100 = 100% 
10 10 4  22  15  6  2
a th
107. (c) 22  30  18  8 78
=  = 1.59
Required difference = (45 – 40) × 100 = 500 49 49
115. (b)
ty a

108. (d)
Required difference = (15 – 5)% = 10% of 100000 79.2
Required amount = × 56800 = 12496
360
di M

= 10,000
A

Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3

 DATA INTERPRETATION
(BAR GRAPH)
Direction (1-3) :Number of male and female 3. What is the ratio of number of males
members in different organizations A, B, C, D working in organizations C, D and E taken
and E are given in the bar graph. together to that of females working in
fuEu ckj xzkiQA,esaB, C, D vkSjE fofHkUu laxBuksa organizations A, B and C taken together?
esa iq:"k vkSj efgyk lnL;ksa dh la[;k nh xbZ gSA laxBuksa
C, D vkSjE eas dk;Zjr iq:"kksa dh dqy la
vkSj laxBuksa
A, B vkSjC esa dk;Zjr efgykvksa d
dqy la[;k ds chp vuqikr Kkr djsaA
SSC CGL 2020

r
(a) 11 : 10 (b) 10 : 11

si
(c) 46 : 49 (d) 49 : 46
4. The given bar graph shows the imports and

an by exports (in crore Rs.) of streel for 5 years

from 2014 to 2018.

n
fn, x;k naM vkjs[k 2014 ls 2018 rd 5 o"kZ e
bLikr dk vk;kr vkSj fu;kZr (djksM+ esa) n'kkZr
ja
R s
1. What is the ratio of average number of
a th

females in all the five organizations to the

average number of males in all the five
organizations?
ty a

lHkh ikap laxBuksa eas efgykvksa dh vkSlr la[;k dh

lHkh ikap laxBuksa eas iq:"kksa dh vkSlr la[;k dk vuqikr
di M

Kkr djsaA
SSC CGL 2020
(a) 49 : 51
What is the ratio of average export to
(b) 46 : 49
average import over the five years?
(c) 49 : 46
ikap o"kkZsa esa vkSlr fu;kZr vkSj vkSlr vk;kr dk
(d) 51 : 49 Kkr djsaA
2. For which organisation, difference between SSC CGL 2020
the number of males and the average (a) 109 : 247
A

number of females of all the organisation (b) 218 : 247

is minimum? (c) 247 : 109
(d) 247 : 218
fdl laLFkku eas iq:"kksa dh la[;k vkSj lHkh laLFkkuksa esa
efgykvksa dh vkSlr la[;k ds chp varj U;wure gS\Direction (5-6) : Cough syrup of three different
flavors - A, B and C (in lakh bottles)
SSC CGL 2020
manufactured by a medicine company over a
(a) C (b) B period of five years from 2010 to 2014 has been
(c) D (d) A shown in the bar graph.
 ckj xzkiQ esa 2010 ls 2014 rd ikap o"kZ dh vof/ esa
fdlh nok daiuh }kjk fufeZr
A, B vkSjC rhu fHkUu &fHkUu
Loknksa ds diQ fljiQ dh ek=kk (yk[k ckssry esa) n'kkZbZ xbZ gSA

7. What is the ratio of the total numbers of

students enrolled in institutes B in 2015
and 2017 to that of students enrolled in
5. The ratio of the average production of all institute A in 2014 and 2016?
flavors in 2012 to the difference of the 2015 vkSj 2017 esa laLFkku
B esa ukekafdr Nk=kk
dqy la[;k vkSj 2014 vkSj 2016 esa laLFkku
A esa

r
average production of flower A in 2012,
2013 and 2014 and the average production ukekafdr Nk=kksa dh dqy la[;k dk vuqikr Kkr d

si
of flavor C in 2012, 2013 and 2014 is : SSC CGL 2020
2012 esa lHkh Loknksa ds vkSlr mRiknu dk 2012] 2013
vkSj 2014 esa Lokn an by
A ds vkSlr mRiknu vkSj 2012]
(a) 92 : 137
(b) 91 : 111

n
2013 vkSj 2014 esa Lokn C ds vkSlr mRiknu ds (c) 111 : 81
varj ls vuqikr Kkr djsaA (d) 137 : 92

ja 8. The total number of students enrolled in

R s
SSC CGL 2020
institute B during 2014, 2016 and 2018 is
a th

(a) 26 : 15 what percent of the total number of

(b) 15 : 26 students enrolled in institute A during the
(c) 11 : 3 five Years?
ty a

2014] 2016 vkSj 2018 ds nkSjku laLFkku

B esa ukekafd
(d) 3 : 11
Nk=kksa dh dqy la[;k] ikap o"kksZa ds nkSjku
A esa
di M

6. Production of flavor A in 2012 is what ukekafdr Nk=kksa dh dqy la[;k dk fdruk izfr'kr g
percent less than the average production SSC CGL 2020
of flavor B in all the years (correct to 2 (a) 49%
decimal places)? (b) 66%
2012 ls LoknA dk mRiknu lHkh o"kksZa esa Lokn (Ýysoj)
(c) 57%
B ds vkSlr mRiknu ls fdruk izfr'kr de gS\ (d) 55%
(n'keyo) ds nks LFkkuksa rd lgh 9. The average number of students enrolled
SSC CGL 2020 in institute A during 2014, 2016 and 2018
is what percent less than the number of
(a) 3.87 (b) 6.98
students enrolled in institute B during
A

(c) 5.66 (d) 4.66

2017 (correct to two decimal places)?
Direction (7- 9): The bar graph shows the number 2014] 2016 vkSj 2018 ds nkSjku laLFkku
A esa ukekafd
of students enrolled for a science course in Nk=kksa dh vkSlr la[;k 2017 ds nkSjku BlaLFk
esa
institutes A and B during 5 years from 2014 to ukekafdr Nk=kksa dh la[;k ls fdrus izfr'kr de (B
2018. nks n'keyo LFkku rd) gS\
fuEu ckj xzkiQ 2014 ls 2018 rd 5 o"kZ ds nkSjku
A SSC CGL 2020
vkSjB laLFkkuksa esa foKku ikB~;Øe ds fy, ukekafdr Nk=kksa
(a) 22.46% (b) 29.17%
dh la[;k dks n'kkZrk gSA
(c) 26.15% (d) 32.75%
10. The following Bar Graphs represent the 11. The following Bar Graph represents the
Export of Tea (in lakh tonnes) by two Production of Fertilizers by Company A and
companies A and B during the years 2010 Company B (in 10000 tonnes) Over the
to 2015. Years from 2015 to 2010. The X-axis
fuEu ckj xzkiQ] nks daifu;ksa
A vkSjB }kjk o"kZ 2010 represents the years, and the Y-axis
ls 2015 rd pk; ds fu;kZr dks (yk[k Vu esa) fu:fir represents the Production of Fertilizers (in
djrk gSAX- v{k o"kksZa dks fu:fir djrk gS Y- vkSj 10000 tonnes).
v{k fu;kZr dh xbZ pk; dh ek=kk (yk[k Vu esa) dks fuEukafdr ckj xzkiQ dk vè;;u djsa vkSj mlds c
fu:fir djrk gSA fn, x, iz'u dk mÙkj nsaA\
Study the chart and answer the question ckj xzkiQ] daiuh
A dks daiuhB }kjk 2005 ls 2010
written below: rd moZjdksa ds mRiknu dks (10000 Vu esa fu
pkVZ dk vè;;u djsa vkSj mlds ckn fn, x, iz'u djrk gSAX v{k o"kksZa dks fu:fir djrk gS YvkS
dk mÙkj nsaA v{k moZjdksa ds mRiknu (10000 Vu eas) dks
djrk gSA

r
(Note : The data shown below is only for
(Note : The data shown below is only for

si
mathematical exercise. They do not
mathematical exercise. They do not
represent the actual figures).

an by
(uksV% fn;k x;k MsVk dsoy xf.krh; vH;kl ds fy,
represent the actual figures of the
companies)
gSA os daifu;ksa ds okLrfod vkadM+ksa dks fu:fir ugha

n
djrs gSA) (uksV% fn;k x;k MsVk dsoy xf.krh; vH;kl ds fy, gS
daifu;ksa ds okLrfod vkadM+ksa dks fu:fir ugha djrs

ja
Export of Tea (in lakh tonnes) by two
R s
companies A and B during 2010 to 2015
a th
ty a
di M

What is the average production (in 10000

tonnes) of fertilizer in 2008, 2009 and
2010 of Company A?

What is the ratio of the total exports of

2008] 2009 vkSj 2010 esa] daiuh
A ds moZjdksa d

company B in 2011 and 2014 to the total

vkSlr mRiknu (10000 Vu esa) Kkr djsaA
exports of company A in 2012 and 2015? SSC CGL 2020

2011 vkSj 2014 esa daiuh

B ds dqy fu;kZr dk 2012 (a) 590 (b) 600
A

vkSj 2015 esa daiuh

A ds dqy fu;kZr ls vuqikr Kkr djsaA (c) 620 (d) 570
SSC CGL 2020 12. The following bar graph shows the demand
(a) 55 : 68 and production (in Lakhs) of motor cycles
of five different companies A, B, C, D and
(b) 29 : 37
E in 2020.
(c) 68 : 55
fuEu ckj xzkiQ 2020 A, esaB, C, D vkSjE ikap
(d) 37 : 29 fofHkUu daifu;ksa dh eksVj lkbfdyksa dh ekax vk
dks (yk[kksa esa) n'kkZrk gSA
 2013 ls 2015 rd dPps eky dh [kjhn ds fy,
fuosf'kr dqy jkf'k vkSj 2014] 2016 vkSj 2017
rS;kj eky dh dqy fcØh dk vuqikr Kkr djsaA
SSC CGL 2020
(a) 56 : 27 (b) 64 : 37
(c) 37 : 64 (d) 27 : 56
14. The bar graph given below shows the sales
of Newspapers (in lakh number) from six
What is the ratio of total production of motor
branches of a Media Publication Company
cycles of companies A, B, C, D and E, to that
during two consecutive years 2017 and
of the total demand of motor cycles of all
2018.
the companies during the five years?
uhps fn;k x;k ckj xzkiQ 2017 vkSj 2018 nks Øe
daiuh A, B, C, D vkSjE ds eksVj lkbfdyksa ds dqy
o"kZ ds nkSjku fdlh ehfM;k izdk'ku daiuh dh
mRiknu vkSj ikap o"kZ ds nkSjku lHkh daifu;ksa ds eksVj

r
'kk[kkvksa ls lekpkj i=kksa dh fcØh dks (yk[
lkbfdyksa dh dqy ekax dk vuqikr Kkr djsaA

si
la[;k esa) n'kkZrk gSA (uksV % uhps n'kkZ;k x
SSC CGL 2020 dsoy xf.krh; vH;kl ds fy, gSA os okLrfod vkadM
(a) 50:53
an by (b) 53:50 dks fu:fir ugha djrs gSa)

n
(c) 60:61 (d) 61:60 Sales of Newspapers (in thousand numbers)
13. The following bar graph shows the amount from Six Branches - P, Q, R, S T and U of a

ja Media Publishing Company in 2017 and 2018

R s
(in Lakh Rs.) invested by a Company in
purchasing raw material over the years and
a th

the values (in Lakh Rs.) of finished goods

sold by the Company over the years
fuEu ckj xzkiQ o"kksZa esa fdlh daiuh }kjk dPps eky dh
ty a

[kjhn esa fuosf'kr jkf'k (:i;s yk[k esa) vkSj o"kksZa esa
daiuh }kjk csps x, rS;kj eky ds ewY;ksa (:i;s yk[k esa)
di M

dks n'kkZrk gSA

Total Sales of U for both the years is what

percent (correct to one place of decimal)
of the combined Sales of the branches Q
and R for 2017 and 2018?
nksuksa o"kZ esaU dh
'kk[kk
dqy fcØh 2017 vkSj 2018
esa 'kk[kk
Q vkSjR dh la;qDr fcØh dk fdruk izfr'kr
gS\ (Bhd ,d n'keyo LFkku rd)
A

SSC CGL 2020

(a) 41.0% (b) 48.6%

The ratio of total amount invested for (c) 67.1% (d) 44.4%
purchasing raw material from 2013 to 2015 Direction (15-17): Study the given graph and
to the total sales of finished goods in 2014, answer the question that follows.
2016 and 2017 is: fuEukafdr xzkiQ dk vè;;u djsa vkSj mlds ckn f
x, iz'u dk mÙkj nsaA
 Direction (18-20): Study the given graph and an-
swer the question that follows.
fuEukafdr xzkiQ dk vè;;u djsa vkSj mlds ckn fn, x
iz'u dk mÙkj nsaA

