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CONCEPT-01 20
(a) 20 (b)
(BODMAS Rule) 3
25
This rule is the basic principle of solving (c) (d) 25
algebraic or numerical expressions. 3
4. Simplify
;g fu;e chtxf.krh; ;k la[;kRed O;atdksa dks gy djus fuEufyf[kr dk eku Kkr djsaA
dk ewy fl¼kar gSA 2.5 × [144 ÷ 198 × {121 × 81 ÷ (11 × 9)}]
Order to solve/gy djus ds Øe SSC CGL 17/07/2023 (Shift-01)
r
(a) 180 (b) 175
B Bracket (dks"Bd)
(c) 185 (d) 190
si
O of (dk) 5. Simplify./fuEu dk eku Kkr djsaA
D Division an by
M Multiplication
(Hkkx)
(xq.kk)
325 + 276 ÷ [150 – {9 × 9 + (83 – 4 × 15)}]
SSC CGL 20/07/2023 (Shift-03)
n
(a) 332 (b) 333
A Addition (tksM+) (c) 334 (d) 331
ja (?kVko) 6. If (48 ÷ 72 × 3) – [15 ÷ 8 × (40 – 32) – 10] + 2P
R s
S Subtraction
= 6 ÷ 2, then find the value of P?
Types of Bracket & Solving order
a th
;fn (48 ÷ 72 × 3) – [15 ÷ 8 × (40 – 32) – 10] +
dks"Bd ds çdkj ,oa gy djus ds Øe
2P = 6 ÷ 2, rksP dk eku Kkr dhft,\
(i) Vinculum/Line/Bar bracket (js[kk SSC CPO 03/10/2023 (Shift-3)
ty a
dks"Bd) (a) 2 (b) 4
(c) 1 (d) 3
(NksVk dks"Bd)
di M
(ii) ( ) Small bracket
7. Find the value of the given expression.
(iii) { } Curly bracket (ea>yk dks"Bd) uhps fn, x, O;atd dk eku Kkr dhft,A
(iv) [ ] Square bracket (cM+k dks"Bd)
1 1 4 3 1 1
1. The value of 11 × 11 + 11 ÷ 11 – 11 × 11 + 11 4 3 3 3 1 5 3 4 1 2 1 3
+ 11 × 11 – 11 – 11 × 11 is:
2 5 2
11 × 11 + 11 ÷ 11 – 11 × 11 + 11 + 11 × 11 –
3 6 3
11 – 11 × 11 dk eku D;k gS\
SSC CPO 03/10/2023 (Shift-01) SSC CHSL 10/08/2023 (Shift-2)
(a) 121 (b) 0 3 3
(c) 11 (d) 1 (a) 11 (b) 10
8 8
A
2. Evaluate the following 5 – [96 ÷ 4 of 3 – (16 –
55 ÷ 5)]. 3 5
(c) 14 (d) 16
5 – [96 ÷ 4 of 3 – (16 – 55 ÷ 5)] dk eku Kkr dhft,A 8 8
(a) 0 (b) 3 8. Simplify the following expression
(c) 2 (d) 4
fn, x, O;atd ljy dhft,A
3. Simplify the given expression.
fn, x, O;atd dk eku Kkr dhft,A
25 – 16 – 14 – 18 – 8 3
18 ÷ 3 of 2 × 5 + 72 ÷ 18 of 2 × 3 – 4 ÷ 8 × 2
(a) 16 (b) 18
SSC CGL 14/07/2023 (Shift-4) (c) 15 (d) 20
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9. Simplify: 13. Simplify the given expression.
fuEu dk eku Kkr dhft,A fn, x, O;atd dk lyhdj.k djsaA
1 1 1 1 1 (326 222)2 (326 222)2
3 4 3 2 (326 222)
3 3 3 3 3
SSC CGL 18/07/2023 (Shift-01)
SSC CPO 04/10/2023 (Shift-01)
(a) 1 (b) 4
8 1 (c) 3 (d) 2
(a) – (b)
3 3 14. The value of/dk eku Kkr djasA
2 1 0.325 × 0.325 + 0.175 × 0.175 + 25 × 0.00455 0.5
(c) (d) +
3 3 5 × 0.0065 × 3.25 – 7 × 0.175 × 0.025 1.5
1 1 47 47 SSC CPO 24/11/2020 (Shift-1)
10. If A = 3 4 ÷ 34 – and B = 11
4 4 32 16 (a) (b) 3
3
1 1 11
2 5 ÷ 55 – then what is the value of 7
r
2 2 10 (c) 0 (d)
3
A – B?
si
15. The value of
1 1 47 47
;fn A = 3 4 ÷ 34 –
an by rFkkB = 4.669 4.669 – 9 (0.777)²
4 4 32 16 is (1 – k),
(4.669)² (2.331)² 14(0.667)(2.331)
1 1 11
n
2 5 ÷ 55 – gks] rks
A – B dk eku D;k gS\ where k = ?
2 2 10
4.669 4.669 – 9 (0.777)²
5 ja dk eku (1
R s
(a) (b) 1 (4.669)² (2.331)² 14(0.667)(2.331)
8
– k) gS] ftlesak = ?
a th
3
(c) 0 (d) SSC CPO 11/12/2019 (Shift-02)
8 (a) 0.666 (b) 0.647
CONCEPT-02 (c) 0.467 (d) 0.768
ty a
(A) (0.13)² (0.21)²
16. The value of
2
a² + b² = (a + b) – 2ab (0.39)² 81(0.07)²
di M
a² + b² = (a – b)2 + 2ab (2.4)4 3 (11.52) 9
lies between:/dk
(2.4)6 6(2.4)4 3 (17.28)
a² – b² = (a + b)(a – b)
eku fdlds chp fLFkrgS\
11. Simplify the following expression. SSC CPO 12/12/2019 (Shift-01)
fuEufyf[kr O;atd dk eku Kkr dhft,A (a) 0.7 and 0.8 (b) 0.4 and 0.5
(c) 0.6 and 0.7 (d) 0.5 and 0.6
7.35 7.35 2.25 2.25
0.24
(B)
a + b = (a + b)(a – ab + b2)
3 3 2
SSC CGL 27/07/2023 (Shift-3)
a3 – b3 = (a – b)(a2 + ab + b2)
(a) 204 (b) 320 17. Simplify the given expression.
A
(c) 225 (d) 304 fn, x, O;atd dk ljyhdj.k djsaA
12. Simpify:
432 432 247 247 432 247
fuEu dks ljy dhft,A 432 432 432 247 247 247
(379 276)2 (379 – 276)2 SSC CGL 19/07/2023 (Shift-01)
379 379 276 276 1 1
(a) (b)
SSC CHSL 11/08/2023 (Shift-2) 259 185
(a) 2 (b) 655 1 1
(c) (d)
(c) 103 (d) 1 679 450
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(0.83)3 – (0.1)3 23. Simplify/ljy djsa%
18. Simplify: =?
(0.83)2 0.083 0.01 (3.321)3 (2.681)3 (1.245)3
(3.321)2 (2.681)2 (1.245)2
(0.83)3 – (0.1)3
lehdj.k (0.83)2 0.083 0.01
=? 3 3.321 2.681 1.245
(3.321 2.681) (2.681 1.245) (1.245 3.321)
SSC CHSL 14/08/2023 (Shift-2) SSC CHSL 04/08/2023 Shift-01
(a) 0.98 (b) 0.27 (a) 6.125 (b) 8.645
(c) 0.93 (d) 0.73 (c) 7.247 (d) 10.245
24. Simplify the given expression.
19. The value of/dk eku Kkr dhft,A
(0.013)³ (0.007)(0.000049) (80 80 80) (70 70 70) (50 50 50) – 840000
6400 4900 2500 – 5600 – 3500 – 4000
(0.007)² 0.013(0.013 – 0.007)
fn, x, O;atd dks ljy dhft,A
SSC CPO 13/12/2019 (Shift-02) SSC CHSL 10/08/2023 (Shift-01)
r
(a) 0.07 (b) 0.02 (a) 100 (b) 200
(c) 0.06 (d) 0.04 (c) 400 (d) 300
si
25. Simplify the following.
675 675 675 325 325 325 fuEufyf[kr dk ljyhdj.k dhft,A
20. an by
67.5 67.5 32.5 32.5 – 67.5 32.5
to:
is equal
0.01 0.01 0.01 0.003 0.003 0.003
n
0.05 0.05 0.015 0.05 0.015 0.015
675 675 675 325 325 325
SSC CGL 24/07/2023 (Shift-3)
ja
67.5 67.5 32.5 32.5 – 67.5 32.5
R s
13 13
fuEufyf[kr esa ls fdlds cjkcj gS% (a) 103 (b) 10 3
25 15
a th
(a) 100 (b) 10,000
(c) 1,000 (d) 1,00,000 13 13
(c) 103 (d) 10 3
15 25
2513 2493
ty a
21. The value of CONCEPT-03
25.1 25.1 – 624.99 24.9 24.9
is 5 × 10k, where the value of k is ____. (Bar Type Questions/ckj okys iz'u
)
di M
26. Convert 0.7777......... into fraction
2513 2493 n'keyo 0.7777......... dks fHkUu esa cnysa
dk eku 5 × 10k,
25.1 25.1 – 624.99 24.9 24.9
7 7
gS] tgk¡
k dk eku ____ gSA (a)
9
(b)
3
(a) 4 (b) 5
7 77
(c) 3 (d) 6 (c) (d)
10 99
(C) 27. Convert 0.535353............into fraction
If a + b + c = 0 n'keyo 0.535353............ dks fHkUu esa cnysa
53 53
then a³ + b³ + c³ = 3abc (a)
99
(b)
49
What is the value of/dk eku Kkr dhft,A
A
22. 53 53
(c) (d)
100 59
0.74 1.23 0.13 28. Convert it into vulgar fraction
(0.37)3 (0.41)3 – 8(0.39)3
0.5 87
SSC CPO 11/12/2019 (Shift-01) CISF HCM 30/10/2023 Shift-01
–1 93 97
(a) (b) 1 (a) (b)
3 167 165
1 95 91
(c) –1 (d) (c) (d)
3 167 165
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29. Correct expression of 0.0654 . (the bar 35. Find the value of/dk eku Kkr dhft;s
indicates repeating decimal)
0.2 0.3 0.32
0.0654 dk lgh O;atd gS (ckj n'keyo dh iqujko`frÙk (a) 0.77 (b) 0.82
dks n'kkZrk gS)%
(c) 0.86 (d) 0.87
NTPC CBT-2 19/01/2017 (Shift-1)
36. Find the value of/dk eku Kkr dhft;s
654 654
(a) (b) 324.786 – 10.193
1000 10000
18 18 (a) 314.59345 (b) 314.59445
(c) (d)
275 277
(c) 314.59345 (d) 314.59445
30. Correct expression of 1.427 . (the bar indicates
37. If A = 0.312 , B = 0.415 and C = 0.309 , then
repeating decimal)
what is the value of A + B + C?