15. The percentage increase in the sale of mo-

torcycles in 2020 as compared to that in
2019 is below 15% for the dealer:
Mhyj ------ ds fy, 2020 eas eksVjlkbfdyksa dh fcØh esa
18. The average exports of country ABC in
gqbZ izfr'kr o`f¼ 2019 dh rqyuk esa 15» ls de gSA
2015,2017 and 2018 are what percentage

r
Selection Post - Phase IX (08 Feb 2022) of the total imports from 2015 to 2018

si
(a) A (correct to one decimal place)?
(b) D 2015] 2017 vkSj 2018 esa ns'k
ABC dk vkSlr fu;kZr]
(c) E
an by 2015 ls 2018 rd ds dqy vk;kr dk fdruk izfr'kr
gS (n'keyo ds ,d LFkku rd lgh)\

n
(d) B
Selection Post - Phase IX (08 Feb 2022)
16.

ja
The number of motorcycles sold by dealer
R s
D in 2020 is what percentage more than (a) 20.6%
a th

the number of motorcycles sold by dealer (b) 22.8%

C in 2019 ? (c) 22.1%
2020 esa MhyjD }kjk csph xbZ eksVjlkbfdyksa dh (d) 24.4%
ty a

la[;k 2019 esa Mhyj C }kjk csph xbZ eksVjlkbfdyksa

19. In 2020, if the imports increased from
dh la[;k ls fdrus izfr'kr vf/d gS\ 2019 by the same percentage as in 2019
di M

Selection Post - Phase IX (09 Feb 2022) over its previous year, then what were the
amount of imports (in Rs crores) in 2020?
(a) 48% (b) 44%
2020 esa] ;fn 2019 dh rqyuk es vk;kr esa of¼ 201
(c) 30.5% (d) 33.3%
esa fiNys o"kZ dh rqyuk ds leku izfr'kr o`f¼ gq
17. The average number of motorcycles sold by 2020 esa vk;kr dh jkf'k (:i;s djksM+ esa) D;k Fk
dealers B and D in 2019 is what percent-
Selection Post - Phase IX (09 Feb 2022)
age less than the average number of mo-
(a) 450 (b) 500
torcycles sold by dealers A, C and E in
2020 (correct to one decimal place)? (c) 420 (d) 480

2019 esa MhyjB vkSjD }kjk csph xbZ eksVjlkbfdyksa 20. The exports of country ABC in 2019 are
A

dh vkSlr la[;k] 2020 esa Mhyj A, C vkSjE }kjk what percentage more than its imports in
2017 ?
csph xbZ eksVjlkbfdyksa dh vkSlr la[;k ls fdrus izfr'kr
de gS (n'keyo ds ,d LFkku rd lgh)\ 2019 esa ns'k ABC dk fu;kZr] 2017 esa mlds vk;k
ls fdrus izfr'kr vf/d gS\
Selection Post - Phase IX (10 Feb 2022)
Selection Post - Phase IX (10 Feb 2022)
(a) 11.10% (b) 14.8%
(a) 40% (b) 28%
(c) 14.3% (d) 18.1%
(c) 12% (d) 30%
Direction (21-23): Study the given graph and 23. The average production of fertilisers by
answer the question that follows. country Z in 2017, 2018 and 2020 is what
fuEukafdr ikbZ pkVksaZ dk vè;;u djs vkSj mlds cknpercentage more than the average
production of fertilisers by country X in
fn, x, iz'u dk mÙkj nsaA
2018 and 2020?
2017] 2018 vkSj 2020 esa ns'k Z }kjk moZjdksa
vkSlr mRiknu] 2018 vkSj 2020 esaXns'k}kjk moZjdk
ds vkSlr mRikni ls fdruk izfr'kr vf/d gS\
SSC CGL MAINS 2020
(a) 48%
(b) 32.4%
(c) 49.6%
(d) 45%

r
Direction (24-26):Study the given graph and
answer the question that follows.

si
fuEukafdr xzkiQ dk vè;;u djsa vkSj mlds ckn f
21.

an by
The total production of fertilisers by
country Y in 2017 and 2019 and by
x, iz'u dk mÙkj nsaA

n
country X in 2016 is what percentage of
the total production of fertilisers by

ja
country Z in 2016, 2018 and 2020?
R s
2017 vkSj 2019 esa Y }kjk moZjdksa dk dqy mRiknu
a th

vkSj 2016 eas ns'k

X moZjdksa dk dqy mRiknu 2016]
2018 vkSj 2020 esa ns'k
Z }kjk moZjdksa ds dqy mRiknu
dk fdruk izfr'kr gS\
ty a

SSC CGL MAINS 2020

di M

(a) 77%
(b) 70%
24. In 2020, the production of cement by
(c) 69% company C increased by the same
(d) 60% percentage as in 2019, over its previous
22. What is the ratio of total production of year. The production (in million tonnes) of
fertilisers by country X in 2017 and cement by company C in 2020 (correct to
country Y in 2020 to the production of one decimal place) was:
fertilisers by country Z in 2019? daiuh C }kjk 2020 esa lhesaV ds mRiknu esa m
2017 esa ns'k Y }kjk moZjdksa dk izfr'kr o`f¼ dh xbZ ftruh fiNys o"kZ dh rqyuk
X vkSj 2020 esa ns'k
Z }kjk moZjdksa ds mRiknu 2019 esa dh xbZ FkhA 2020 esa
dqy mRiknu 2019 esa ns'k C daiuh
}kjk fd;k x;k
A

dk vuqikr Kkr djsaA lhesaV dk mRiknu (fefy;u Vu esa] n'keyo ds ,d

LFkku rd lgh) fdruk Fkk\
SSC CGL MAINS 2020
SSC CGL MAINS 2020
(a) 19 : 12
(a) 454.6
(b) 3 : 2
(b) 455.8
(c) 27 : 20
(c) 457.1
(d) 4 : 3
(d) 452.4
25. The ratio of the production of cement by 26. The average production of cement by
company A is 2016 and company C in 2018 company B in 2015, 2016 and 2018 is what
to the total production of cement by percentage less than the average
company B in 2017 and 2019 is: production of cement by company C in
daiuh A }kjk2016 esa vkSj daiuh
C }kjk2018 esa 2015 and 2017.
fd, x, lhesaV esa dqy mRiknu dk] daiuh
B }kjk daiuh B }kjk2015, 2016 vkSj2018 fd;k x;k
2017 vkSj2019 esa fd, x, lhesaV ds dqy mRiknu lhesaV dk vkSlr mRiknu] daiuh
C }kjk 2015 vkSj
ls vuqikr fdruk gS\ 2017 esa fd, x, lhesaV ds vkSlr mRiknu ls fdru
SSC CGL MAINS 2020 çfr'kr de gS\
(a) 9 : 8 SSC CGL MAINS 2020
(b) 7 : 6 3 1
(a) 7 % (b) 7 %
(c) 8 : 7 7 7
(d) 10 : 9
1 2
(c) 5 % (d) 6 %

r
3 3

si
an by
n
Answer Key
ja
R s
1.(b) 2.(d) 3.(b) 4.(d) 5.(c) 6.(c) 7.(d) 8.(b) 9.(c) 10.(a)
a th

11.(d) 12.(b) 13.(c) 14.(d) 15.(c) 16.(b) 17.(c) 18.(c) 19.(b) 20.(d)
ty a

21.(b) 22.(b) 23.(a) 24.(c) 25.(a) 26.(d)

di M
A
 DATA INTERTRETATION
(PIE CHART)
Eg1. Savings is what % of total income? 3. The paper cost is approximately what per
(cpr dqy vk; dk fdrus izfr'kr gS\) cent of the printing expenditure?
dkxt dh ykxr yxHkx NikbZ ds •pZ dk fdruk
Clothes Savings çfr'kr gS\
10% ?? (a) 25.2% (b) 30%
(c) 28.6% (d) 32.5%
Rent
20%
Food 4. If other expenditure is Rs 8000, then
24% binding expenditure is how much more
Education than royalty?

r
20% ;fn vU; O;; :i;s 8000 gS] rks ckbafMax O;; jkW;
ls fdruk vf/d gS\

si
Eg2. Amount spent on food is what % more than
the amount spent on education? (a) Rs.9750 (b) Rs.6500
(•kus ij O;;] f'k{kk ij O;; ls fdruk çfr'kr vf/d
gS\)
an by 5.
(c) Rs.8000 (d) Rs.6000
What is the central angle for the sector of

n
Directions (Q.1–5) : Study the following chart advertisement?
carefully and answer the questions given below foKkiu ds {ks=k ds fy, dsaæh; dks.k D;k gS\
it:

ja
(v/ksfyf•r xzkiQ dk lko/kuhiwoZd vè;;u djsa] vkSj uhps
(a) 90° (b) 45°
R s
(c) 76° (d) 64.8°
fn, x, ç'uksa dk mÙkj nsaA)
6. The given Pie-Chart shows the degree wise
a th

Expenditure incurred in publication and

sales of books breakup of expenditure of a family in a
(fdrkcksa ds çdk'ku vkSj fcØh dk O;;) month. Total income of a family is Rs. 43200.
fn;k x;k ikbZ vkjs[k] ,d eghus esa fdlh ifjokj d
Oth 4%
Paper
e rs

oy 10%

ty a

O;; dk fMxzh&okj fooj.k n'kkZrk gSA ifjokj dh

% y
15 a l t

vk; 43200 :i;s gSA

R

di M

Printing
35% Adverti-
sement
18%
Binding
18%
1. If printing expenses was 24500, then what
was the expenditure on royalty?
;fn eqæ.k O;; 24500 :i;s Fkk] rks jkW;YVh ij O;;
D;k Fkk\
(a) Rs 8750
(b) Rs 10500
(c) Rs 9500
A

The amount spent on food is what percent

(d) Rs 12000
2. The central angle of the sector of printing of the savings and miscellaneous
expenditure is how much larger than the expenses?
angle of the sector of advertisement [kk| ij [kpZ dh xbZ jkf'k] cpr vkSj iQqVdj O;;
expenditure?
dk fdruk izfr'kr gS\
eqæ.k O;; ds {ks=k dk dsaæh; dks.k foKkiu O;; ds
dsUæh; dks.k ls fdruk cM+k gS\ SSC CGL 2020
(a) 61.2° (b) 70° (a) 84% (b) 75%
(c) 54.8° (d) 72° (c) 90% (d) 60%
7. The pie chart shows the percentage 9. Pie-Chart shows the degree wise breakup
distribution of a total of 800 employees in of expenditure of a family in a month.
different departments of a company. Total income of the family is Rs 144000.
fn;k x;k ikbZ vkjs[k fdlh daiuh ds vyx&vyx fn;k x;k ikbZ vkjs[k] ,d eghus esa fdlh ifjokj d
foHkkxksa esa dqy 800 deZpkfj;ksa ds izfr'kr forj.k dks
O;; dk fMxzh&okj fooj.k n'kkZrk gSA ifjokj dh
n'kkZrk gSA vk; 144000 :i;s gSA

r
si
How many employees are working in the
field of marketing? What is the expenditure (in Rs) on

an by
ekdsZfVax ds {ks=k esa fdrus deZpkjh dk;Zjr gS\ education?
f'k{kk ij O;; (:i;s esa) Kkr djsaA

n
SSC CGL 2020
(a) 240 (b) 120 SSC CGL 2020
(c) 200
ja (d) 176 (a) 24,000 (b) 20,000
R s
8. The given Pie-Chart shows the degree wise (c) 12,000 (d) 36,000
a th

breakup of expenditure of a family in a 10. The pie graph shows the distribution of
month. Total income of a family is Rs. employees working in five departments A,
43200. B, C, D and E of a company. Total number
ty a

fn;k x;k ikbZ vkjs[k] ,d eghus esa fdlh ifjokj ds of employees = 900
O;; dk fMxzh&okj fooj.k n'kkZrk gSA ifjokj dh dqy ikbZ vkjs[k] fdlh daiuh ds
A, B, C, D vkSjE ikap
di M

vk; 43200 :i;s gSA foHkkxksa esa dk;Zjr deZpkfj;ksa ds (fMxzh o

dks n'kkZrk gSA deZpkfj;ksa dh dqy= 9000
la[;k
Distribution (degree wise) of the employ
yees working in five departments A,B, C,
D and E in a company
A

Expenditure on food is what percent more

than expenditure on rent?
fdjk, ij O;; dh rqyuk esa [kk| ij O;; fdruk
izfr'kr vf/d gS\ If the number of employees working in
SSC CGL 2020 department A is x and the total number of
(a) 200/3% (b) 50% employees working in departments C and
E is y, then the value of y-2x is:
(c) 50/3% (d) 100/3%
 ;fn foHkkxA esa dk;Zjr deZpkfj;ksa dh la[;k
x gS (a) 27.2 (b) 26.4
vkSj foHkkxC vkSjE esa dk;Zjr deZpkfj;ksa dh dqy (c) 26.9 (d) 25.8
la[;k y gS] rks
y – 2x dk eku Kkr djsaA 13. The ratio of the number of boys and girls
SSC CGL 2020 who appeared in the examination from
(a) 915 (b) 850 institutes S and T are 11:12 and 5:4,
(c) 725 (d) 1000 respectively. What is the difference
between the total number of boys who
Direction (11-13): Study the given pie chart and
appeared from institutes S and T and the
answer the question that follows.
total number of girls who appeared from
The pie chart shows the distribution
these two institutes?
(degree-wise) of the students who appeared in
the annual examination from institutes P, Q, R, laLFkkuS vkSjT ls ijh{kk esa mifLFkr yM+dk
S and T in 2020. The total number of students yM+fd;ksa dh la[;k dk vuqikr Øe'k% 11 % 12
who appeared is 3000. 5 % 4 gSA laLFkkuS vkSjT ls mifLFkr yM+dksa
fuEukafdr ikbZ pkVZ dk vè;;u djsa vkSj mlds ckn dqy la[;k vkSj bu nksuksa laLFkkuksa ls mifLF
fn, x, iz'u dk mÙkj nsaA ikbZ pkVZ] 2020 esaP,laLFkku dh dqy la[;k ds chp dk varj fdruk gS\

r
Q,
R, S vkSjT ls okf"kZd ijh{kk esa mifLFkr gksus okys Nk=kksa ds SSC PHASE IX 2022

si
forj.k (fMxzh ckj) dks n'kkZrk gSA mifLFkr Nk=kksa dh(a)dqy
24 (b) 20
la[;k 3000 gSA

an by (c) 23 (d) 17
Direction (14 - 15): Study the given graph and

n
answer the question that follows.
fn, x, xzkiQ dk vè;;u djsa vkSj uhps fn, x, ç'u dk
ja
R s
mÙkj nsaA
a th

Breakup (degreewise) of the number of

employees working in departments
A, B, C, D and E of company
ty a

11. The average number of students who E A

75.6° 93.6°
di M

appeared in the examination from the

institutes Q, S and T exceeds the number B
D
of students who appeared from institute R 64.8°
72°
by x. The value of x lies between: C
laLFkkuksa
Q, S vkSjT ls ijh{kk esa cSBus okys Nk=kksa dh 54°
vkSlr la[;k] laLFkkuR ls mifLFkr gksus okys Nk=kksa
14. Number of employees in Department B = 432
dh la[;k ls X vf/d gSA X dk eku buds chp gS%
The total number of employees working in
SSC PHASE IX 2022 offices A and E is what percentage more
(a) 9 to 14 (b) 14 to 18 than the total number of employees
(c) 23 to 27 (d) 18 to 22 working in offices B and C?
12. The number of students who appeared in dk;kZy;A vkSjE esa dk;Zjr deZpkfj;ksa dh dqy la[
A

the examination from institute P is what dk;kZy;B vkSjC esa dk;Zjr deZpkfj;ksa dh dqy la
per cent of the total number of students ls fdrus izfr'kr vf/d gS\
who appeared from the institutes Q, R and
SSC PHASE IX 2022
S (correct to one decimal place)?
laLFkkuP ls ijh{kk esa cSBus okys Nk=kksa dh la[;k] 14 14
laLFkkuksa
Q, R vkSjS (,d n'keyo LFkku rd lgh) (a) 42 % (b) 41 %
33 33
ls mifLFkr gksus okys Nk=kksa dh dqy la[;k dk fdruk
çfr'kr gS\ 14 14
(c) 40 % (d) 45 %
SSC PHASE IX 2022 33 33
15. If the ratio of the male and female (a) 18 : 23
employees working in department A is (b) 24 : 23
5:11, and 40% of the employees in
(c) 23 : 18
department C are females, then what is the
ratio of the number of female employees (d) 23 : 24
in department A to that of male employees 17. The following Pie chart represents the
in department C? percentage-wise distribution of 800
;fn foHkkxA esa dk;Zjr iq#"k vkSj efgyk deZpkfj;ksa students of class XII in a school in six
dk vuqikr 5 % 11 gS] vkSj foHkkx
C esa 40» deZpkjh different sections A, B, C, D, E and F.
efgyk,a gSa] rks foHkkx
A esa efgyk deZpkfj;ksa dh la[;k
fuEu ikbZ pkVZ
A, B, C, D, E vkSjF Ng fHkUu&fH
dk foHkkxC esaiq#"k deZpkfj;ksa dh la[;k ls vuqikr
oxksZaa esaXII
d{kk
ds 800 fo|kfFkZ;ksa ds izfr'kr o
fdruk gS\ forj.k dks fu:fir djrk gSA
SSC PHASE IX 2022
(a) 100 : 91 (b) 113 : 75
(c) 111 : 124 (d) 143 : 72

r
16. The following Pie chart represents the

si
percentage-wise distribution of 300
students of class X in a school in six

fuEu ikbZ pkVZ an by

different sections A, B, C, D, E and F.
A, B, C, D, E vkSjF N% fHkUu&fHkUu

n
oxksZaa esa fdlh fo|ky; esaXd{kk ds 300 fo|kfFkZ;ksa
ds izfr'kr okj forj.k dks fu:fir djrk gSA
ja
R s
a th

The table given below shows the number

ty a

of girls of class XII in six different sections

A, B, C, D, E and F.
di M

nh xbZ rkfydkA, B, C, D, E vkSjF Ng fHkUu&fH

oxksZa esaXII
d{kkdh yM+fd;ksa dh la[;k n'kkZrh g
The table given below shows the number
of boys of class X in six different section A B C D E F
A, B, C, D, E and F. 102 80 104 98 0 60
nh xbZ rkfydkA, B, C, D, E vkSjF N% fHkUu&fHkUuThe total number of girls in sections B, C
oxksZa esaXd{kk
ds yM+dksa dh la[;k dks n'kkZrh gSA and D together is what percent more than
A B C D E F the total number of boys in sections A, B
36 26 34 28 x 20 and D together?
If in section E, the ratio of the number of ,d lkFk oxZ A, B vkSjD eas yM+dksa dh dqy la
A

boys to the number of girls is 3 : 4, then dh rqyuk eas ,d lkFk oxZ B, C vkSjD esa yM+fd;
the ratio of number of boys in section E dh la[;k dqy fdrus izfr'kr vf/d gS\
to the number of girls in section C is: SSC CGL 2020
;fn oxZ E esa] yM+dksa dh la[;k vkSj yM+fd;ksa dh
(a) 76.25%
la[;k dk vuqikr 3 % 4 gS] rks oxZ
E esa yM+dksa dh
(b) 50%
la[;k vkSj oxZC esa yM+fd;ksa dh la[;k ds chp dk
vuqikr Kkr djsaA (c) 80%
SSC CGL 2020 (d) 65.75%
18. The following pie charts represent the
distribution of candidates who were
enrolled for a competitive examination,
and the candidates (out of those enrolled)
who passed the exam from five different
institutes P, Q, R, S and T.
fuEu ikbZ vkjs[k] ikap vyx&vyx laLFkkuksa
P, Q, R,
S vkSjT ls fdlh izfr;ksxh ijh{kk ds fy, ukekafdr
mEehnokjksa vkSj (ukekafdr esa ls) ijh{kk mÙkh.kZ djus
Figure (ii) Total number of candidates
okys mEehnokjksa ds forj.k dks fu:fir djrk gSA
passed the examination from five
institutes = 4000
Figure (ii) ikap fHkUu&fHkUu laLFkkuksa ls i
djus okys mEehnokjksa dh dqy=la[;k
4000

r
si
What is the ratio of the total number of

an by
candidates enrolled in institutes Q, R and
S together, to the number of candidates

n
passed from the institutes Q, R and S
together?

ja
R s
laLFkkuksa
Q, R vkSjS esa ukekafdr mEehnokjksa dh dqy
la[;k vkSj laLFkkuksa
Q, R vkSjS ls mÙkh.kZ mEehnokjksa
a th

What is the pass percentage for institute

dh dqy la[;k dk vuqikr Kkr djsaA
Q? (correct upto one decimal place.)
SSC CGL 2020
laLFkku
Q ls mÙkh.kZ gksus okys Nk=kksa dk
(a) 15 : 71
ty a

djsaA (n'keyo ds ,d LFkku rd)

(b) 71 : 15 SSC CGL 2020
di M

(c) 44 : 75 (a) 42.7% (b) 80%

(d) 75 : 44 (c) 48% (d) 71.1%
19. The following Pie charts represent the Directions (20-22):- Answer the questions
on the basis of the information given
distribution of candidates who were below:
enrolled for competitive examination and uhps nh xbZ tkudkjh ds vk/kj ij ç'uksa ds mÙkj
the candidates (out of those enrolled) who In the following pie-charts the percentage
passed the exam from five different of employees of a company working in 8
institutes P, Q, R, S and T. different countries has been given.
fuEufyf•r ikbZ&pkVZ esa 8 fofHkUu ns'kksa e
fuEu ikbZ pkVZ mu mEehnokjksa dk forj.k dks fu:fir djrkokyh daiuh ds deZpkfj;ksa dk çfr'kr fn;k x;k gSA
gS tks fdlh izfr;ksxh ijh{kk ds fy, ukekafdr Fks vkSj os
A

Total number of employees = 86000

mEehnokj (mu ukekafdrksa esa ls)P,ftUgksaus
Q, R, S vkSj
T ikap fHkUu&fHkUu laLFkkuksa ls ijh{kk mÙkh.kZ dh FkhA VIII
VII 8% I
Figure (i) Total number of candidates 7% 18%

enrolled in five different institutes = 7500 VI II

16% 12%
Figure (i) ikap fHkUu&fHkUu laLFkkuksa esa ukekafdr III
mEehnokjksa dh dqy la[;k
= 7500 V
17% IV
8%
14%
 Total number of male employees = 56000 Course Number of boys
VIII B.Sc Maths 40%
8% I
VII
15% B.Sc Physics 68%
9%
II B.Sc Chem 58%
VI 10%
7% B.Sc C.Sci 80%
III
V 12% B.Com 75%
18% IV
18% BBA 65%
What is the ratio of number of girls in B.Sc.
Note: Roman digits show the countries
20. What is the ratio between the number of Maths to number of boys in B.Sc. C.Sci.?
male employees and female employees in ch-,l lh xf.kr esa yM+fd;ksa dh la[;k vkSj ch-,ll
country II? dEI;wVj foKku esa yM+dksa dh la[;k dk vuqikr K
ns'kII esa iq#"k deZpkfj;ksa vkSj efgyk deZpkfj;ksa dh SSC CGL 2020
la[;k ds chp vuqikr D;k gS\ (a) 2 : 7 (b) 3 : 5
(a) 70 : 53 (b) 70 : 31

r
(c) 5 : 3 (d) 7 : 2
(c) 70 : 59 (d) 53 : 70
24. The pie-chart given below shows the

si
21. What is the approximate average number
of male employees in countries I, II and number of laptops in an office provided by
III?

an by
I] II vkSj III ns'kksa esa iq#"k deZpkfj;ksa dh
six different companies in the percentage
vuqekfur
of total number of laptops. The central

n
vkSlr la[;k D;k gS\ angles given in the pie chart are not
(a) 9670 (b) 6970 accurate to any scale.
(c) 6907
ja
(d) 6977
R s
uhps fn;k x;k ikbZ&pkVZ Ng vyx&vyx daifu;ks
22. By what percent is the total number of
miyC/ djk, x, ySiVkWi dh dqy la[;k ds çfr'kr
a th

employees in countries V, VI and VII more

than the number of male employees esa ,d dk;kZy; esa ySiVkWi dh la[;k dks n'kk
working in the countries II, III and IV? ikbZ pkVZ esa fn, x, dsaæh; dks.k fdlh Hkh iS
ns'kksa]
V] VI vkSjVII esa dqy deZpkfj;ksa dh la[;k lVhd ugha gSaA
ty a

II] III vkSj IV ns'kksa esa dk;Zjr iq#"k deZpkfj;ksa dh

la[;k ls fdrus çfr'kr vf/d gS\
di M

(a) 50.2% (b) 53.6%

(c) 55% (d) 48%
23. Pie-chart shows the distribution of
percentage of students in various courses.
ikbZ vkjs[k] fofHkUu ikB~;Øeksa esa Nk=kksa ds izfr'kr forj.k
dks n'kkZrk gSA
Total number of students is 1400
To what degree is the corresponding
BBA central angle (to one decimal place) of
13% B.Sc
laptops provided by company C6?
A