1.427 dk lgh ljyhdj.k gS (ckj n'keyo dh iqujko`fÙk
;fn A 0.312 , B 0.415 rFkk C 0.309 gS] rks
r
dks n'kkZrk gS)%
A B C dk eku fdruk gS\
si
NTPC CBT-2 17/01/2017 (Shift-3)
1211 1043
(a)
1427
1000
an by (b)
157
110
(a)
1100
(b)
1100
n
1427 157 1097 1141
(c) (d) (c) (d)
10000 111 1100 1100
ja Find the value of/dk eku Kkr dhft;s
R s
31. 2.8768 ? 38.
a th
878 9 22.4 11.567 – 33.59
(a) 2 (b) 2
999 10 SSC CGL TIER - II 11/09/2019
(a) 0.412 (b) 0.31
ty a
292 4394
(c) 2 (d) 2 (c) 0.412 (d) 0.32
333 4995
di M
32. Find the Value of x/x dk eku Kkr dhft;s 39. Find the value of/dk eku Kkr dhft;s
0.3 0.4 0.5 0.6 x 0.57 – 0.432 0.35
SSC CGL TIER - II 16/11/2020
(a) 3 (b) 5
(a) 0.494 (b) 0.498
(c) 2 (d) 8
33. The value of 0.56 0.43 0.89 is (c) 0.498 (d) 0.494
dk eku gS 40. Find the value of/dk eku Kkr dhft;s
0.56 0.43 0.89
NTPC CBT-1, 23/02/2021 (Shift-01) 0.47 0.503 – 0.39 0.8
SSC CGL TIER - II 13/09/2019
(a) 1.98 (b) 1.87
A
(a) 0.615 (b) 0.615
(c) 1.89 (d) 1.88
(c) 0.625 (d) 0.625
34. 3.245 1.234 2.12 is equal to: 41. Find the value of/dk eku Kkr dhft;s
3.245 1.234 2.12 cjkcj gS & 0.56 – 0.723 0.39 0.7
ICAR Mains, 08/07/2023 (Shift-3) SSC CGL TIER - II 12/09/2019
(a) 2.358 (b) 2.437 (a) 0.154 (b) 0.154
(c) 2.243 (d) 2.536 (c) 0.158 (d) 0.158
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42. Find the value of/dk eku Kkr dhft;s 48. Which of the following is the largest fraction?
fuEu esa ls lcls cM+h fHkUu dkSu lh gS\
2.4 0.6 3 0.16 0.27 0.83 0.16
8 6 4 13
SSC CGL TIER - II 15/11/2020 , , ,
9 11 9 15
(a) 0.814 (b) 0.11
8 6
(c) 1.1 (d) 1.36 (a) (b)
9 11
43. 2.75 3.78
4 13
(a) 1.03 (b) 1.53 (c) (d)
9 15
(c) 4.53 (d) 5.53 LCM Method
5 49. Find the greatest among
44. If 0. ab 0. ba , find the value of a + b.
9 fuEufyf•r fHkUuksa esa ls lcls cM+h fHkUu Kkr dhft,A
5
;fn 0. ab 0. ba ] rksa + b dk eku Kkr dhft;sA 1 5 3 6
9 , , &
2 7 4 7
r
(a) 5 (b) 6
(c) 7 (d) 8 1 5
si
7 (a) (b)
45. If 0.xy = , find x² + y² =? 2 7
11
;fn 0.xy =
7
11
an by
, rc x² + y² =? (c)
3
4
(d)
6
7
n
(a) 36 (b) 44 50. Find the smallest among
(c) 45 (d) 55
ja fuEufyf[kr esa ls dkSu&lk fHkUu lcls NksVk gS\
R s
CONCEPT-04
2 8 10 16
of fraction/fHkUuksa dh) rqyuk
, , &
a th
(Comparison 3 9 27 9
Cross Multiplication Method 2 8
(a) (b)
46. Which fraction among the following is the 3 9
ty a
least?
10 16
fuEufyf[kr esa ls dkSu&lk fHkUu lcls NksVk gS\ (c) (d)
di M
27 9
5 7 8 9
, , , Proper Fractions
11 12 13 17
SSC CGL MAINS (08/08/2022) Numerator of the fraction is less then
denominator or we can say value of the
8 5 fraction is less than 1.
(a) (b)
13 11 va'k dk eku gj ls NksVk gks vFkok fHkUu dk eku 1 ls
9 7 de gksA
(c) (d)
17 12 1 2 4 7 12
, , , , etc.
47. Find the greatest of the following fractions. 2 3 5 11 23
fuEufyf•r fHkUuksa esa ls lcls cM+h fHkUu Kkr dhft,A To compare/rqyuk ds fy,%
A
8 15 4 13 Step 1: Take the difference of Nr and Dr of each
, , ,
11 19 5 21 of the fractions. /izR;sd fHkUu ds va'k rFkk gj dk
CRPF HCM 23/02/2023 (Shift - 01) varj Kkr djsaA
Step 2: Difference must be same. If the given
13 15 difference is not same, make them same by
(a) (b)
21 19 taking LCM of each difference./varj leku gksuk
4 8 pkfg,A ;fn varj leku ugh gSa rks izR;sd varj dk y-l-i-
(c) (d)
5 11 ysdj mls leku dj ysaA
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Step 3: Fraction with smaller Nr will be least Step 2: Difference must be same. If the given
and fraction with greater Nr will be greatest./ difference is not same, make them same by
bl izdkj izkIr U;wure va'k okyh fHkUu lcls NksVh taking LCM of each difference./varj leku gksuk
rFkk
vf/dre va'k okyh fHkUu lcls cM+h gksxhA pkfg,A ;fn varj leku ugh gSa rks izR;sd varj dk y-l-i-
ysdj mls leku dj ysaA
Ex: Compare/rqyuk djsaA
Step 3: Fraction with smaller N r will be
4 6 13 11 greatest and fraction with greater Nr will be
, , ,
5 7 14 12 smallest./bl izdkj izkIr U;wure va'k okyh fHkUu
lcls
Ex: The greatest value among the fractions
cM+h rFkk vf/dre va'k okyh fHkUu lcls NksVh gksxhA
Ex: Compare/rqyuk djsaA
lcls cM+k vkSj lcls NksVk fHkUu Kkr djsaA
23 37
2 1 5 3 ,
, , , 18 32
7 3 6 4 Ex: Find smallest and greatest fraction
51. What is difference between the largest and the lcls NksVk vkSj lcls cM+k fHkUu Kkr djsaA
5 7 8 11 16 20 25 35
smallest fractions among , , and ? , , ,
r
9 11 15 17 15 19 24 34
Ex: Find smallest and greatest fraction
si
esa ls lcls cM+s vkSj lcls NksVs lcls NksVk vkSj lcls cM+k fHkUu Kkr djsaA
5 7 8 11
, , vkSj
9 11 15 17
an by
fHkUu dk varj D;k gS\
15 8 11 7
, , ,
16 3 12 8
n
CRPF HCM 24/02/2023 (Shift - 02)
Base Method
29 8
(a) (b) (i) When denominator is equal.
255 ja 99
R s
tc gj cjkcj gksA
1 17
a th
(c) (d) 2 14 9 25
45 165 , , ,
17 17 17 17
52. What is the difference of the largest and
smallest of the given fractions? Fraction with greater numerator will be
ty a
greatest
nh xbZ fHkÂksa esa ls lcls cM+h vkSj lcls NksVh fHkUu dk varj and vice-versa.
D;k gS\ cM+s va'k okyk fHkUu lcls cM+k gksxk vkSj blds foi
di M
Hkh lgh gksxkA
5 5 3 6 (ii) When Numerator is equal.
, , ,
11 7 8 11 tc va'k cjkcj gksA
SSC CHSL 13/03/2023 (Shift-01) 9 9 9 9
, , ,
17 19 4 7 10 13
(a) (b)
56 56 Fraction with smaller denominator will be
1 23 greatest and vice-versa.
(c)
7
(d)
56 NksVs gj okyk fHkUu lcls cM+k gksxk vkSj bldk foij
Improper Fractions lgh gksxkA
Numerator is greater than denominator or value (iii) If we increase N r and decrease D r , then
of the fraction is greater than 1. resultant fraction will be greater.
A
va'k dk eku gj ls cM+k gks vFkok fHkUu dk eku 1 ls vf/ ;fn ge N dks c<+krs gSaDvkSj dks ?kVkrs gSa] rks ifj.kkeh
r r
d gksA fHkUUk vf/d gksxkA
3 5
3 13 6 27 e.g. (i) <
, , , etc. 7 6
2 4 5 17
101 103
To compare//rqyuk ds fy,% (ii) <
236 234
Step 1: Take the differecne of Nr and Dr of each
339 347
of the fractions./izR;sd fHkUu ds va'k rFkk gj dk varj (iii) <
237 231
Kkr djsaA
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(iv) If we decrease Nr and increase Dr then resultant
247 228
fraction will be smaller. (iii) &
437 387
;fn ge Nr dks ?kVk,¡ vkSj
Dr dks c<+k,¡ rks ifj.kkeh
fHkUu NksVk gks tk,xkA – 19 < 10%
7 6
e.g. (i) >
11 13
247 228
101 99 Sol:
(ii) > 437 387
236 247
334 329
(iii) > – 50 > 10%
229 235
(v) If we increase N r and D r together or we
Here % decrease in Dr dominates.
decrease Nr and Dr together than the resultant
fraction can be increase or decrease or will Resultan fractions will be greater.
have no change that can be determined by 247 228
using percentage change.
437 387
r
;fn ge Nr vkSjDr dks ,d lkFk c<+krs gSa ;kNge r
vkSj
743 691
si
Dr dks ,d lkFk ?kVkrs gSa rks ifj.kkeh fHkUu c<+ ;k ?kV
(iv) &
829 789
ldrk gS ;k blesa dksbZ ifjorZu ugha gksxk ftls çfr'kr
an by
ifjorZu dk mi;ksx djds fu/kZfjr fd;k tk ldrk gSA
– 52 > 5%
n
123 137
e.g. (i) &
237 267
+14 ja
11% Sol: 743 691
R s
829 789
a th
123 137
Sol: – 40 < 5%
237 267
ty a
743 691
+30 13% 829 789
di M
Here % increase in Dr dominates.
CONCEPT-05
Resultant fraction will be smaller.
(Ladder fractions)
123 137
1
237 267 53. 1 =?
1
1
4
423 492 1
(ii) & 5
322 389
21 17
(a) (b)
+69 < 20% 17 13
23 23
(c) (d)
14 15
A
423 492 1
Sol: 54. The value of 1 +
322 389 1
1+
1
1+
1
+ 67 > 20% 1+
2
1+
Here % increase in Dr dominates. 3
Resultant fraction will be smaller. 21 17
(a) (b)
13 2
423 492 34 8
(c) (d)
322 389 21 5
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55. Find the value of x in the following equation: 59. What will the value after simplifying this
continued fraction?
fuEufyf•r lehdj.k esa x dk eku Kkr dhft,%
bl fujarj fHkUu dks ljy cukus ds ckn eku D;k gksxk\
1
1
1 6 x 2+
1 1 = 1
1 11 2 3+
1 1
1 2+
1 4
5 NTPC CBT-1, 03/02/2021 (Shift-02)
NTPC CBT-2 17/06/2022 (Shift-3) 43 43
(a) (b)
(a) 2 (b) 1 5 19
5 19
1 2 (c) (d)
(c) (d) 19 43
2 3 1 9
60. If A = , then the value of A is:
1 10
1 5 1
1
r
56. If = , then what is the value of x? 2
1 8 3
1+
si
1 1 9
1+ ;fn A , gS] rksA dk eku gS%
1 =
1+ 1 10
an by x 1
2
1
3
n
1 5 SSC CPO 03/10/2023 (Shift-3)
;fn 1
= , gks] rks
x dk eku D;k gS\
8
1+ 3 2
ja 1 (a) (b)
R s
1+ 10 5
1
1+
a th
x 1 1
(c) (d)
10 5
(a) 1 (b) 2
(c) 3 (d) 4 1
61. Simplify: 15 +
ty a
1
1 6+
57. Find the value of 1 – 1
1 8+
di M
1– 10
1
1– CRPF HCM 01/03/2023 (Shift - 02)
2
1–
3 81 71
(a) 15 (b) 15
496 186
2 1
(a) (b)
3 3 81 31
(c) 15 (d) 15
1 2 472 374
(c) (d)
3 3
1
62. 2+ =?
1 1
58. 1 =? 2–
1 1
A
1– 3–
1 1
1 4–
1 4
1–
1
1 CRPF HCM 28/02/2023 (Shift - 01)
3
41 15
1 11 (a) (b) 2
(a) (b) 67 41
2 7
41 15
3 9 (c) 2 (d)
(c) (d) 67 41
4 4
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63. Find the value of the following 1 17
68. find a + b + c + d =?
fuEufyf[kr dk eku Kkr dhft,A a
1 60
1
5 b
4– 1
1 c
1+ d
1
3+
1 (a) 11 (b) 12
2+
4
(c) 13 (d) 14
1 1
(a) (b) 1 30
4 8 69. find a + b + c + d = ?
1 43
2 3 a
(c) (d) 1
4 7 2b
1
3c
64. The value of/dk eku fdruk gksxk\ d
(a) 10 (b) 11
1 1
3 3
r
7 2 1 (c) 7 (d) None
is :
1 1 1
3 3 3
si
4 7 1 70. If a, b, c, d are integers such that
2
1
an by 2
2
a
1
1
29
154
, then a + b + c + d = ?
n
ICAR Mains, 10/07/2023 (Shift-2) 1
b
(a) 213.50 (b) 209.25 1
c
(c) 225.15 ja (d) 232.35 d
R s
65. 5 =?
a th
3 (a) 12 (b) 13
3
5
7 (c) 14 (d) 15
1
5
45 1
ty a
9 13 71. If , where a, b and c are
(a) (b) 53 1
13 9 a
1
di M
b
11 19 2
(c) (d) c–
2 5 5
Positive integers, then what is the value of
(4a + b + 3c)?
1
66. 2 =? SSC CGL TIER - II 15/11/2020
1
3
1 (a) 5 (b) 4
4
1
5 (c) 6 (d) 7
7 4
(a) (b) 72. If 1 29 , where x, y and y are
4 7
1 79
x
A
11 12 2
(c) (d) y
14 5 1
z
4
1 13
67. find a – b + c=? natural numbers, then the value
1 29
a (2x + 3y – z) is:
1
b
c SSC CGL TIER - II 16/11/2020
(a) 1 (b) 2 (a) 1 (b) 4
(c) 0 (d) 3 (c) 0 (d) 2
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CONCEPT-06 79.
1
1
1
+.....+
1
1 2 3 2 3 4 3 4 5 98 99 100
(Continuous Fraction Series/lrr fHkUukRed Js.kh
)
4949 1980
1 1 1 1 1 1
(a) (b)
73. 1 – 1 – 1 – 1 –
19800 49490
....
1 –
1 –
3 4 5 6 99 100
9898 1980
(c) (d)
2 1 19800 47490
(a) (b)
99 25 80. Which of the following statement is true?
1 1 1 1 1 1 5
(c) (d) I. .........
50 100 2 6 12 110 6
74. The sum of 1 1 1 1 7
II. .........
1 1 1 1 3 15 35 143 13
+ + + ..... +
2 6 12 n n +1 is: SSC CHSL 13/03/2023 (Shift-04)
(a) Only I
1 1 1 1 (b) Both I and II
r
+ +
2 6 12
+ ..... +
n n +1 dk ;ksx gS& (c) Only II
(d) Neither I nor II
si
NTPC CBT-1, 02/03/2021 (Shift-03) 1 1 1 1
81. If x = + + ......+ , y =
n 1
(a)
n
an by (b)
n 1
2n 1
12.13 13.14 14.15
1 1 1
23.24
x
n
+ + .......+ then is
n n 1 n 36.37 37.38 38.39 71.72 y
(c) (d)
2 n 1 equal to:
ja
R s
1 1 1 1 1 1
75. ....... (a) (b)
11 12 12 13 13 14 80 81 3 24
a th
69 70 1
(a) (b) (c) (d) 3
890 891 72
1 1 1
ty a
71 72 ......
(c) (d) 82.
790 891 1 3 5 3 5 7 9 11 13
di M
1 1 1 1 35 35
76. ....... (a) (b)
1 4 4 7 7 10 97 100 429 439
33 34 25 25
(a) (b) (c) (d)
100 99 329 329
1 1 1
35 37 83. ......
(c) (d) 123 4 23 4 5 6789
99 100
83 84
1 1 1 1 (a) (b)
77. +....+ 1512 1513
3 7 7 11 11 15 899 903
83 84
21 18 (c) (d)
(a) (b) 1415 1413
509 409
A
1 1
25 29 84. +.................+
(c) (d) 1 3 5 7 3 5 7 9
301 31
1
1 1 1 1
78. ....... 11 13 15 17
4 9 9 14 14 19 99 104
20 22
7 9 (a) (b)
(a) (b) 1991 1989
104 100
25 27
5 8 (c) (d)
(c) (d) 1990 1991
104 105
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1 1 1 1 1 1 1
85. + ....+ 91. 2 2 .....
4 11 18 11 18 25 18 25 32 2
2 –1 4 –1 6 –1 202 – 1
1 9 10
67 74 81 (a) (b)
19 19
425 425
(a) (b) 11 10
263736 253737 (c) (d)
19 21
424 425
(c) (d) 1 1 1 1
253737 253736 92. 2 2
2 2
2 2
....
7 –3 13 – 3 19 – 3 49 – 32
2
1 1
86. + + ..........+ 1 3
1 2 3 4 5 23456 (a) (b)
26 52
1
10 11 12 13 14 1 3
(c) (d)
13 26
10009 10009
(a) (b) CONCEPT-07
960960 960970
r
10019 10018 b
Types of numbers/izdkj dh la[;k,a)
si
(c) (d) (a
960961 960961 c
5 7
an by 9 11 13 (a) If denominator of a number same as multiplier
87. + + + + +
22.32 32.42 4 2.52 52.62 62.72
;fn fdlh la[;k dk gj mlds xq.kt ds leku gks rks
n
15 17 19
+ 2 2 + 2 is equal to. 95
72.82 8 .9 9 .102 93. The value of 99 99 is
99
1 ja 6
R s
(a) (b) (a) 9798 (b) 9997
100 25
(c) 9898 (d) 9896
a th
101
(c) (d) 1 98
100 94. 999 99 is equal to:
99
4 6 8 10 12 (a) 98999 (b) 99899
ty a
88. +
3 7 7 13 13 21 21 31 31 43
(c) 99989 (d) 99998
39 40
di M
(a) (b) 994
128 129 95. 999 999
999
41 42 (a) 908999 (b) 999099
(c) (d)
130 135 (c) 998995 (d) 989095
1 1 1 1 1 (b) If difference between numerator and denomi-
89. + +
1 3 5 1 4 3 5 7 4 7 579 nator is 1.