Maths
20%
B.Sc
daiuhC6 }kjk miyC/ djk, x, ySiVkWi dk laxr dsaæh
B. Com
20% Phy dks.k (,d n'keyo LFkku rd) fdl fMxzh rd gS\
10%
B.Sc SSC MTS- 2 August 2019 (Afternoon)
Chem
12%
(a) 51.7°
B.Sc C.Sci
25% (b) 45.1°
(c) 39.6°
Percentage-Wise distribution of number of
(d) 52.8°
boys:/yM+dksa dh la[;k dk izfr'kr okj forj.k%
25. The pie-chart given below shows the num- If the total number of laptops in office is
ber of laptops in an office provided by six 3800, then find the difference between the
different companies in the percentage of laptops of company C2 and C5 ?
total number of laptops. The central angles ;fn dk;kZy; esa ySiVkWi dh dqy la[;k 3800 gS
given in the pie chart are not accurate to daiuhC2 vkSjC5 ds ySiVkWi ds chp dk varj Kk
any scale.
dhft,\
uhps fn;k x;k ikbZ&pkVZ Ng vyx&vyx daifu;ksa }kjk SSC MTS- 2 August 2019 (Afternoon)
miyC/ djk, x, ySiVkWi dh dqy la[;k ds çfr'kr
(a) 382 (b) 362
esa ,d dk;kZy; esa ySiVkWi dh la[;k dks n'kkZrk gSA
(c) 342 (d) 322
ikbZ pkVZ esa fn, x, dsaæh; dks.k fdlh Hkh iSekus ij
27. The given pie chart shows the percentage
lVhd ugha gSaA
of students enrolled i nto the colleges A, B,
C, D, E and F in a c ity, and the table shows
the ratio of boys to girls in the college.
fn;k x;k ikbZ pkVZ 'kgj esa dkWyst

r
A, B, C, D, E
vkSjF esa ukekafdr Nk=kksa dk çfr'kr fn•krk gS

si
rkfydk dkWyst esa yM+dksa vkSj yM+fd;ksa

an by dks n'kkZrh gSA

n
What is the ratio of the number of laptops
of company C1 to those of company C3 ?

ja
daiuh C1 ds ySiVkWi dh la[;k dk daiuh
C3 ds ySiVkWi dh
R s
la[;k ls vuqikr fdruk gS\
a th

SSC MTS - 2 August 2019 (Afternoon)

(a) 2 : 1 (b) 12 : 11
(c) 2 : 3 (d) 6 : 11
ty a

26. The pie-chart given below shows the num-

di M

ber of laptops in an office provided by six A 9 : 4

different companies in the percentage of B 5 : 9
total number of laptops. The central angles C 3 : 4
given in the pie chart are not accurate to D 7 : 2
any scale. E 1 : 4
uhps fn;k x;k ikbZ&pkVZ Ng vyx&vyx daifu;ksa }kjk F 3 : 2
miyC/ djk, x, ySiVkWi dh dqy la[;k ds çfr'kr
Based on this information, if the total
esa ,d dk;kZy; esa ySiVkWi dh la[;k dks n'kkZrk gSA
number of students is 9800, then the
ikbZ pkVZ esa fn, x, dsaæh; dks.k fdlh Hkh iSekus ij
number of girls in the college B is:
lVhd ugha gSaA
bl tkudkjh ds vk/kj ij] ;fn Nk=kksa dh dqy la[;k
A

9800 gS] rks dkWyst

B esa yM+fd;ksa dh la[;k gS%
CHSL 18-03-2020 (Morning shift)
(a) 560
(b) 280
(c) 1008
(d) 504
28. Study the pie-chart and the table carefully
and answer the questions that follow. The
percentage distribution of lecturers in five
different subjects in a university is shown
in the pie-chart. The total n umber of lec-
turers is 500.
ikbZ&pkVZ vkSj rkfydk dk è;kuiwoZd vè;;u djsa vkSj
uhps fn, x, ç'uksa ds mÙkj nsaA ,d fo'ofo|ky; esa Ratio of male to female lecturers:
ikap vyx&vyx fo"k;ksa esa O;k[;krkvksa dk çfr'kr iq#"k ls efgyk O;k[;krkvksa dk vuqikr%
forj.k ikbZ&pkVZ esa fn•k;k x;k gSA O;k[;krkvksa dh
Lecturers Male : Female
dqy la[;k 500 gSA
Mathematics 7: 3
Physics 2: 3
Chemistry 4: 1

r
Botany 3: 5
Zoology 2: 5

si
Find the n umber of male lecturers in

an by Physics
HkkSfrdh esa iq#"k O;k[;krkvksa dh la[;k Kkr d

n
Ratio of male to female lecturers: CHSL 15-10-2020 (Morning shift)
iq#"k ls efgyk O;k[;krkvksa dk vuqikr%

ja (a) 42
R s
Lecturers Male : Female (b) 46
Mathematics 7:3
a th

(c) 44
Physics 2:3
(d) 40
Chemistry 4:1
30. The given pie chart shows the percentage
Botany 3:5
ty a

of students enrolled for the courses, A, B,

Zoology 2:5
C, D and E in a univesity and the table
What is the difference in the number of
di M

shows the percentage of students that

female lecturers in chemistry and
passed, out of the enrolled students.
Mathematics?
jlk;u foKku vkSj xf.kr esa efgyk O;k[;krkvksa dh fn;k x;k ikbZ pkVZ ,d fo'ofo|ky; esa ikBÔØe
A, B, C, D, E vkSjF ds fy, ukekafdr Nk=kksa
la[;k esa fdruk varj gS\
CHSL 19-10-2020 (Morning shift)
çfr'kr n'kkZrk gS vkSj rkfydk ukekafdr Nk=kk
(a) 30 (b) 22
mÙkh.kZ Nk=kksa dk çfr'kr n'kkZrh gSA
(c) 20 (d) 25
29. Study the pie-chart and the table carefully
and answer the questions that follow. The
A

percentage distribution of lecturers in five

different subjects in a university is shown
in the pie-chart. The total n umber of
lecturers is 500.
ikbZ&pkVZ vkSj rkfydk dk è;kuiwoZd vè;;u djsa vkSj
uhps fn, x, ç'uksa ds mÙkj nsaA ,d fo'ofo|ky; esa
ikap vyx&vyx fo"k;ksa esa O;k[;krkvksa dk çfr'kr
forj.k ikbZ&pkVZ esa fn•k;k x;k gSA O;k[;krkvksa dh
dqy la[;k 500 gSA
 32. If the population below poverty line in
Courses % passed out village D is 42,000, then the total
A 76 population in all the six villages taken
B 82 together is:
C 80 ;fn xk¡o D esa xjhch js•k ds uhps dh tula[;k 42]000
D 90 gS] rks lHkh Ng xk¡oksa dh dqy tula[;k dks feykd
E 75 SSC PHASE IX 2022
(a) 3,00,000 (b) 3,50,000
If the total number of students is 60000,
then the total number of students who did (c) 2,80,000 (d) 2,95,000
not pass in the courses A, C is : Direction (33-35): The given pie-chart shows the
;fn Nk=kksa dh dqy la[;k 60000 gS] rks ikBÔØeksa
numberesaof tourists for the year 2015, travelling
ikfjr djus okys Nk=kksa dh dqy A,
la[;k
C gS% from India and to India. /fn;k x;k ikbZ pkVZ o"kZ 2
CHSL 14 - 10-2020 (Afternoon Shift) esa Hkkjr ls vkSj Hkkjr dh ;k=kk djus okys i;ZVdksa
n'kkZrk gSA

r
(a) 7852 (b) 4992
(c) 8254 (d) 7628

si
Direction (31 - 32) : Study the given pie chart

an by
that represents the percentage population of six
villages A, B, C, D, E and F in 2020, and answer

n
the question that follows.
fn, x, ikbZ pkVZ dk vè;;u djsa tks 2020 esa Ng
xkaoksa
A, B, C, D, E vkSjF dh izfr'kr vkcknh dks n'kkZrk

ja
R s
gS] vkSj uhps fn, x, iz'u dk mÙkj nsaA
a th

33. In the given pie-chart, if 1657850 is the

total number of tourists visiting India, how
ty a

many visited from Australia:

fn, x, ikbZ&pkVZ esa] ;fn 1657850 Hkkjr vkus
di M

i;ZVdksa dh dqy la[;k gS] rks vkWLVªsfy;k

i;ZVd vk,%
SSC CPO 15 March 2019 (Morning)
Village % of population below poverty low
A 30 (a) 563669
B 45 (b) 589320
C 55 (c) 457602
D 60
(d) 331570
E 58
F 40 34. In the given pie-chart, from which coun-
31. The difference between the central angles try tourists have come to India more than
A

of the sectors representing the percentage Indians going to that country:

population of villages D and F is: fn, x, ikbZ&pkVZ esa] ml ns'k esa tkus okys H
xk¡oD vkSj F dh çfr'kr tula[;k dk çfrfuf/Ro ls vf/d i;ZVd fdl ns'k ls Hkkjr vk, gSa%
djus okys lsDVj ds eè; dks.k ds chp dk varj gS% SSC CPO 15 March 2019 (Morning)
SSC PHASE IX 2022 (a) Australia and Africa
(b) Europe
(a) 15° (b) 18°
(c) USA
(c) 25° (d) 22°
(d) Australia only
35. In the given pie-chart, if the number of Direction (37-38): Study the following pie-chart
tourists visiting India is 21,35,600 and the and table to answer the question
number from India to other countries is ç'u dk mÙkj nsus ds fy, fuEufyf•r ikbZ&pkVZ
20,45,450. How many more people visit rkfydk dk vè;;u djsa
the USA from India than from the USA to Total number of students admitted in a
India? university in various fields = 5000
fn, x, ikbZ&pkVZ esa] ;fn Hkkjr vkus okys i;ZVdksa Distribution of the number of students into
dh la[;k 21]35]600 gS vkSj Hkkjr ls vU; ns'kksa esavarious fields:
vkus okys i;ZVdksa dh la[;k 20]45]450 gSA la;qÙkQ jkT;
,d fo'ofo|ky; esa fofHkUu {ks=kksa esa ços'k y
vesfjdk ls Hkkjr dh rqyuk esa Hkkjr ls fdrus vf/d Nk=kksa dh dqy la[;k ¾ 5000
yksx la;qÙkQ jkT; vesfjdk tkrs gSa\
fofHkUu {ks=kksa esa Nk=kksa dh la[;k dk forj
SSC CPO 15 March 2019 (Morning)
(a) 303833 (b) 358097
(c) 342675 (d) 287698

r
36. The given bar graph shows the biscuit ex-

si
ports of India over a period of five years.
Study the graph and answer the question
that follows.
an by
fn;k x;k ckj xzkiQ ikap o"kks± dh vof/ esa Hkkjr ds

n
fcLdqV fu;kZr dks n'kkZrk gSA xzkiQ dk vè;;u djsa
vkSj uhps fn, x, ç'u dk mÙkj nsaA

ja
R s
a th
ty a
di M

37. What is the average number of boys in CS,

ECE and EEE Fields?
CS, ECE vkSjEEE iQhYM esa yM+dksa dh vkSl
D;k gS\
CGL 2019 Tier-II (18-11-2020 )
(a) 406 (b) 506
(c) 514 (d) 516
38. What is the difference between the num-
ber of girls in IT and number of girls in
In which year, the percentage increase in
A

ECE?
exports was maximum from its preceding
IT esa yM+fd;ksa dh la[;k ECE
vkSj esa yM+fd;ksa
year:
la[;k ds chp fdruk varj gS\
fdl o"kZ esa] fu;kZr esa çfr'kr o`f¼ mlds iwoZorhZ o"kZ
CGL 2019 Tier-II (18-11-2020)
ls vf/dre Fkh%
(a) 25
CHSL 19-03-2020 (afternoon shift)
(b) 21
(a) 2009 (b) 2006
(c) 20
(c) 2008 (d) 2007
(d) 30
Direction (39 - 41): Study the given graph and ;fn laLFkkuA ls mÙkh.kZ gksus okys yM+dksa
answer the question that follows. dk] mlh laLFkku ls mÙkh.kZ gksus okyh yM+fd;
fuEukafdr xzkiQ dk vè;;u djsa vkSj mlds ckn fn, 5 : 6 gS vkSj laLFkku
D ls mÙkh.kZ gksus okys N
x, iz'u dk mÙkj nsaA ls 40% yM+ds gSa] rkss laLFkku
A ls mÙkh.kZ gksus
Distribution (degree wise)
of students appeared in the
Distribution of students
(percent wise) passed in the
yM+dksa dh la[;k dk] laLFkku
D ls mÙkh.kZ yM+dk
examination from institute examination from institute la[;k ls vuqikr fdruk gksxk\
A, B, C, D & E A, B, C, D & E SSC CGL MAINS 03 Feb 2022
laLFkkuA, B, C, D vkSj
E ls ijh{kk laLFkkuA, B, C, D vkSj
E ls ijh{kk (a) 25 : 24
esa 'kkfey gksus okys Nk=kksa dk forj.k
esa mÙkh.kZ gksus okys Nk=kksa dk forj.k
(b) 4 : 3
(fMxzh okj) (çfr'kr okj)
(c) 5 : 4
(d) 3 : 2
A A
B B Direction (42-44): Study the given pie charts and
98° 22%
112° 30%
40° E 10% E answer the question that follows.