1 ;fn va'k vkSj gj ds chp dk varj 1 gks rks
+.....+ upto 20 terms
7 10
1 791
6179 6070 96. + 999 × 99
(a) (b) 8 792
15275 14973 (a) 89000 (b) 88900
(c) 95900 (d) 99000
A
7191 5183
(c) (d)
15174 16423 1 494
97. Find the value of 999 99
1 1 1 5 495
90. a1 , a2 , a3 t h e n , (a) 90000 (b) 99000
25 58 8 11
a1 + a2 + .... + a100 + ? (c) 90900 (d) 99990
25 30 1 692
(a) (b) 98. 999 99 is equal to:
151 157 7 693
1 9 (a) 1 (b) 99000
(c) (d)
4 55 (c) 99800 (d) 99900
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(c) Series Type/Js.kh izdkj • Sum of the cubes of first 'n' natural numbers.
1 2 4 izFken izkÑfrd la[;kvksa ds ?kuksa dk ;ksxiQy
99. 999 + 999 + ........... + 999
5 5 5 2
(n + 1)
(a) 3798 (b) 3998 1 + 2 + 3 +...........+n = n
3 3 3 3
2
(c) 3899 (d) 9939
• Sum of even integers/le iw.kkZadksa dk ;ksxiQy
100. Find the value of
2 + 4 + 6........... + 2n = n (n + 1)
1 2 3 4 • Sum of odd integers/fo"ke iw.kkZadksa dk ;ksxiQy
777 777 777 777
5 5 5 5
1 + 3 + 5 +............(2n – 1) = n2
1 2 3 4
777 777 777 777 dk eku Kkr dhft,A 101. What is sum of odd numbers from 1 to 50?
5 5 5 5
1 ls 50 rd fo"ke la[;kvksa dk ;ksx D;k gS\
NTPC CBT-1, 03/03/2021 (Shift-01)
(a) 625 (b) 650
(a) 3110 (b) 3018
(c) 667 (d) 670
r
(c) 3000 (d) 3108
102. What is sum of first 50 odd numbers?
si
CONCEPT-08 çFke 50 fo"ke la[;kvksa dk ;ksx D;k gS\
Some Standard Formulae for Addition
an by (a) 2500 (b) 2600
tksM+ ds fy, dqN lkekU; lw=k (c) 2700 (d) 2800
n
• Sum of first 'n' natural numbers 103. 72 + 8² + ....+ 12² =?
izFken izkÑfrd la[;k dk ;ksxiQYk (a) 459 (b) 559
ja
R s
n(n +1) (c) 567 (d) 570
1 + 2 + 3 +.........+ n =
2 104. Find the value of 21 + 222 + 232 ........ + 30²
2
a th
• Sum of the squares of first 'n' natural numbers. 212 + 222 + 232 ........ + 30² dk eku Kkr dhft,A
izFken izkÑfrd la[;kvksa ds oxksaZ dk ;ksxiQy (a) 6855 (b) 6585
n(n +1)(2n +1) (c) 5865 (d) 8565
ty a
12 + 22 + 32 +..........+ n2 =
6 105. 9³ + 10³ +.....+ 14³ + 15³
Sum of square of n odd/even number/n fo"ke@le
di M
• (a) 12104
n(n +1)(n + 2) (b) 12105
la[;kvksa ds oxksaZ =dk ;ksx where n is last (c) 13104
6
odd/even number/tgk¡n vafre fo"ke@le la[;k gSA (d) 14104
A
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ANSWER KEY
1.(d) 2.(c) 3.(a) 4.(a) 5.(d) 6.(d) 7.(d) 8.(a) 9.(a) 10.(d)
11.(a) 12.(a) 13.(b) 14.(a) 15.(a) 16.(c) 17.(c) 18.(d) 19.(b) 20.(d)
21.(a) 22.(a) 23.(c) 24.(b) 25.(d) 26.(a) 27.(a) 28.(b) 29.(c) 30.(b)
31.(c) 32.(c) 33.(c) 34.(a) 35.(d) 36.(c) 37.(d) 38.(c) 39.(c) 40.(d)
41.(a) 42.(c) 43.(c) 44.(a) 45.(c) 46.(b) 47.(c) 48.(a) 49.(d) 50.(c)
51.(a) 52.(b) 53.(c) 54.(c) 55.(a) 56.(b) 57.(c) 58.(d) 59.(d) 60.(d)
61.(a) 62.(c) 63.(b) 64.(b) 65.(d) 66.(b) 67.(a) 68.(c) 69.(c) 70.(c)
r
si
71.(a) 72.(d) 73.(c) 74.(d) 75.(b) 76.(a) 77.(c) 78.(c) 79.(a) 80.(d)
81.(d) an by
82.(a) 83.(a) 84.(b) 85.(a) 86.(a) 87.(b) 88.(b) 89.(b) 90.(a)
n
91.(d) 92.(a) 93.(d) 94.(a) 95.(c) 96.(d) 97.(b) 98.(b) 99.(b) 100.(a)
101.(a) 102.(a) ja103.(b) 104.(b) 105.(c)
R s
a th
ty a
di M
A
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SIMPLIFICATION/ljyhdj.k
(Practice Sheet With Solution)
1. What value should come in the place of 6. Simplify the following
question mark in the following equation? fuEufyf[kr dks gy dhft,A
fuEufyf[kr lehdj.k ds iz'uokpd fpÉ ds LFkku ij 3
dkSu&lk eku vkuk pkfg,\ 814 + [(20 ÷ 5 of 3 × 6)+{(8 ÷ 24 of 3)× 4} – 10
–2
(0.006 ÷ 0.01 ) + (0.008 ÷ ?)–(0.003 ÷ 0.03)
= 0.6
1 5
÷ 5] –
32
UPSI 1/12/2021 (Shift-01)
(a) 0.08 (b) 0.001 SSC CPO 09/11/2022 (Shift-01)
r
(c) 0.8 (d) 0.008 1 1
(a) 24 (b) 21
si
2. Find the value approximate to two decimals. 4 9
nks n'keyo rd vuqekfur eku Kkr dhft,A 4 4
an by
(44.6+346.33+3346.333+ 33346.3333) ÷ 50 = ?
UPSI 13/11/2021 (Shift-02)
(c) 27
5
(d) 29
9
n
7. Solve the following equation.
(a) 742.67 (b) 740.67
fuEufyf[kr lehdj.k dks gy dht,A
(c) 743.67 ja (d) 741.67
R s
3. What is the value of 123 × (162 – 142 – 40) ÷ 2 – 94 = ?
SSC CPO 09/11/2022 (Shift-03)
a th
7 5 14 10 10 (a) 17280 (b) 6561
× + ?
7– 5 14 – 10 5 (c) 10719 (d) 986
8. If 65% of 350 – ?% of 250 + 40% of 120 = 158,
7 5 14 10 10
dk eku D;k gS\
ty a
× + then find the value of ?
7– 5 14 – 10 5
;fn 350 dk 65% – 250 dk ?% + 120 dk 40%
SSC CGL MAINS (08/08/2022) = 158 gks] rks
? dk eku Kkr djsaA
di M
(a) 2 +1 (b) 2 2 +2 SSC CGL 19/07/2023 (Shift-03)
(c) 2+ 2 (d) 2 2 +1 (a) 57 (b) 63
4. What value should come in the place of (c) 47 (d) 54
question mark (?) in the following equation? 9. Simplify: [0.08 – {3.5 – 4.9 – (12.5 – 7.8 – 4.6)})
fuEufyf•r lehdj.k esa ç'u fpÉ (\) ds LFkku ij D;k [0.08 – {3.5 – 4.9 – (12.5 – 7.8 – 4.6)}) dk eku
eku vkuk pkfg,\ Kkr dhft,A
(0.006 ÷ ?) + (0.004 ÷ 0.04) + (0.03 ÷ 0.3) SSC CGL 20/07/2023 (Shift-04)
= 0.3 (a) 1.58
UPSI 14/11/2021 (Shift-03) (b) 0.08
A
(a) 0.006 (b) 0.6 (c) 2.58
(c) 0.001 (d) 0.06 (d) 12.58
5. If (48 ÷ 72 × 3) – [15 ÷ 8 × (40 – 32) – 10] + 2P 0.04
= 6 ÷ 2, then find the value of P 10 Find the value of of
0.05
;fn (48 ÷ 72 × 3) – [15 ÷ 8 × (40 – 32) – 10] +
2P = 6 ÷ 2, rksP dk eku Kkr dhft, 1 1 1 1
3 – 2 of 1
3 2 2 4
SSC CPO 03/10/2023 (Shift-3)
1 1 1
(a) 2 (b) 4 of
3 5 9
(c) 1 (d) 3
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15. Simplify/ljy dhft,A
1 1 1 1
3 – 2 of 1 3 3
–2744× –216
0.04 3 2 2 4
of 1 1 1 dk eku Kkr dhft,A 64
0.05 of 3
3 5 9 729
SSC CPO 03/10/2023 (Shift-01) SSC CPO 09/11/2022 (Shift-01)
(a) 5 (b) 0.4 (a) 164 (b) 512
(c) 3 (d) 0.03 (c) 189 (d) 156
11. The square root of is:/dk oxZewy gS% 16. Simplify/ljy dhft,A
1 1 1 1 36 3 76 19
456 – 76 +152 + of ÷
4 9 25 49 121 4 18 72 × 24
SSC CPO 04/10/2023 (Shift-01) SSC CPO 10/11/2022 (Shift-02)
(a) 443 (b) 256
11 1
(a) (b) (c) 356 (d) 401
12.60 1260
17. Simplify/ljy dhft,A
r
11 1260
(c) (d) 3 76 19
1260 11 456 – 76 +152 +
si
of ÷
12. The value of 4 18 72 × 24
1
2 3 1
5 4 4 2
an by
3 1
1 1 1
1 3 1
2 3 3
(a) 17
SSC CPO 10/11/2022 (Shift-03)
(b) 21
n
(c) 19 (d) 12
18. Simplify/ljy dhft,A
2 3 1 1 1 1
1 3 1 1 3 1
5 4 4 2 ja
2 3 3 dk eku 5 3 5 3
R s
– – – of 8.8 –1.2
fdruk gksxkA 8 8 8 8
a th
SSC CPO 04/10/2023 (Shift-01)
1 1 3 3
4 ÷2.5 × 2 ÷ of 60 + –
(a) 1 (b) 3 6 6 4 8
(c) 2 (d) 0 SSC CPO 11/11/2022 (Shift-01)
ty a
13. Simplify/ljy dhft,A 22 23
(a) 5 (b) 3
1 7 3 2 1 43 67
di M
2 ÷1 ÷ 9 ÷11 of 44 4
2 8 8 3 8 (c) 4 (d) 4
85 5
SSC CHSL 09/08/2023 Shift-04
19. Find the value of/dk eku Kkr djsaA
33 11
(a) (b) 8 7 1 1 2 1
135 135 6 ÷ of 1 +5 + ÷7
15 9 10 5 5 5
28 57 SSC CPO 11/11/2022 (Shift-02)
(c) (d)
135 135
25 5
14. Simplify/ljy dhft,A (a) (b)
16 14
1 1 4 3 1 1 25 5
4 3 3 3 1 5 3 4 1 2 1 3
A
(c) (d)
18 18
2 5 2 20. What is the positive value of the following
3 6 3 expression?
SSC CHSL, 10/08/2023 (Shift-2)
fuEufyf[kr O;atd dk /ukRed eku D;k gksxk\
3 3 25 × 4 ÷ 4 of
(a) 11 (b) 10
8 8 36 ÷ 15 of 2 of 29 – 8 – 11 ÷
3 5 9 × 5 ÷ 5 of 3
(c) 14 (d) 16
8 8 SSC CPO 11/11/2022 (Shift-02)
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25. Simplify the following expression.
5 1
(a) 1 (b) 1 fuEu O;atd dk eku Kkr dhft,A
6 5
72
67
48 36
12 5 71 [(51 4 13) (13 12 7)]
4 3 16
11
(c) 2 (d) 2
5 5 232
21. Simplify/ljy dhft,A SSC CGL 19/07/2023 (Shift-02)
–31 31
16+ 28 of 7÷22 (a) (b)
2 3 233 233
25 +8÷2 – 2 2 1
– 18 ÷12 of 41 31
8 (c) (d)
232 232
SSC CPO 11/11/2022 (Shift-03)
(a) 626 (b) 529 x 24
26. If 1+ = , then the value of x is:
(c) 721 (d) 579 529 23
r
22. Find the value of/dk eku Kkr djsaA x 24
;fn 1+ = gS] rks
x dk eku Kkr dhft,A
si
3 529 23
40 – of 32
4 SSC CPO 11/11/2022 (Shift-02)
3
37 – of (34 – 6)
4
an by (a) 15 (b) 27
n
(c) 47 (d) 30
SSC CPO 23/11/2020 (Shift-1) 27. Simplify the given expression.
(a) 1 ja (b) 0 y + 2x – [(y – (y – x + y) – (x + y) + y] – 2y.
R s
fn, x, O;atd dk eku Kkr dhft,A
1 1
a th
(c) – (d) y + 2x – [(y – (y – x + y) – (x + y) + y] – 2y.
2 2
SSC CGL TIER I 17/07/2023 (Shift-01)
23. Find the value of/dk eku Kkr djsaA (a) – y (b) – 2x
ty a
1 3 1 1 7 9 (c) Y (d) 2x
5 ÷ of ÷ 5 – 7 ÷ 9 28. What is the value of the given expression ?
4 7 2 9 8 20
di M
1 4 a 4 –5 × 4 a 2
11
× – 5 ÷ 2 of 15 × 4 a –22 × 4 a
21 2
fn, x, O;atd dk eku D;k gS\
SSC CPO 23/11/2020 (Shift-2)
4 a 4 –5 × 4 a 2
35
(a) 0 (b) 15 × 4 a –22 × 4 a
24
SSC CGL (PRE) 24/07/2023 (Shift-1)
15
(c) –2 (d) (a) 16 (b) 64
28
24. Find the value of/dk eku Kkr djsaA (c) 20 (d) 24
4 –2 –7 5
A
1 1 3 1 29. Arrangement of the fractions , , , in
5 ÷ 2 of + 5 ÷ of 3 9 8 12
2 4 7 2
ascending order is
1 7 9 11
÷ 5 – 7 ÷ 9 × 4 –2 –7 5
9 8 20 21 fHkUuksa
, , , dks vkjksgh Øe esa O;ofLFkr djuk gS
3 9 8 12
SSC CPO 24/11/2020 (Shift-1)
7 2 5 4 7 2 4 5
(a) – ,– , , (b) – ,– , ,
35 15 8 9 12 3 8 9 3 12
(a) (b)
24 28 2 7 5 4 2 7 4 5
(c) – ,– , , (d) – ,– , ,
(c) –2 (d) 8 9 8 12 3 9 8 3 12
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11 7 3
30. Sum of thre fractions is 2 . On dividing the (a) (b)
24 9 7
7
largest fraction by the smallest fraction is 5 7
6 (c) (d)
9 10
1
obtained which is greater than the middle 34. A fraction having denominator 30 and lying
3
fraction. The smallest fraction is 5 7
between and is
8 11
11
rhu fHkUuksa dk2;ksx gSA lcls cM+h fHkUu dks lcls
24 5 7
,d fHkUu ftldk gj 30 gS vkSj tks vkSj ds chp gS
7 8 11
NksVh fHkUu ls foHkkftr djus çkIr
ij gksrk gS tks fd
6 18 19
1 (a) (b)
eè; fHkUu ls vf/d gSA lcls NksVh fHkUu gSA 30 30
3
20 21
5 3 (c) (d)
r
(a) (b) 30 30
8 4
si
35. The greatest number among 0.7 0.16 ,
5 3
(c) (d)
6 7 0.6
31.
an by
Which of the largest of the following fraction?
fuEufyf•r esa ls dkSu lk fHkUu lcls cM+k gS\
1.02 –
24
, 1.2 × 0.83 and 1.44
n
0.6
8 3 0.7 0.16 , 1.02 – 24 , 1.2 × 0.83 vkS j
(a) (b)
11 ja 5
esa ls lcls cM+h la[;k
R s
1.44
11 2
(c) (d) (a) 0.7 0.16 (b)
a th
17 3 1.44
32. Which of the following fractions does not lie 0.6
(c) 1.2 × 83 (d) 1.02
5 8 24
between and ?
ty a
6 15
4 9
5 8 36. The least number among , , and (0.8)2 is
fuEufyf•r esa ls dkSu lk fHkUuvkSj ds chp ugha 9 49 0.45
di M
6 15
gS\ 4 9
, , vkSj(0.8)2 esa lcls NksVh la[;k gS
2 3 9 49 0.45
(a) (b)
3 4
4 9
4 6 (a) (b)
(c) (d) 9 49
5 7
(c) 0.45 (d) (0.8)2s
9
33. A fraction becomes
11
, if 2 is added to both 37. Find the value of/dk eku Kkr dhft,A
the numerator and the denominator. If 3 is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ..... ..... ... + 10 × 11
added to both the numerator and the (a) 4329 (b) 5826
A
5 (c) 4290 (d) 3815
denominator it becomes . What is the
6 38. If 5 3 + 75 = 17.32 then the value of
fraction?
14 3 + 108 is:
9
,d fHkUu gks tkrh gS] ;fn mlds va'k vkSj gj nksuksa ;fn 5 3 + 75 = 17.32 gS] rks14 3 + 108 dk
11
eku Kkr djsaA
esa 2 tksM+ fn;k tk,A ;fn va'k vkSj gj nksuksa esa 3 tksM+
SSC CGL 20/04/2022 (Shift-03)
5
fn;k tk, rks ;g gks tkrk gS fHkUu D;k gS\ (a) 32.46 (b) 35.64
6 (c) 34.64 (d) 33.86
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39. Find the value of/dk eku Kkr dhft,A 45. Simplify/ljy dhft,A
1 1 1 1 1 1 1 1 1 2 3 4 5 6
999 + 999 + 999 + 999 + 999 + 999
2 6 12 20 30 42 56 72 7 7 7 7 7 7
7 3 (a) 5997 (b) 5979
(a) (b)
9 4 (c) 5994 (d) 2997
6 8 999
(c) (d) 46. 999 7 is equal to/fdlds cjkcj gS%
5 9 1000
40. Find the value of/dk eku Kkr dhft,A
7 7
1 1 1 1 1 (a) 6993 (b) 7000
1000 1000
20 30 42 72 90
7 993
1 3 (c) 6633 (d) 6999
(a) (b) 1000 1000
10 5
Find the value of/dk eku Kkr dhft,A
r
47.