C
°
60 50°
C 1 8% 20% fuEukafdr ikbZ pkVksZa dk vè;;u djsa vkSj m
D D fn, x, iz'u dk mÙkj nsaA

r
Total Number of Students Total Number of Students

si
Distribution (degree wise) of Distribution of students (percent wise)
appeared = 1800 passed = 1200 students appeared in the passed in the examination from institutes
'kkfey gq, dqy Nk=k
= 1800 mÙkhZ.k gq, dqy=Nk=k
1200 examination from institutes A, B, C, D & E

39.
an by
The number of students who passed from
institute C exceeds the number of students
A, B, C, D & E

B A
B
18%
A
15%

n
54° 72° C
who appeared from institute E is x. The C 30%
9% E
90° 39° E
D
value of x lies between: D 28%

laLFkku
ja
C ls mÙkh.kZ gksus okys Nk=kksa dh la[;k laLFkku
105°
R s
Total Number of students
E ls mifLFkr gksus okys Nk=kksaxdh
ls la[;k
vf/d Total Number of students passed = 900
a th

appeared = 1200
gSAx dk eku buds chp gS\
SSC CGL MAINS 03 Feb 2022 42. Which institute has the second highest
(a) 18 and 22 (b) 14 and 18 percentage of students who passed to the
ty a

(c) 10 and 14 (d) 22 and 26 students who appeared from that

40. The number of students who appeared from institute?
di M

institute B is what percentage more than fdl laLFkku dk ml laLFkku ls ijh{kk eas 'kkfey
the total number of students who passed okys Nk=kksa dh rqyuk esa mÙkh.kZ gksus
from institutes A and C? izfr'kr nwljk loksZPp gS\
laLFkkuB ls mifLFkr gksus okys Nk=kksa dh la[;k] laLFkku SSC CGL MAINS 29 Jan 2022
A vkSjC ls mÙkh.kZ gksus okys Nk=kksa dh dqy la[;k (a) lsC
(B) E
fdrus izfr'kr vf/d gS\
(c) D
SSC CGL MAINS 03 Feb 2022
(d) B
2 1 43. The number of students who passed the
(a) 16 % (b) 15 %
3 3 examination from institute C is what
percentage of the total number of students
A

1 2 who appeared from institutes D and E?

(c) 14 % (d) 7 %
7 7 laLFkku
C ls ijh{kk mÙkh.kZ djus okys Nk=kksa
41. If the ratio of the number of boys to that laLFkkuD vkSjE ls mifLFkr gksus okys Nk=kksa
of the girls who passed from institute A is
la[;k dk fdruk izfr'kr gS\
5:6, and 40% of the students who passed
SSC CGL MAINS 29 Jan 2022
from institute D are boys, then the ratio
(a) 56.25%
of the number of boys who passed from
(b) 54.25%
institute A to that of boys who passed from
(c) 58.3%
institute D is:
(d) 52.1%
44. The number of students who passed the laLFkku
D ls ijh{kk mÙkh.kZ djus okys Nk=kksa
examination from institute D exceeds the laLFkku
A ls ijh{kk esa 'kkfey gksus okys Nk=kksa
number of students who appeared from ls x vf/d gSA x dk eku fuEufyf[kr esa ls fdlds
institute A is x. The value of x lies
eè; gksxk\
between:
SSC CGL MAINS 29 Jan 2022
(a) 8 and 11 (b) 11 and 14
(c) 5 and 8 (d) 14 and 17

Answer Key
1. (b) 6. (a) 11. (d) 16. (a) 21. (c) 25. (c) 29. (c) 33. (a) 37. (b) 41. (c)

r
2. (a) 7. (d) 12. (c) 17. (a) 22. (b) 26. (c) 30. (b) 34. (a) 38. (b) 42. (c)

si
3. (c) 8. (b) 13. (b) 18. (d) 23. (b) 27. (c) 31. (b) 35. (a) 39. (b) 43. (a)
4. (d) 9. (a)
an by
14. (a) 19. (a) 24. (c) 28. (d) 32. (b) 36. (d) 40. (a) 44. (b)

n
5. (d) 10. (a) 15. (d) 20. (c)

ja
R s
a th
ty a
di M
A
 DATA INTERTRETATION
(LINE GRAPH)
Directions (1 – 4) : Study the following graph to Directions (5) : Study the following graph to an-
answer the question given below: swer the question given below:
v/ksfyf•r lkj.kh dk lko/kuhiwoZd vè;;u djsa] vkSj uhps v /ksfyf•r lkj.kh dk lko/kuhiwoZd vè;;u djsa] vkS
fn, x, ç'uksa dk mÙkj nsaA uhps fn, x, ç'uksa dk mÙkj nsaA
Export over the years in Rs. (crore)
Amount (in Rs crore)

1000 Profit earned by a company over the years

900
800 (Profit in crore rupees)
700
600
500 fn;sx, o"kks± esa daiuh }kjk vftZr ykHk (djksM+ #i,
400

r
300 80
200

Profit (in Rs or)

100 70

si
0 60
1995 1996 1997 1998 1999 2000 2001
Years 50
40
1.
cent of the export in 1998?
an by
Export in 1997 is approximately what per- 30
20
10

n
1997 es fu;kZr] 1998 es fu;kZr dk yxHkx fdrus 0 1998 1999 2000 2001 2002 2003 2004 2005
çfr'kr gS \ Years

(a) 145
ja (b) 135 5. In which year is the percent increase in the
R s
(c) 150 (d) 300 profit from that in the previous year the
a th

2. What is difference in exports in 1997 and highest?

1998? fdl o"kZ fiNys o"kZ dh rqyuk es] ykHk es
1997 rFkk 1998 ds fu;kZr es fdrus #i, dk varj gS \ o`f¼ vf/dre gS\
ty a

(a) Rs. 150 Cr. (a) 1998 (b) 2000

(c) 2002 (d) 2004
di M

(b) Rs. 1500 Cr.

(c) Rs. 15 Cr. 6. The line graph shows electricity consump-
tion (in units) for three households A, B
(d) Rs. 100 Cr.
and C for months February to May.
3. What is the total export (in crore rupees)
in the given years?
ykbu xzkiQ iQjojh ls ebZ ds eghuksa ds fy, rhu
A] B vkSjC ds fy, fctyh dh •ir (bdkb;ksa esa)
fn, x, o"kksZa esa] dqy fu;kZr (djksM+ :i;s esa) D;k gS\
dks n'kkZrk gSA
(a) 4100 (b) 3700
95
(c) 3900 (d) 3950
62
4. Which year has the highest per cent in-
crease/decrease in exports as compared to
A

107 C
the previous year? 61
100
54 B
fdl o"kZ fu;kZr es fiNys o"kZ dh rqyuk es çfr'kr
o`f¼@deh vf/dre gS\ 40
A

55 93
(a) 1997 84
68
(b) 1998 42

(c) 2001
February March April May
(d) Can't be determined
 In the given graph, the percentage increase 8. In the given line graph, what was the
in electricity consumption of ‘B’ between average temperature on first sunday ?
March and May is: fn, x, js•k xzkiQ esa] igys jfookj dk vkSlr rkieku
fn, x, xzkiQ esa] ekpZ vkSj ebZ ds‘B’chp
dh fctyh fdruk Fkk\
•ir esa çfr'kr o`f¼ gS% SSC CPO 15 March 2019 (Evening)
SSC CPO 13 March 2019 (Morning) (a) 21.2
(a) 95% (b) 85% (b) 24
(c) 167.5% (d) 98% (c) 25.4
7. The line graph shows electricity consump- (d) 23
tion (in units) for three households A, B 9. In the given line graph, when was the
and C for months February to May. maximum temperature recorded in
ykbu xzkiQ iQjojh ls ebZ ds eghuksa ds fy, A rhu
] B?kjksa Chennai?
vkSjC ds fy, fctyh dh •ir (bdkb;ksa esa) dks n'kkZrk gSA fn, x, ykbu xzkiQ esa] psUubZ esa vf/dre rki
95 dc ntZ fd;k x;k Fkk\

r
62 SSC CPO 15 March 2019 (Evening)

si
(a) 4th Sunday
107 C
100
(b) 3rd Sunday
61

an by
54 B

A
(c) 1st Sunday
(d) 2nd Sunday

n
40
55
84
93 10. In the given line graph, what is the differ-
68 ence between the temperature of Delhi and
42

ja
R s
Chennai on the 3rd Sunday ?
February April May fn, x, js•k vkys• esa] rhljs jfookj dks fnYyh vkS
a th

March
In the given graph, the difference between psUubZ ds rkieku esa fdruk varj gS\
the total electrical consumption between SSC CPO 15 March 2019(Evening)
months of February and April is: (a) 21 (b) 13
ty a

fn, x, xzkiQ esa] iQjojh vkSj vçSy ds eghuksa ds chp (c) 17 (d) 8
dqy fctyh dh •ir ds chp dk varj gS%
di M

Direction (11 – 13): The line graph shows the

SSC CPO 13 March 2019 (Morning) production of product A and B (in thousands)
(a) 97 (b) 74 during the period 2004 to 2009 and the second
(c) 121 (d) 15 line Graph shows the percentage sale of these
Directions (8 - 10): The line graph shows the products.
temperature on four Sundays of three cities. funZs'k (11 & 13)% ykbu xzkiQ 2004 ls 2009 dh v
js•k xzkiQ rhu 'kgjksa ds pkj jfookjksa dks rkieku fn•krk gSA mRikn
ds nkSjku A vkSjB (gtkjksa esa) ds mRiknu dks n'k
34
32
31 gS vkSj nwljh iafÙkQ dk xzkiQ bu mRiknksa dh fcØ
30
28 29
28
28
n'kkZrk gSA
26 26 PRODUCTION OF PRODUCT A(in’000)
24 24.2
22 PRODUCTION OF PRODUCT B(in’000)
A

21.2
20 16
18 18.9 18
14 13 14
16 14 16 12
14 12 11
12 12
10 10 10 8.5 11
8 8 9
6 7 8 7
4 6
2
0 4
1st 2nd 3rd 4th 2
Sunday Sunday Sunday Sunday
0
Mumbai 2004 2005 2006 2007 2008 2009
 90 80
75 80
80 75
70 65
60 70
60 75
50
56 55
40 55 50
30
20
10
0
2004 2005 2006 2007 2008 2009

11. In the given line graph, what is the total

sale of Products A and B in the year 2007?
fn, x, ykbu xzkiQ esa] o"kZ 2007 esaAmRikn
vkSjB
dh dqy fcØh fdruh gS\
The table given below represents the
SSC CPO 16 March 2019 (Evening)
respective ratio of the production (in
(a) 10290 tonnes) of Company A to the production (in
(b) 13460 tonnes) of Company B, and the respective

r
ratio of the sales (in tonnes) of Company
(c) 11500
A to the sales (in tonnes) of Company B.

si
(d) 12490
uhps nh xbZ rkfydk daiuh A ds mRiknu (Vu) esa
12. In the given line graph, what is the total daiuh B ds mRiknu (Vu esa) ls lacaf/r vuqikr vk
an by
sale of Product B in the year 2004 and daiuhA dh fcØh (Vu esa)] daiuhB dh fcØh (Vu

n
2008 together? esa) ls lacaf/r vuqikr dk çfrfuf/Ro djrh gS A
fn, x, ykbu xzkiQ esa] o"kZ 2004 vkSj 2008 esa mRikn

ja
B dh dqy fcØh fdruh gS\
R s
Years Production Sales
SSC CPO 16 March 2019 (Evening) 2006 5:4 2:3
a th

(a) 11950 2007 8:7 11 : 12

2008 3:4 9 : 14
(b) 12500
2009 11 : 12 4:5
(c) 14600
ty a

2010 14 : 13 10 : 9
(d) 11825 2011 13 : 14 1:1
di M

13. In the given line graph, what is the total

sale of Product A in the year 2005 and 14. What is the approximate percentage
2009 taken together? increase in the production of Company A
fn, x, ykbu xzkiQ esa] o"kZ 2005 vkSj 2009 esa mRikn
(in tonnes) from the year 2009 to the
production of Company A (in tonnes) in the
A dh dqy fcØh dks feykdj dqy fdruh gS\
year 2010?
SSC CPO 16 March 2019 (Evening)
o"kZ 2009 esa daiuh
A ds mRiknu (Vu esa) ls o"
(a) 16400
2010 esa daiuhA ds mRiknu (Vu esa) esa vuqek
(b) 14600
çfr'kr o`f¼ D;k gS\
(c) 17500
(a) 18 (b) 38
(d) 18500
A

(c) 23 (d) 27
Directions (14-18):- Answer the questions on the
15. The sales of Company A in the year 2009
basis of the information given below:/uhps nh xbZ
was approximately what percent of the
tkudkjh ds vk/kj ij ç'uksa ds mÙkj nsa% production of Company A in the same
The graph given below represents the year?
production (in tonnes) and sales (in tonnes)
of company A from 2006-2011. o"kZ 2009 esa daiuh
A dh fcØh mlh o"kZ daiuh
A ds

uhps fn;k x;k xzkiQ daiuh

A ds 2006&2011 esa mRiknu
mRiknu dk yxHkx fdruk çfr'kr Fkh\
(Vu esa) vkSj fcØh (Vu esa) dk çfrfuf/Ro djrk gS (a) 65 (b) 73
(c) 79 (d) 83
16. What is the average production of company (a) 81 : 64 (b) 64 : 55
B (in tonnes) from the year 2006 to the (c) 71 : 81 (d) 81 : 65
year 2011?
18. What is the respective ratio of production
o"kZ 2006 ls o"kZ 2011 rd daiuh
B dk vkSlr mRiknu of Company B (in tonnes) in the year 2006
(Vu esa) fdruk gS\ to production of Company B (in tonnes) in
(a) 574 (b) 649 the year 2008?