3 7
(c) (d)
995
si
20 20
999 × 999
41. Find the value of/dk eku Kkr dhft,A 999
1
1
1
1
1
an by
1
1
1
(a) 990809
(c) 999824
(b) 998996
(d) 998999
n
20 30 42 56 72 90 110 132
1 1 998
(a) (b) 48. 999 × 999 is equal to/fdlds cjkcj gSA
8 ja 7 999
R s
(a) 998999 (b) 999899
1 1
a th
(c) (d) (c) 989999 (d) 999989
6 10
42. Find the value of/dk eku Kkr dhft,A 49. Simplify/ljy dhft,A
1 1 1 1 1
ty a
1
15 35 63 99 143 2
1
2
di M
5 4 1
(a) (b) 2
39 39 1
2
5
2 7
(c) (d)
39 39 137 157
43. The simplified value of/dk ljyhÑr eku gSA (a) (b)
85 65
1 1 1 1 1 138 183
1 – 1 – 1 – ....... 1 – 1 – (c) (d)
3 4 5 99 100 72 95
2 1 50. Simplify/ljy dhft,A
(a) (b)
99 25
1
A
2
1 1 1
(c) (d) 1
50 100 1
3
1
44. The value of/dk eku gS 2
5
1 1 1 1
1 1 1 ....... 1 136 127
2 3 4 120 (a) (b)
49 36
(a) 30 (b) 40.5
153 189
(c) (d)
(c) 60.5 (d) 121 64 81
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51. Simplify/ljy dhft,A (a) 1 (b) 4
2 (c) 3 (d) 2
2
3 5 1
5
4 56. If A
3
and B
a tahen the value
1 1
3 3–
5 2 1
1– 2–
25 141 3 5
(a) (b) 7
11 69
of A + B
138 125
(c) (d)
60 65 5 1
52. Simplify/ljy dhft,A ;fn A 3 rFkkB 1 rksA + B dk
3 3–
9 2 1
3 1– 2–
3 3 5
2 7
7
1 eku D;k gS\
8
297 531 7 7
r
(a) (b) (a) (b)
54 65 3 6
si
413 217 13 8
(c) (d) (c) (d)
91 98 12 5
53.
1 1
an by
On simplification, the expression
57. If A = 1
1
and B
1
, then what
n
4 –2 1 2
7 7 1 1 3
1 1 1 1 1
3 1 2 1 2
2 7 ja 1 is equal to
9 2
R s
2 is the value of 19 (A + B)?
1
5–
5 1
a th
1
1 1 ;fn A 1 1 rFkkB rks19 (A +
4 –2 1 2
7 7 1 1 3
1 1
1 1 1 9 2
3 1 2 2
ty a
ljyhdj.k djus ij O;atd 2 7
2
1 ds B) dk eku D;k gS\
1
5– (a) 34 (b) 200
di M
5 (c) 30 (d) 25
cjkcj gS
1
28 24
(a) (b) 58. If A 1 then what will be the value
65 53 3
1
56 14 1
(c) (d) 1
53 65 2
4
54. The value of/dk eku gSA of 24A?
a 1
1– ;fn A rks 24A dk Ekku D;k gS\
1 1
1 3
a 1
1 1
1– a 1
A
2
(a) a (b) 1 – a 4
(c) 1 (d) 0 13
(a)
55. The value of/dk eku gSA 4
1 1
13
4 –2 (b)
7 4 1 40
1 1 1 13
3 1 2 (c)
2 7 1 12
2
1 13
5–
5 (d)
2
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59. What is the value of /Ekku D;k gS\ 65. Simplify the given expression.
1 fn, x, O;atd dks ljy dhft,A
1 (4.2)3 – 0.008
5–
1
3– ? (4.2)2 0.84 0.04
1
5– SSC CHSL, 14/08/2023 (Shift-2)
4
(a) 4 (b) –2
45 53
(a) (b) (c) 2 (d) –4
291 246 66. Simplify the expression:
48 96 fuEufyf[kr O;atd dks ljy dhft,A
(c) (d)
297 281 143 143 143 139 139 139
60. The value of/dk eku gSA 143 143 143 – 139 139 139
SSC CHSL 17/08/2023 (Shift-3)
(157 157) (157 133) (133 133)
(157 157 157) (133 133 133) 1
(a) (b) 282
2
r
SSC CPO 04/10/2023 (Shift-02)
1
si
1 (c) (d) 4
(a) 24 (b) 4
290
an by 67. The value of/dk eku Kkr djsaA
1
(c) 290 (d) 6.35 × 6.35 × 6.35 + 3.65 × 3.65 × 3.65
24
n
6.35 × 6.35 + 3.65 × 3.65 – 6.35 × 3.65
61. Simplify the given expression.
SSC CPO 23/11/2020 (Shift-1)
fn, x, O;atd dks ljy dhft,A
ja (a) 0.01 (b) 10
R s
0.09 0.09 0.04 0.04 0.16 0.16 2 0.09
(c) 1 (d) 0.1
0.04 2 0.04 0.16 2 0.09 0.16
a th
68. Simplify the given expression.
0.3 0.3 0.2 0.2 0.4 0.4
SSC CHSL 02/08/2023 Shift-04 a 2 – b2 – 2bc – c2
(a) 0.38 (b) 0.24 a 2 b2 2ab – c2
ty a
(c) 0.32 (d) 0.29 fuEufyf[kr O;atd dk eku D;k gS\
4913 343
di M
62. Simplify the given expression . a 2 – b2 – 2bc – c2
289 49 119
a 2 b2 2ab – c2
4913 343
fn, x, O;atd dks ljy dhft,A SSC CPO 03/10/2023 (Shift-02)
289 49 119
a– bc
SSC CHSL 03/08/2023 (Shift-02) (a)
a bc
(a) 24 (b) 26 a bc
(c) 22 (d) 20 (b)
a– b–c
63. Simplify/ljy djsa% a b–c
3.3213 2.6813 1.2453 3 3.321 2.681 1.245
(c)
a– b–c
3.3212 2.6812 1.2452 3.321 2.681 2.681 1.245 1.245 3.321
a–b–c
SSC CHSL 04/08/2023 Shift-01 (d)
A
ab–c
(a) 6.125 (b) 8.645
69. Simplify the following expression. (3x + 5)² + (3x
(c) 7.247 (d) 10.245
– 5)²
64. Simplify the given expression.
fn, x, O;atd dks ljy dhft,A fuEufyf[kr O;atd dk ljyhdj.k djsaA
(80 80 80) (70 70 70) (50 50 50) – 840000 (3x + 5)² + (3x – 5)²
6400 4900 2500 – 5600 – 3500 – 4000 SSC CGL 17/07/2023 (Shift-01)
SSC CHSL 10/08/2023 (Shift-01)
(a) 500x (b) 450x
(a) 100 (b) 200
(c) 9x² + 50 (d) 2(9x2 + 25)
(c) 400 (d) 300
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256x 4 16y 4 74 The value of 1801 ×1801 is:
70. Simplify 80x 2 20y 2 16 x 2 4y 2 .
1801 × 1801 dk eku --------- gSA
SSC CPO 09/11/2022 (Shift-02)
256x 16y4 4 (a) 3423601 (b) 3243601
80x 20y 16x 4y dks ljyhÑr dhft,A
2 2 2 2 (c) 2343601 (d) 3243106
75. Convert 0.18 into vulgar fraction
SSC CGL (PRE) 25/07/2023 (Shift-2)
0.18 dks vf'k"V fHkUu esa ifjofrZr djsa
1
(a) 5 (b) 17 18
20 (a) (b)
90 99
1 2 20 16
(c) (d) (c) (d)
5 5 99 90
76. Convert 0.43213 into vulgar fraction
s 2 + t2 + 2st - u 2
71. Simplify the expression , dks vf'k"V fHkUu esa ifjofrZr djsa
s 2 - t2 - 2tu - u 2 0.43213
provided (s + t + u) 0.
r
4316 4317
2 2 2
(a) (b)
s + t + 2st - u 9999 9990
si
O;atd , dk eku dhft,] ;fn (s + t +
s 2 - t2 - 2tu - u 2 3217 2553
an by (c) (d)
u) 0. fn;k x;k gksA 9990 9999
SSC CGL PRE 25/07/2023 (Shift-4) 77. The difference of 5.76 and 2.3 is
n
s+t– u s+t+u vkSj 2.3 ds chp varj gSA
(a) (b) 5.76
s–t–u s – t+u
ja (a) 2.54 (b) 3.73
R s
s–t–u s–tu
(c) (d)
s+t– u stu (c) 3.46 (d) 3.43
a th
72. Simplify
78. 0.142857 ÷ 0.285714 is equal to/ds cjkcj gSA
1 1 1 x (a) 10 (b) 2
+ + , when p =
2 + 2p 2 + 2q 2 + 2r yz
1 1
ty a
y z (c) (d)
2 3
q= and r =
z+x x +y
di M
x z
79. 0.11 0.22 × 3 is equal to/ds cjkcj gSA
y
;fn p = y z gS] vkSj
q= vkSjr =
x +y
gS rks (a) 3 (b) 1.9
z+x
1 1 1 (c) 1 (d) 0.3
+ +
2 + 2p 2 + 2q 2 + 2r
dks ljyhd`r dhft,A
80. The vulgar fraction of 0.39 is:
(a) 1 (b) x + y + c
0.39 dh vf'k"V fHkUu gSA
1
(c) 2 (d) 15 11
2 (a) (b)
33 39
x– y
73. Simplify the expression , where x = 2 17 13
x+ y (c) (d)
A
39 33
and y = 3.