(c) 675 (d) 593 o"kZ 2006 esa daiuh

B ds mRiknu (Vu esa) dk lacaf
17. What is the respective ratio of the total r vuqikr o"kZ 2008 esa daiuh
B dk mRiknu (Vu esa)
production (in tonnes) of Company B to the ls vuqikr D;k gS\
total sales (in tonnes) of Company B? (a) 2 : 5 (b) 4 : 5
daiuh B ds dqy fcØh (Vu esa) daiuhB ds dqy (c) 3 : 4 (d) 3 : 5
mRiknu (Vu esa) dk lacaf/r vuqikr D;k gS\

r
Answer Key

si
1.
2.
(b)
(a)
3.
4.
(c)
(a) an by
5.
6.
(b)
(c)
7.
8.
(c)
(b)
9. (b)
10. (a)
11. (a)
12. (d)
13. (a)
14. (d)
15. (b)
16. (c)
17. (d) 18. (c)

n
ja
R s
a th
ty a
di M
A
 DATA INTERTRETATION
(LINE GRAPH)
Directions (1 – 4) : Study the following graph to Directions (5) : Study the following graph to an-
answer the question given below: swer the question given below:
v/ksfyf•r lkj.kh dk lko/kuhiwoZd vè;;u djsa] vkSj uhps v /ksfyf•r lkj.kh dk lko/kuhiwoZd vè;;u djsa] vkS
fn, x, ç'uksa dk mÙkj nsaA uhps fn, x, ç'uksa dk mÙkj nsaA
Export over the years in Rs. (crore)
Amount (in Rs crore)

1000 Profit earned by a company over the years

900
800 (Profit in crore rupees)
700
600
500 fn;sx, o"kks± esa daiuh }kjk vftZr ykHk (djksM+ #i,
400

r
300 80
200

Profit (in Rs or)

100 70

si
0 60
1995 1996 1997 1998 1999 2000 2001
Years 50
40
1.
cent of the export in 1998?
an by
Export in 1997 is approximately what per- 30
20
10

n
1997 es fu;kZr] 1998 es fu;kZr dk yxHkx fdrus 0 1998 1999 2000 2001 2002 2003 2004 2005
çfr'kr gS \ Years

(a) 145
ja (b) 135 5. In which year is the percent increase in the
R s
(c) 150 (d) 300 profit from that in the previous year the
a th

2. What is difference in exports in 1997 and highest?

1998? fdl o"kZ fiNys o"kZ dh rqyuk es] ykHk es
1997 rFkk 1998 ds fu;kZr es fdrus #i, dk varj gS \ o`f¼ vf/dre gS\
ty a

(a) Rs. 150 Cr. (a) 1998 (b) 2000

(c) 2002 (d) 2004
di M

(b) Rs. 1500 Cr.

(c) Rs. 15 Cr. 6. The line graph shows electricity consump-
tion (in units) for three households A, B
(d) Rs. 100 Cr.
and C for months February to May.
3. What is the total export (in crore rupees)
in the given years?
ykbu xzkiQ iQjojh ls ebZ ds eghuksa ds fy, rhu
A] B vkSjC ds fy, fctyh dh •ir (bdkb;ksa esa)
fn, x, o"kksZa esa] dqy fu;kZr (djksM+ :i;s esa) D;k gS\
dks n'kkZrk gSA
(a) 4100 (b) 3700
95
(c) 3900 (d) 3950
62
4. Which year has the highest per cent in-
crease/decrease in exports as compared to
A

107 C
the previous year? 61
100
54 B
fdl o"kZ fu;kZr es fiNys o"kZ dh rqyuk es çfr'kr
o`f¼@deh vf/dre gS\ 40
A

55 93
(a) 1997 84
68
(b) 1998 42

(c) 2001
February March April May
(d) Can't be determined
 In the given graph, the percentage increase 8. In the given line graph, what was the
in electricity consumption of ‘B’ between average temperature on first sunday ?
March and May is: fn, x, js•k xzkiQ esa] igys jfookj dk vkSlr rkieku
fn, x, xzkiQ esa] ekpZ vkSj ebZ ds‘B’chp
dh fctyh fdruk Fkk\
•ir esa çfr'kr o`f¼ gS% SSC CPO 15 March 2019 (Evening)
SSC CPO 13 March 2019 (Morning) (a) 21.2
(a) 95% (b) 85% (b) 24
(c) 167.5% (d) 98% (c) 25.4
7. The line graph shows electricity consump- (d) 23
tion (in units) for three households A, B 9. In the given line graph, when was the
and C for months February to May. maximum temperature recorded in
ykbu xzkiQ iQjojh ls ebZ ds eghuksa ds fy, A rhu
] B?kjksa Chennai?
vkSjC ds fy, fctyh dh •ir (bdkb;ksa esa) dks n'kkZrk gSA fn, x, ykbu xzkiQ esa] psUubZ esa vf/dre rki
95 dc ntZ fd;k x;k Fkk\

r
62 SSC CPO 15 March 2019 (Evening)

si
(a) 4th Sunday
107 C
100
(b) 3rd Sunday
61

an by
54 B

A
(c) 1st Sunday
(d) 2nd Sunday

n
40
55
84
93 10. In the given line graph, what is the differ-
68 ence between the temperature of Delhi and
42

ja
R s
Chennai on the 3rd Sunday ?
February April May fn, x, js•k vkys• esa] rhljs jfookj dks fnYyh vkS
a th

March
In the given graph, the difference between psUubZ ds rkieku esa fdruk varj gS\
the total electrical consumption between SSC CPO 15 March 2019(Evening)
months of February and April is: (a) 21 (b) 13
ty a

fn, x, xzkiQ esa] iQjojh vkSj vçSy ds eghuksa ds chp (c) 17 (d) 8
dqy fctyh dh •ir ds chp dk varj gS%
di M

Direction (11 – 13): The line graph shows the

SSC CPO 13 March 2019 (Morning) production of product A and B (in thousands)
(a) 97 (b) 74 during the period 2004 to 2009 and the second
(c) 121 (d) 15 line Graph shows the percentage sale of these
Directions (8 - 10): The line graph shows the products.
temperature on four Sundays of three cities. funZs'k (11 & 13)% ykbu xzkiQ 2004 ls 2009 dh v
js•k xzkiQ rhu 'kgjksa ds pkj jfookjksa dks rkieku fn•krk gSA mRikn
ds nkSjku A vkSjB (gtkjksa esa) ds mRiknu dks n'k
34
32
31 gS vkSj nwljh iafÙkQ dk xzkiQ bu mRiknksa dh fcØ
30
28 29
28
28
n'kkZrk gSA
26 26 PRODUCTION OF PRODUCT A(in’000)
24 24.2
22 PRODUCTION OF PRODUCT B(in’000)
A

21.2
20 16
18 18.9 18
14 13 14
16 14 16 12
14 12 11
12 12
10 10 10 8.5 11
8 8 9
6 7 8 7
4 6
2
0 4
1st 2nd 3rd 4th 2
Sunday Sunday Sunday Sunday
0
Mumbai 2004 2005 2006 2007 2008 2009
 90 80
75 80
80 75
70 65
60 70
60 75
50
56 55
40 55 50
30
20
10
0
2004 2005 2006 2007 2008 2009

11. In the given line graph, what is the total

sale of Products A and B in the year 2007?
fn, x, ykbu xzkiQ esa] o"kZ 2007 esaAmRikn
vkSjB
dh dqy fcØh fdruh gS\
The table given below represents the
SSC CPO 16 March 2019 (Evening)
respective ratio of the production (in
(a) 10290 tonnes) of Company A to the production (in
(b) 13460 tonnes) of Company B, and the respective

r
ratio of the sales (in tonnes) of Company
(c) 11500
A to the sales (in tonnes) of Company B.

si
(d) 12490
uhps nh xbZ rkfydk daiuh A ds mRiknu (Vu) esa
12. In the given line graph, what is the total daiuh B ds mRiknu (Vu esa) ls lacaf/r vuqikr vk
an by
sale of Product B in the year 2004 and daiuhA dh fcØh (Vu esa)] daiuhB dh fcØh (Vu

n
2008 together? esa) ls lacaf/r vuqikr dk çfrfuf/Ro djrh gS A
fn, x, ykbu xzkiQ esa] o"kZ 2004 vkSj 2008 esa mRikn

ja
B dh dqy fcØh fdruh gS\
R s
Years Production Sales
SSC CPO 16 March 2019 (Evening) 2006 5:4 2:3
a th

(a) 11950 2007 8:7 11 : 12

2008 3:4 9 : 14
(b) 12500
2009 11 : 12 4:5
(c) 14600
ty a

2010 14 : 13 10 : 9
(d) 11825 2011 13 : 14 1:1
di M

13. In the given line graph, what is the total

sale of Product A in the year 2005 and 14. What is the approximate percentage
2009 taken together? increase in the production of Company A
fn, x, ykbu xzkiQ esa] o"kZ 2005 vkSj 2009 esa mRikn
(in tonnes) from the year 2009 to the
production of Company A (in tonnes) in the
A dh dqy fcØh dks feykdj dqy fdruh gS\
year 2010?
SSC CPO 16 March 2019 (Evening)
o"kZ 2009 esa daiuh
A ds mRiknu (Vu esa) ls o"
(a) 16400
2010 esa daiuhA ds mRiknu (Vu esa) esa vuqek
(b) 14600
çfr'kr o`f¼ D;k gS\
(c) 17500
(a) 18 (b) 38
(d) 18500
A

(c) 23 (d) 27
Directions (14-18):- Answer the questions on the
15. The sales of Company A in the year 2009
basis of the information given below:/uhps nh xbZ
was approximately what percent of the
tkudkjh ds vk/kj ij ç'uksa ds mÙkj nsa% production of Company A in the same
The graph given below represents the year?
production (in tonnes) and sales (in tonnes)
of company A from 2006-2011. o"kZ 2009 esa daiuh
A dh fcØh mlh o"kZ daiuh
A ds

uhps fn;k x;k xzkiQ daiuh

A ds 2006&2011 esa mRiknu
mRiknu dk yxHkx fdruk çfr'kr Fkh\
(Vu esa) vkSj fcØh (Vu esa) dk çfrfuf/Ro djrk gS (a) 65 (b) 73
(c) 79 (d) 83
16. What is the average production of company (a) 81 : 64 (b) 64 : 55
B (in tonnes) from the year 2006 to the (c) 71 : 81 (d) 81 : 65
year 2011?
18. What is the respective ratio of production
o"kZ 2006 ls o"kZ 2011 rd daiuh
B dk vkSlr mRiknu of Company B (in tonnes) in the year 2006
(Vu esa) fdruk gS\ to production of Company B (in tonnes) in
(a) 574 (b) 649 the year 2008?