81. The vulgar fraction of 2.3 49
x – y
;fn x = 2 vkSjy = 3 gS] rks x + y O;atd dks gy
2.3 49 dh vf'k"V fHkUu gSA
dhft,A
2326 2326
SSC CGL PRE 25/07/2023 (Shift-4) (a) (b)
999 990
(a) 2 6 – 6 (b) 6 –5
2347 2347
(c) 5 – 2 6 (d) 2 6 – 5 (c) (d)
999 990
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82. What value should come in the place of ques- 87. A and B have together three times what B and
tion mark (?) in the following question? C have, while A, B, C together have thirty
rupees more than that of A. If B has 5 times
fuEufyf•r ç'u esa ç'u fpÉ (\) ds LFkku ij dkSu & lk
that of C, then A has
eku vkuk pkfg,\
A vkSj B ds ikl dqy feykdj B vkSj C ls rhu xquk
0.537 – 0.335 0.234 = ? vf/d gS] tcfd A] B] C ds ikl dqy feykdj A ls
UPSI 14/11/2021 (Shift-02) rhl #i;s vf/d gSaA ;fn B ds ikl C ls 5 xquk vf/d
(a)
422
(b)
412 gS] rks
A ds ikl gSaA
990 990 (a) Rs 60 (b) Rs 65
442 432
(c) (d) (c) Rs 75 (d) Rs 45
990 990
83. Natu and Buchku each have certain number of 88. 252 m of pant cloth and 141 m of shirt cloth
orange. Natu says to Buchku. "If you give me are available in a cloth store. To stitch one
10 of your oranges. I will have twice the 1 3
number of oranges left with you" Buchku pant and one shirt, 2 m and 1 m of cloth
2 4
replies, "If you give me 10 of your oranges. I
r
will have the same number of oranges as left are needed respectively. Then the approximate
with you". What is the number of oranges with number of pants and shirts that can be made
si
Natu and Buchku, respectively? out of it are
ukVw vkSj cqpdw çR;sd ds ikl fuf'pr la[;k esa larjs gSaA ukVw]
an by ,d diM+s dh nqdku esa 252 ehVj iSaV dk diM+k vkSj 14
cqpdw ls dgrk gS- ;fn vki eq>s vius 10 larjs ns nsa rks esjs ehVj 'kVZ dk diM+k miyC/ gSA ,d iSaV vkSj ,d 'kV
ikl vkids ikl cps gq, larjksa ls nksxqus larjsa gksaxs cqpdw mÙkj
n
1 3
nsrk gS] ;fn vki eq>s vius 10 larjs ns nsaxs rks esjs ikl flyus ds fy, Øe'k%2 2 ehVj vkSj1 4 ehVj diM+sa dh
vkids ikl cps gq, larjksa ds leku la[;k gksxhA ukVw vkSj
ja vko';drk gksrh gSA fiQj blls cukbZ tk ldus okyh iSaV
R s
cqpdw ds ikl Øe'k% larjksa dh la[;k fdruh gS\
vkSj 'kVZ dh vuqekfur la[;k gS
(a) 50, 20 (b) 70, 50
a th
(c) 20, 50 (d) 50, 70 (a) (80, 100) (b) (100, 80)
84. In an exam the sum of the scores of A and B (c) (10, 90) (d) (90, 80)
is 120, that of B and C is 130 and that of C
89. There are 50 boxes and 50 persons. Person 1
ty a
and A is 140. Then the score of C is:
keeps 1 marble in every box. Person 2 keeps
,d ijh{kk esaA vkSjB ds vadksa dk ;ksx 120 BgS]
vkSj 2 murpbles in every 2nd box person 3 keeps
C ds vadksa dk ;ksx 130 gS vkSj
C vkSj A ds vadksa dk
di M
3 marbles in every third box. This process goes
;ksx 140 gSA Crks
dk vad gS% on till person 50 keeps 50 marbles in the 50th
(a) 65 (b) 75 box. Find the total number of marbles kept in
(c) 70 (d) 60 the 50th box.
85. A number is doubled and 9 is added. If the ogk¡ 50 cDls vkSj 50 O;fÙkQ gSaA O;fÙkQ 1 çR;sd fM
resultant is trebled, it becomes 75. What is 1 ekcZy j•rk gSA O;fÙkQ 2 çR;sd nwljs ckWDl esa 2 ek
that number?
j•rk gS] O;fÙkQ 3 çR;sd rhljs ckWDl esa 3 ekcZYl j•rk
,d la[;k dks nksxquk fd;k tkrk gS vkSj 9 tksM+k tkrk gSA
gSA ;g çfØ;k rc rd pyrh jgrh gS tc rd fd 50
;fn ifj.kke dks frxquk dj fn;k tk,] rks ;g 75 gks O;fÙkQ 50osa fMCcs esa 50 ekcZy u j• ysA 50osa fM
tkrk gSA og la[;k D;k gS\ j•s x, dapksa dh dqy la[;k Kkr dhft,A
(a) 6 (b) 35
A
(a) 43 (b) 78
(c) 8 (d) None of these
86. The sum of two number is 8 and their product (c) 6 (d) 93
is 15. the sum of their reciprocals 90. The sum of a two digit number and the number
nks la[;kvksa dk ;ksx 8 gS vkSj mudk xq.kuiQy 15 gSA
obtained by reversing its digits is a square
muds O;qRØeksa dk ;ksx number. How many such numbers are there?
8 15 nks vadksa dh la[;k vkSj mlds vadksa dks myVus ij çkIr la[;
(a) (b) dk ;ksx ,d oxZ la[;k gksrh gSA ,slh fdruh la[;k,¡ gSa\
15 8
15 (a) 5 (b) 6
(c) 23 (d)
8 (c) 7 (d) 8
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Answer Key
1.(a) 2.(d) 3.(a) 4.(d) 5.(d) 6.(d) 7.(c) 8.(c) 9.(a) 10.(c)
11.(c) 12.(a) 13.(c) 14.(d) 15.(c) 16.(a) 17.(c) 18.(c) 19.(c) 20.(b)
21.(d) 22.(a) 23.(c) 24.(d) 25.(d) 26.(c) 27.(d) 28.(a) 29.(a) 30.(b)
31.(b) 32.(d) 33.(a) 34.(b) 35.(b) 36.(b) 37.(c) 38.(c) 39.(d) 40.(c)
41.(c) 42.(a) 43.(c) 44.(c) 45.(a) 46.(d) 47.(b) 48.(a) 49.(b) 50.(a)
51.(c) 52.(a) 53.(c) 54.(d) 55.(a) 56.(b) 57.(a) 58.(d) 59.(b) 60.(d)
r
61.(d) 62.(a) 63.(c) 64.(b) 65.(a) 66.(c) 67.(d) 68.(d) 69.(d) 70.(c)
si
71.(a) 72.(a) 73.(d) 74.(b) 75.(a) 76.(b) 77.(d) 78.(c) 79.(c) 80.(d)
81.(b)
an by
82.(d) 83.(b) 84.(b) 85.(c) 86.(a) 87.(b) 88.(b) 89.(d) 90.(d)
n
ja
R s
a th
ty a
di M
A
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SOLUTIONS
1. (a) 6. (d)
0.006 0.008 0.003 Using the simplification rule BODMAS
= 0.6
0.01 x 0.03
20 5 of 3 6 –2
0.008 3 1 5
0.6 + – 0.1 = 0.6
x
81 8 24 of 3 4 –
4
32
0.008
=x –10 5
0.1
x = 0.08 –2
4
3 20 15 6 1 5 5
2. (d) 3 4 –
8 72 4 –10 5
2
Given that,
(44.6+346.33+3346.333+ 33346.3333) ÷ 50
r
3 20 8
3 6 4 – 2 – 4
37083.55963 15 72
741.67
si
50
3. (a) 4
7 5
14 10
an by
10
27 8
9
– 2 – 4
n
7 5 14 10 5
4 4
= 27 + 8 – 2 – 4 + =29
9 9
ja
7 5 2 35 14 10 2 140
R s
2 4 7. (c)
= 10
a th
Solve using the simplification rule BODMAS
5 123 × (162 – 142 – 40) 2 – 94
140 = 1728 × (256 – 196 – 40) 2 – 6561
= 6 35 6
2 = 1728 × 20 2 – 6561
ty a
2
= 17280 – 6561
= 6 35 6 35 2
= 10719
di M
8. (c)
= (36 – 35) + 2= 2 +1
65% of 350 – ?% of 250 + 40% of 120 = 158
4. (d) 227.5 + 48 – 158 = ? % of 250
4 × 8 + 3 = 50 – 30 ÷ 2 117.5 = ? % of 250
32 + 3 = 50 – 15 117.5
35 = 35 [satisfied] ?% =
250
(0.03 ÷ 0.3) = 0.3 117.5×100
(0.006 ÷ x) + 0.1 + 0.1 = 0.3 ?= = 47
250
(0.006 ÷ x) = 0.1 9. (a)
x = 0.06 [(0.08 – {3.5 – 4.9 – (12.5 – 7.8 – 4.6)}]
A
5. (d) 8
= – –1.4 – 12.5 –12.4
(48 ÷ 72 × 3) – [15 ÷ 8 × (40 – 32) – 10] + 2P = 6 ÷ 2 100
8
2 15
3 –
8 – 10 + 2P = 3 = – –1.4 – 0.1
3 8 100
8
2 – 5 + 2P = 3 = – –1.5
100
2P = 6
8
P = 3 = 1.5 = 0.08 1.5 = 1.58
100
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10. (c)
4 75 24
1 1 1 1 3 ÷ 8 × 35
3 3 – 2 2
2 of 1 4
0.04
of 1 1 1 4 45
0.05 of ÷
3 5 9
3 7
10 5 1 5
3 – 2 2 of 4 4 7
×
28
4 =
of 1 1 3 45 135
5
3 45 14. (d)
5 5 1 1 4 3 1 1
4 +3 ×1 ÷3 × 1 +1
4 6 8 4 4 45 3 3 5 4 2 3
of 16 of
5 5 3 16 2 5 2
45 3 ÷ 6 × 3
4 15
of =3
5 4 13 10 9 15 3 4
r
+ × ÷ × +
11. (c) 3 3 5 4 2 3
2 6 2
si
1 1 1 1 36 × ×
3 5 3
4 9 25 49 121
=
1 1 1 1 11
an by 13 10 9 4 17
+ × × ×
3 3 5 15 6
n
2 3 5 7 6
8
11 15
=
1260 ja
R s
12. (a) 13 68
a th
+ 133 15 133 5
2 3 1 1 1 1 3 15 = = = 16
3 1 8 15 8 8 8
1 1 3 1
5 4 4 2 2 3 3 15
ty a
15. (c)
7 15 5 1 3 10 4
–
5 4 4 2 2 3 3 Given,
di M
3
–2744 × 3 –216
7 15 5 1 3 10 3 =
–
5 4 4 2 2 3 4 3 64
729
7 15 5 1 15 3
– 14 × –14 × – 14
–
5 4 4 2 4 3
× – 6× – 6 × – 6
=
7 15 5 15 4×4×4
– 3
5 4 2 4 9×9×9
7 15 75 –14 × – 6
– =
5 4 8 4
9
7 2 5
A
– = =1 –14 × – 6 × 9
5 5 5 =
4
1 7 3 2 1 = 189
13. (c) 2 2 1 8 9 8 11 3 of 8 ?