(c) 675 (d) 593 o"kZ 2006 esa daiuh

B ds mRiknu (Vu esa) dk lacaf
17. What is the respective ratio of the total r vuqikr o"kZ 2008 esa daiuh
B dk mRiknu (Vu esa)
production (in tonnes) of Company B to the ls vuqikr D;k gS\
total sales (in tonnes) of Company B? (a) 2 : 5 (b) 4 : 5
daiuh B ds dqy fcØh (Vu esa) daiuhB ds dqy (c) 3 : 4 (d) 3 : 5
mRiknu (Vu esa) dk lacaf/r vuqikr D;k gS\

r
Answer Key

si
1.
2.
(b)
(a)
3.
4.
(c)
(a) an by
5.
6.
(b)
(c)
7.
8.
(c)
(b)
9. (b)
10. (a)
11. (a)
12. (d)
13. (a)
14. (d)
15. (b)
16. (c)
17. (d) 18. (c)

n
ja
R s
a th
ty a
di M
A
 DATA INTERTRETATION
(HISTOGRAM )
Direction (01) : The given bar graph represents Direction (03) : The given graph shows the
the number of teachers in different weight weights of students in a school on a particular
groups. Study the graph and answer the question day.
that follows. fn;k x;k xzkiQ fdlh fo'ks"k fnu Ldwy esa Nk=kksa ds otu dk
fn;k x;k naM vkys• fofHkUu Hkkj lewgksa esa f'k{kdksa dh la[;k
dks n'kkZrk gSA xzkiQ dk vè;;u djsa vkSj uhps fn, x, ç'u dk 60
55
mÙkj nsaA
50
45
9

Number is Students
9 40
8 40
7 35
7
6 30
6 30
5 5 25
5
4 4
3 20
3
2 10
1
0
40 45 50 55 60 65 70 75 0
40 45 50 55 60 65 70

1. In which of the following weight groups is Weight (kg)

the number of teachers the maximum? 3. The number of students weighing less than
fuEufyf•r esa ls fdl Hkkj lewg esa f'k{kdksa dh la[;k 50 kg is what percent less than the number
lcls vf/d gS\ of students weighing 55 kg or more?
CHSL 19-03-2020 (afternoon shift) 50 fdxzk ls de otu okys Nk=kksa dh la[;k 55 fdxzk
(a) 60-65 (b) 45-50 vf/d otu okys Nk=kksa dh la[;k ls fdruk çfr'kr de gS\
(c) 65-70 (d) 40-45 SSC CGL Tier II-13 September 2019
Direction (02) : Study the graph and answer the (a) 44 (b) 40
following questions (c) 55 (d) 30
xzkiQ dk vè;;u djsa vkSj fuEufyf[kr ç'uksa ds mÙkj nsaA
Direction (04) : The given graph shows the marks
32 obtained by students in an examination.
32
fn;k x;k xzkiQ ,d ijh{kk esa Nk=kksa }kjk çkIr vadksa d
28 22
Number of students

70
20 60
60
16
Number of Students

50 45
12
8 8 40
8 6 40
4 35 35
4 30
30

0 1 2 3 4 5 6 7 20
No. of hours playing
10
mobile games per day
2. How many students spend 5 hours or more 0
150 200 250 300 350 400 450
than 5 hours playing mobile games per day?
Marks
fdrus Nk=k çfrfnu eksckby xse osyus esa 5 ?kaVs4.
;k 5 ?kaVs
The number of students who obtained less
ls vf/d le; O;rhr djrs gSa\ than 300 marks is what percent more than
CHSL 16-10-2020 (Morning shift) the number of students who obtained 350
(a) 46 (b) 14 or more marks?
(c) 8 (d) 6
 300 ls de vad çkIr djus okys Nk=kksa dh la[;k 350 6.
;k The number of patients aged 10 or more
vf/d vad çkIr djus okys Nk=kksa dh la[;k ls fdruk years but below 40 years is what percent
less than the number of patients aged 50
çfr'kr vf/d gS\ or more years but below 80 years ?
SSC CGL Tier II- 12 September 2019
(a) 80% (b) 28%
10 ;k vf/d o"kZ ysfdu 40 o"kZ ls de vk;q ds jksfx;
(c) 44.4% (d) 22.7% dh la[;k 50 ;k vf/d o"kZ ysfdu 80 o"kZ ls de vk;q
Direction (05): Study the graph and answer the ds jksfx;ksa dh la[;k ls fdruk çfr'kr de gS\
question that follows. CGL 2019 Tier-II (15-11-2020 )
xzkiQ dk vè;;u djsa vkSj uhps fn, x, ç'u dk mÙkj nsaA (a) 30.2 (b) 25
(c) 27.5 (d) 34
70 65 Direction (07 - 09): The given histogram shows
60
60 the height of the student.
55
fn;k x;k fgLVksxzke Nk=k dh ÅapkbZ n'kkZrk gSA
Number of Workers

50 45
20
40 15

Number of students
35 15 13 14
30 10 12
30 10 6
20 5
0

150-155

155-160

160-165

165-170

170-175

175-180
10

0
400 450 500 550 600 650 700
Daily wages (in Rs)
5. What is the ratio of the total number of Height of the students in cm
workers whose daily wages are less than Rs
500 to the total number of workers whose 7. The difference between the number of
daily wages are Rs. 600 and above? students whose height is between 150-
155cm and the number of students whose
500 #i;s ls de nSfud etnwjh okys Jfedksa dh dqy
height lies between 175-180cm is:
la[;k dk mu Jfedksa dh dqy la[;k ls vuqikr D;k gS
mu Nk=kksa dh la[;k ftudh ÅapkbZ 150& 155
ftudh nSfud etnwjh #i;s 600 vkSj Åij gS\
SSC CGL Tier II-11 September 2019
chp gS vkSj ftudh ÅapkbZ 175&180 lseh ds chp g
(a) 5 : 6 chp dk varj gS%
(b) 6 : 7 CPO 23-11-2020 (Morning shift)
(c) 3 : 4 (a) 3 (b) 8
(d) 15 : 11 (c) 9 (d) 7
Direction (06): Study the given graph and answer 8. What is the percentage of students whose
the question that follows. height is in the class interval 160-170?
fn, x, xzkiQ dk vè;;u djsa vkSj uhps fn, x, ç'u dk mÙkj nsaA (correct to the nearest integer)
d{kk varjky 160&170 esa ÅapkbZ okys fo|kfF
DISTRIBUTION (AGE WISE) OF PATIENTS BEING çfr'kr fdruk gS\
TREATED IN A HOSPITAL IN A CITY (fudVre iw.kkZad ds fy, lgh)
CPO 23-11-2020 (Morning shift)
40
34 (a) 39 (b) 25
Number of patients

35 (c) 34 (d) 51
32
30 26 9. The number of students whose height is in
24 the class interval 170-175 is what percent
25 22
20 less than the number of students whose
20
14 height is in the interval 165-170?
15
10 (correct to the nearest integer)
10 10 d{kk varjky 170&175 esa ÅapkbZ okys Nk=kk
5 165&170 ds varjky esa ÅapkbZ okys Nk=kksa
0
10 20 30 40 50 60 70 80 90
fdrus çfr'kr de gS\
(fudVre iw.kkZad ds fy, lgh)
 CPO 23-11-2020 (Morning shift) 11. The number of cars with speed between
(a) 17.3% 70km/hr and 80km/hr is what percentage
(b) 11.5% more than the number of the cars with
(c) 14.3% speed between 50km/h and 60km/h ?
(d) 19.5% (correct to one decimal place)
Direction (10): Study the graph and answer the
70 fdeh@?kaVk vkSj 80 fdeh@?kaVk ds chp x
question that follows
dkjksa dh la[;k 50 fdeh@?kaVk vkSj 60 fdeh@
xzkiQ dk vè;;u djsa vkSj uhps fn, x, ç'u dk mÙkj nsa chp xfr okyh dkjksa dh la[;k ls fdruk çfr'kr vf/
d gS\ (,d n'keyo LFkku rd lgh)
10 9 CPO 23-11-2020 (Evening shift)
Number of Weeks

8
8 (a) 22.2% (b) 15.5%
7
6 6 (c) 29.7% (d) 28.6%
6 5
4 4 Direction (12): Study the given histogram and
4
3 answer the question that follows.
2 fuEukafdr vk;rfp=k dk vè;;u djsa vkSj mlds ckn fn, x,
150 160 170 180 190 200 210 220 230 240 iz'u dk mÙkj nsaA
Cost of living index Distribution of persons (weight-wise)
10. The number of weeks, in which the cost of 60
living was160 or more but less than 190, is 52

Number of persons
50
approximately what percent more than the 48
40
number of weeks in which the cost of living 40
35 38
index was 200 or more but less than 220 30
30 25
(correct to one decimal places) 18

gÝrksa dh la[;k] ftlesa jgus dh ykxr 160 ;k vf/d 20

13
9
Fkh] ysfdu 190 ls de Fkh] mu gÝrksa dh la[;k ls 10
5
yxHkx fdruk çfr'kr vf/d gS ftlesa thou lwpdkad 0
50 55 60 65 70 75 80 85 90 95 100 105
dh ykxr 200 ;k vf/d Fkh ysfdu 220 ls de Fkh Weight (in kg)
12. The number of persons weighing 55 kg or
(,d n'keyo LFkkuksa rd lgh)
more but less than 75 kg is what percentage
more than the number of persons weighting
CGL 2019 Tier-II (16-11-2020) 80 kg or ore but less than 100 kg (correct
(a) 44.4 (b) 36.8 to one decimal place?)
(c) 58.3 (d) 60.6 55 fdxzk ;k vf/d ysfdu 75 fdxzk ls de otu okys
Direction (11): The given histogram shows the O;fÙkQ;ksa dh la[;k 80 fdxzk ;k v;Ld ysfdu 1
frequency distribution of the speed of cars
fdxzk ls de otu okys O;fÙkQ;ksa dh la[;k ls fdr
passing though at a particular spot on a highway.
Study the graph and answer the question that çfr'kr vf/d gS (,d n'keyo LFkku rd lgh\)
follows: SSC CGL MAINS 29 JAN 2022
(a) 66.7% (b) 68.4%
fn;k x;k fgLVksxzke jktekxZ ij fdlh fo'ks"k LFkku ij xqtjus(c) 88.2% (d) 77.8%
okyh dkjksa dh xfr ds vko`fÙk forj.k dks n'kkZrk gSADirection
xzkiQ dk (13): Study the graph and answer the
vè;;u djsa vkSj fuEufyf•r ç'u dk mÙkj nsa% question that follow:
100 90 fuEukafdr xzkiQ dk vè;;u djsa vkSj mlds ckn fn, x, iz'
90 85 dk mÙkj nsaA
Number of cars

80 70 Lifetime (in hours) of neon lamps

70 400 375
60 350
60 50 350 340 325
45 300
50 300 260
275
Hours

40 250 225
30 200 175
20
150 125
10 100
100
0
50-60
40-50

80-90
60-70

90-100

50
70-80

0
1100

1200

1300
200

400

500

600

700

800
900

1000
300

Speed in km/hr
Lifetime (in hours)
13. The total number of neon lamps having Direction (16-17) Study the given
lifetime of 800 or more hours is histogram that shows the marks obtained by
approximately what percentage more than students in an examination and answer the
the total number of neon lamps having question that follow
lifetime of 400 of more hours but less than fn, x, fgLVksxzke dk vè;;u djsa tks ,d ijh{kk e
800 hours?
Nk=kksa }kjk izkIr fd, x, vadksa dks fn[kkrk gS vkSj
800 ;k vf/d ?kaVs ds thoudky okys fu;kWu ySai dh
iz'uksa dk mÙkj Kkr djasA
dqy la[;k 400 ?kaVs ls vf/d ysfdu 800 ?kaVs ls de ds
thoudky okys fu;kWu ySai dh dqy la[;k ls yxHkx 70 60
fdruk çfr'kr vf/d gS\

Number of patients
60
SSC CGL MAINS 03 FEB 2022
50 45 40
(a) 22.7% (b) 12.5%
(c) 32.2% (d) 31.8% 40 35 35
30
Direction (14-15): Study the graph and answer 30
the question that follow: 20
fuEukafdr xzkiQ dk vè;;u djsa vkSj mlds ckn fn, x, iz'u 10
dk mÙkj nsaA 0
14. In the given histogram, in which class 150 200 250 300 350 400 450
interval, the median marks lies? Marks
fn, x, vk;r fp=k esa] ekfè;dk fpÉ fdl oxZ varjky
esa fLFkr gS\ 16. If the total marks obtained by students be
35 32 represented as a pie chart, then the central
30
30 28 angle corresponding to marks 250 or more
Number of students

25 but less than 300 is :(correct to the nearest

20 degree)
16
15 14 ;fn Nk=kksa }kjk izkIr fd, x, dqy vadksa dks ikbZ
10 8 :i esa n'kkZ;k tkrk gS] rks 250 ;k vfèkd ysfdu 300
5 de dk dsanzh; dks.k Kkr djsaA
0
0-15 15-30 30-45 45-60 60-75 75-90
SSC CPO 24/11/2020 (Evening)
Marks of students (a) 88o (b) 128o
o
SSC CPO 13 March 2019 (Morning) (c) 188 (d) 68o
(a) 30-45 17. If the total marks obtained by students be
(b) 45-60 represented as a pie chart, then the central
(c) 60-75 angle of the sector representing mark 200
(d) 15-30 or more but less that 300, is :(correct to
15. In the given histogram, what is the mean one decimal place)
marks of the students, correct to one ;fn Nk=kksa }kjk izkIr dqy vad dks ikbZ pkVZ
deciaml palce? n'kkZ;k tkrk gS] rks 200 ;k vfèkd vkSj 300 ls d
fn, x, vk;rfp=k ds vuqlkj ,d n'keyo LFkku rd vadksa dk izfrfufèkRo djus okys {ks=k dk dsanz
Nk=kksa dk eè; vad D;k gS\ (,d n'keyo LFkku ds fy, lgh)
35 32
30 SSC CPO 25/11/2020 (Evening)
30 28
Number of students

(a) 68o (b) 88o

25
(c) 154o (d) 128o
20
14
16 18. The number of students who obtained less
15
than 350 marks is what percent more than
10 8
the number of students who obtained 400
5
or more marks? (correct to one decimal
0
0-15 15-30 30-45 45-60 60-75 75-90
place)
Marks of students 350 ls de vad izkIr djus okys Nk=kksa dh la[;k 400
SSC CPO 13 March 2019 (Morning) vfèkd vad izkIr djus okys Nk=kksa dh la[;k dh rqy
(a) 51.2 (b) 53.5 izfr'kr vfèkd gSA (,d n'key LFkku rd)
(c) 52.7 (d) 50.6
 SSC CPO 24/11/2020 (Evening) 2012 ls 2016 rd dqy erksa dh la[;k 50» c<+hA ;fn
(a) 100% 2016 esa 16» er oS/ Fks] rks 2016 esa oS/ erksa dh
(b) 375.8% D;k Fkh\
(c) 385.7%
(a) 205 (b) 250
(d) 350%
(c) 120 (d) 140
5. In 2013, if the number of valid votes of
MISSING DI female was 100, what was the respective
Directions (01-05):- Study the following table ratio of number of valid votes of male and
chart carefully and answer the questions given number of valid votes of female in the same
beside. year?
funsZ'k (01&05) % fuEufyf[kr lkj.kh dk lko/kuhiwoZd vè;;u 2013 esa] ;fn efgykvksa ds oS/ erksa dh la[;k 100
djsa vkSj uhps fn, x, iz'uksa ds mÙkj nsaA rks iq#"kksa ds oS/ erksa dh la[;k vkSj mlh o"kZ
Total Respective Ratio of ds oS/ erksa dh la[;k dk vuqikr D;k Fkk\
% of Valid
Year Number Votes Valid votes of male (a) 15 : 19 (b) 13 : 11
of Votes and valid votes of (c) 14 : 5 (d) 13 : 5
female
Directions (06-10) :- This following table is
2011 — — 5:3
related to profit and loss and some values are
2012 500 — 5:4 missing. All the discounts are on the M.P. and
2013 1000 38% — the profit are on C.P.
2014 — 60% funZs'k (06&10)%& fuEufyf•r rkfydk ykHk vkSj g
7:5
2015 2500 40% lacaf/r gS vkSj dqN eku foyqIr gSaA lHkh NwV vafdr
—

1. In 2011, the respective ratio of total ykHk Ø; ewY; ij gSA

number of votes to valid votes was 5 : 4. Ar Cost Profit Marked Discount Selling
Number of valid votes of female in 2011 Price (%) Price
tic
(%) Price
constitutes what percent of the total (Rs.) (Rs.) (Rs.)
le

number of votes in the same year?

Jeans 2280 — 2720 — —
2011 esa] dqy erksa dh la[;k dk oS/ erksa dh la[;k ls Shirt
— 30% — — —
vuqikr 5% 4 FkkA 2011 esa efgykvksa ds oS/ erksa dh
T-shirt — — 2875 14% —
la[;k mlh o"kZ dqy erksa dh la[;k dk fdruk çfr'kr gS\
Suit 2200 — — — 2640
(a) 35% (b) 25%
Saree — 20% 3800 — —
(c) 15% (d) 30%
2. In 2014, if the difference between number 6. The ratio of discount % and profit % of the
of valid votes of male and number of valid T-shirt is 7 : 5. Find CP. of T-shirt?
votes of female was 150, what was the total (approximately)
number of votes in 2014? Vh&'kVZ ij NwV izfr'kr vkSj ykHk izfr'kr dk v
2014 esa] ;fn iq#"kksa ds oS/ erksa dh la[;k vkSj efgykvksa
7% 5 gSA Vh&'kVZ dk Ø; ewY; Kkr dhft,A
ds oS/ erksa dh la[;k dk varj 150 Fkk] rks 2014 esa dqy (a) Rs. 2285 (b) Rs. 2398
erksa dh la[;k D;k Fkh\ (c) Rs. 2552 (d) Rs. 2248
(a) 1845 (b) 1500 7. C.P. of Suit is how much percentage less
(c) 1660 (d) 1600 than the M.P. of Jeans?
3. If the average number of valid votes in 2012 lwV dk Ø; ewY; thUl ds vafdr ewY; ls fdrus çfr'k
and 2015 was 635, what was the number of de gSa\
votes given by female in 2012? (a) 21.2% (b) 19.1%
;fn 2012 vkSj 2015 esa oS/ erksa dh vkSlr la[;k 635 (c) 18.6% (d) 18.4%
Fkh] rks 2012 esa efgykvksa }kjk fn, x, erksaa dh8.la[;kIf the M.P. of jeans is Rs. 140 more than
the C.P. of Shirt and the difference between
fdruh Fkh\
M.P. and S.P. of Shirt is Rs. 780. Find the
(a) 270 (b) 135
discount % of Shirt?
(c) 120 (d) 150
4. The total number of votes increased by 50%
;fn thUl dk vafdr ewY; 'kVZ ds Ø; ewY; ls 14
from 2012 to 2016. If 16% of the votes were :i;s vfèkd gks vkSj 'kVZ ds vafdr ewY; rFkk fo
valid in 2016, what was the number of valid ewY; dk varj 780 :Ik;s gks rks 'kVZ ij NwV izfr'k
votes in 2016? dhft,\
 (a) 20.46% (b) 19.76%
(c) 18.86% (d) 17.56% Sizes Ratio of male of female in the farmers
9. If the discount % and profit % of the Suit who claimed insurance in 2017
A 05 : 03
is same. Find out the M.P. of Suit?
;fn lwV ij NwV izfr'kr vkSj ykHk izfr'kr leku gks rks B
C
04 : 01

lwV dk vafdr ewY; Kkr dhft,\ D 05 :

(a) Rs. 2200 (b) Rs. 2500 E 03 : 02
(c) Rs. 2800 (d) Rs. 3300
10. If the ratio of the C.P. of the Jeans and 11.
The number of farmers who claimed
Saree is 4 : 5, find out the discount % of insurance in state C exceeds the total
Saree? number of farmers (who chimed insurance)
;fn thUl vkSj lkM+h ds Ø;ewY;ksa dk vuqikr 4% 5 gks rks
in state E and state A together by 5280.
lkM+h ij NwV izfr'kr Kkr dhft;s\ What is the percentage of farmers who
(a) 10% (b) 12% claimed insurance in state E?
(c) 15% (d) 20% jkT; C esa chek dk nkok djus okys fdlkuksa dh la
Direction (11) : The following pie-chart and table jkT; E vkSj jkT;A esa dqy feykdj fdlkuksa dh la[;k
show the percentage distribution of farmers in (chek dk nkok djus okys) ls 5280 vf/d gSA jkT;
E
5 states A, B, C, D & E of a country who claimed
esa chek dk nkok djus okys fdlkuksa dk çfr'kr D;k
insurance due to drought in the year 2017 and
SSC CPO 2019
ratio of males to females of the number of
(a) 15% (b) 25%
farmers in states, respectively.
(c) 10% (d) 40%
fuEufyf•r ikbZ&pkVZ ,d ns'k ds 5 jkT;ksa A, B, C, D vkSj E
esa fdlkuksa ds çfr'kr forj.k dks n'kkZrh gS] ftUgksausDirection
o"kZ 2017 (12-15): The following pie chart
represents the percentage-wise distribution of
esa lw•s ds dkj.k chek dk nkok fd;k Fkk vkSj rkfydk jkT;ksa
300 students of class X in a school in six
esa] fdlkuksa dh la[;k esa iq#"kksa rFkk efgykvksa dh la[;k ls
different sections A, B, C, D, E and F.
vuqikr fn[kkrh gSA
Study the pie-chart and the table carefully and fn;k x;k ikbZ&pkVZ fdlh fo|ky; ds d{kk
X ds Ng
answer the following question. vyx&vyx oxks±A, B, C, D, E vkSjF ds 300 Nk=kksa
ikbZ&pkVZ vkSj rkfydk dk è;kuiwoZd vè;;u djsa vkSj
izfr'kr&okj caVu fu#fir djrk gSA
fuEufyf•r ç'uksa ds mÙkj nsaA
Note: Some date are missing in the pie-chart and
table, if required in any question, find the F
missing data first and then answer the question. 12% A
uksV% ikbZ&pkVZ vkSj rkfydk esa dqN frfFk;ka xk;c
(missing) 20%
gSa] ;fn fdlh ç'u esa vko';d gks] rks igys xk;c
(missing) E
B
MsVk •kstsa vkSj fiQj ç'u dk mÙkj nsaA 14%
Total number of farmers who claimed
18%
D C
insurance in the year 2017 = 2,64,000
17%
o"kZ 2017 esa chek dk nkok djus okys fdlkuksa dh dqy
la[;k ¾ 2]64]000
The given table shows the number of boys
of class X in six different sections A, B, C,
E A D, E and F.
24% nh xbZ rkfydk d{kk X ds Ng vyx&vyx oxks±A,
D B, C, D, E vkSjF esa yM+dksa dh la[;k n'kkZrh g
20%
B Section A B C D E F
No. of boys 36 26 34 28 .... 20
C
36% 12. The total number of boys in sections A, B
and D together is what percentage more
than the total number of girls in sections
A, B and D together?
 oxks±A, B vkSjD esa feykdj yM+dksa dh dqy la[;k]
14. The difference between the central angles
oxks±A, B vkSjD esa feykdj yM+fd;ksa dh dqy la[;k of the sectors corresponding to the
ls fdrus izfr'kr vf/d gS\ sections A and F is:
SSC CHSL 06/08/2021 (Shift- 3) vuqHkkxA vkSjF ds vuq:i f=kT;[kaM ds dsaæh; dk
ds chp dk varj Kkr djsa\
(a) 20%
SSC CHSL 13/04/2021 (Shift- 3)
(b) 30%
(a) 28.4° (b) 26.8°
(c) 18%
(c) 38.8° (d) 28.8°
(d) 15% 15. If in section E, the ratio of boys and girls
13. If the total number of students in section is 4 : 3, then the ratio of the number of
C is 57, then the total number of girls in girls in section B to that of the number of
section C and D together will be: girls in section E is:

;fn HkkxC esa dqy Nk=kksa dh57 gS] rks vuqHkkx ;fn oxZ E esa] yM+dksa vkSj yM+fd;ksa dk
la[;k
4 % 3 gS] rks oxZ
C vkSjD esa yM+fd;ksa dh dqy la[;k ------------- gksxhA
B esa yM+fd;ksa dh la[;k dk]Eox
esa yM+fd;ksa dh la[;k ls vuqikr Kkr djsaA
SSC CHSL 11/08/2021 (Shift- 3)
SSC CHSL 13/04/2021 (Shift- 1)
(a) 48 (b) 36 (a) 15 : 8 (b) 13 : 8
(c) 40 (d) 46 (c) 13 : 9 (d) 14 : 9

Answer Key (HISTOGRAM )

1.(a) 2.(b) 3.(a) 4.(a) 5.(a) 6.(c) 7.(c) 8.(c) 9.(c) 10.(c)

11.(d) 12.(d) 13.(b) 14.(b) 15.(d) 16.(a) 17.(c) 18.(c)

(MISSING DI)
1.(d) 2.(b) 3.(c) 4.(c) 5.(c) 6.(d) 7.(b) 8.(c) 9.(d) 10.(a)

11.(c) 12.(a) 13.(d) 14.(d) 15.(d)