16. (a)
5 8 75 35 1 3 76 19
2 ×15 ÷ 8 ÷ 3 of 8 456– 76+152 + of ÷
4 18 72×24
4 75 35 19 19
3 ÷ 8 ÷ 24 = 456– 76+225 + ÷
6 72×24
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19 72×24 98 10 2 5
= 456 – 301+ 6 × 19 = × + ×
15 49 5 36
= 456 – 301 + 288 4 1 24 +1 25
= + = =
= 744 – 301 = 443 3 18 18 18
17. (c) 20. (b)
2 25 × 4 ÷ 4 of
1 7 16 8 2 6×2 +6×
2
36
36 ÷15 of 2 of 29 – 8 –11
÷ 9 × 5 ÷ 5 of 3
2
= 8 1 6 4 6
6
25 × 4 ÷ 4 of
2
= 9+ 30× 36 ÷ 15 of 2 of 29 + 3 ÷
6 =
45 ÷ 15
= 9 + 10 = 19
18. (c)
100 ÷ 4 of
r
5 3 5 3 = 36 ÷15 of 2 of 29 + 3 ÷ 3
– – – of 8.8 –1.2
si
8 8 8 8
1 1 3 3 = 36 ÷15 of 2 of 100 ÷ 4 of 30
4 2.5 2 of60 –
an by
6 6 4 8
= 36 ÷15 of 2 of 100 ÷120
n
5 3 2
8 – 8 – 8 of 8.8 – 1.2 5
= 36 ÷15 of 2 of
= 6
25 ja 6 – 3
÷ 2.5 × 2 ÷10 +
8
R s
6 36 6 1
= = =1
25 5 5
a th
5 1
8 – 8 of 8.8 –1.2 21. (d)
=
25 10 1 3 16+ 28 of 7÷22
× × 2× +
6 25 10 8 2 3
25 +8÷2 – 2
ty a
2 1
–
18 ÷12 of
1 8
of 8.8 – 1.2
di M
2 4.4 – 1.2
= 1 3 = 16+ 28 of 7÷4 –
+ 1 3
3 8 +
3 8 625+8÷8– 1
= 324÷144 of
3.2 × 24 44
8
= 4.4 – 1.2 = =4
8+9 17 85 = [625 + 1 – {16 + 49 – (324 18)}]
24 = [626 – {16 + 49 – 18}]
19. (c) = 626 – 47 = 579
22. (a)
8 7 1 1 2 1
6 ÷ of 1 + 5 + ÷ 7
15 9 10 5 5 5 3
40 –
of 32
4 =?
8 7 1 1 2 1 3
37 – of 34 – 6
A
= 6 ÷ of 1 + 5 + ÷ 7
4
15 9 10 5 5 5
98 7 11+ 52 2 36 3
40 –
× 32
= 15 ÷ 9 of 10 + 5 ÷ 5 ⇒ 4
3
37 – × 28
4
98 7 63 2 36
= ÷ of + ÷
15 9 10 5 5 40 – 24
37 – 21
98 49 2 36
= ÷ + ÷ 16
15 10 5 5 =1
16
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23. (c) 26. (c)
1 3 1 1 7 9 11 1
5 ÷ of ÷ 5 – 7 ÷ 9 5 ÷ 2of x 24
4 7 2 9 8 20 × 21 – 2 1
529 23
=?
Squaring both sides
21 3 1 46 63 189 11
= ÷ of – ÷
4 7 2 ÷ 9 8
×
20 21
– (5 ÷ 1) x 576
1
529 529
21 14
× 529 x 576
4 3 11
×
–5
46 63 20 21 529 529
– ×
9 8 189 x = 576 – 529
49 x = 47
2 11 27. (d)
= × –5
184 – 30 21
36 y + 2x – [(y – (y – x + y) – (x + y) + y]-2y
r
y + 2x –(y – y + x –y) – x – y + y] – 2y
49 × 36 ×11
= –5 y + 2x – [x – y – x – y + y] – 2y
2 ×154 × 21
si
=3–5=–2 y + 2x + y – 2y = 2x
24. (d)
1
an by
1 3 1 1 7 9 11
28. (a)
4a 4 – 5 4a 2
n
5 2 of + 5 of ÷ 5 – 7 9 ×
2 4 7 2 9 8 20 21 15 4a – 22 4a
=?
ja Put, a = 1
R s
21 3 46 63 189 11
5+ ÷ ÷ – ÷ ×
4 14 9 8 20 21 45 – 5 43
=
a th
15 4 – 4 4
49 46 63 20 11
5+ ÷ – × ×
2 9 8 189 21 1024 – 320 704
= = = 16
60 – 16 44
49 46 5 11
ty a
5+ ÷ – ×
2 9 6 21 29. (a)
di M
49 184 – 30 11 4 5
5+ ÷ × Check positive fraction 3 48 > 15
2 36 21 12
49×36 11 –2 –7
5+ × Check negative fraction –16 > – 63
2×154 21 9 8
5 + 3 = 8
–7 –2 5 4
25. (d) Required ascending order
8 9 12 3
72 30. (b)
67
48 36
12 5 71
[(51 4 13) (13 12 7)] Let the three fraction are p, q and r where p > p > r
16 11
p 7 7 1 7–2 5
232 and q –
A
Given,
r 6 6 3 6 6
85 + 4 × –29
232 11
p+q+r= 2
24
85 – 4 × 29
232
7 5 59
85 – 116 r+r+ +r=
6 6 24
232
–31 13r 59 5 39 39 6 3
– r=
6 24 6 24 24 13 4
232
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31. (b) 36. (b)
Check two fraction by cross multiplication 4 3 45 5 4 16
2
Number are , , ,
2 3
9 7 99 11 5 25
10 > 9
3 5
4 3
2 8 Compare 9 28 27
22 < 24 7
3 11
8 11 3 5
136 > 121 33 35
11 17 7 11
8
greatest is 3 16
11 75 112
7 25
32. (d)
Check option by option 3
So is least number
7
5 2 8 5 2 8 37. (c)
(a)
6 3 15 6 3 15
r
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5+.........10 × 11
We know,
si
5 4 8 5 4 8
(b)
6 5 15 6 5 15 n n 1n 2n 3
an by Sum =
4
5 3 8 5 3 8 then,
n
(c)
6 4 15 6 4 15
10 11 12 13
4
5 6 ja 8 8 6
R s
(d) does not lie between 10 × 11 × 39
6 7 15 15 7
390 × 11
a th
5 18 4290
and
6 15 38. (c)
33. (a) 5 3 + 75 = 17.32
ty a
Solve the question by option 5 3 + 5 3 =10 3 =17.32
di M
7 2 9 3 = 1.732
9 2 11
14 3 + 108 = 14 3 + 6 3
7 3 10 5 = 20 3 = 201.732 = 34.64
9 3 12 6
39. (d)
34. (b)
1 1 1 1 1
Multiply by 30 in both fractions .......
1 2 2 3 3 4 4 5 89
5 75 3 7 210 1 9 –1 8
30 18 and 30 19 Sum =
8 4 4 11 11 11 9 9
So, option b lying between fractions 40. (c)
A
35 (b) 1 1 1 1 1 1
20 30 42 56 72 90
0.7 + 0.16 = 0.7 + 0.4 = 1.1
Or
0.6
1.02 – = 1.02 – 0.025 = 0.995 1 1 1 1 1 1
24
4 5 5 6 6 7 7 8 8 9 9 10
1.2 × 0.83 = 0.996
10 – 4 6 3
1.44 1.2 , Hence 1.44 is greatest Sum =
40 40 20
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41. (c) 47. (b)
1 1 1 1 1 1 1 1
995
20 30 42 56 72 90 110 132 999 × 999
999
Or
1 1 1 1 1 1 1 995
999
4 5 5 6 6 7 7 8 8 9 9 10 10 11 999
1 995 4
11 12 = 999 999 × 999 = 999000
–4
998996
12 – 4 8 1
= 48. (a)
48 48 6
42. (a) 998
999 999
999
1 1 1 1 1
15 35 63 99 143
998 1
= 999 –1 998999
× 999 = 999000
Or 999
r
1 1 1 1 1
49. (b)
si
3 5 5 7 7 9 9 11 11 13
10 1 5an by 1 1 27 157
Sum = 2 2 2
39 2 39 1 11 65 65
2 2
5 27
n
43. (c) 2
11
1 1 1 1 1
1 – 1 – 1 – ....... 1 –
ja 1–
3 4 5 99 100 50. (a)
R s
2 3 4 98 99 1 1 38
a th
= ....... 2 2 2
3 4 5 99 100 1 11 49
1 1
5 38
1 1 3
=2× = 11
100 50
ty a
44. (c) 136
1 1 1 1 49
di M
1 1 1 ....... 1
2 3 4 120 51. (c)
3 4 5 121 2 29 138
= ..... 2 2
2 3 4 120 15 60 60
5
9
1 121
= 121 60.5
2 2 52. (a)
45. (a)
9 9 15 162 135 297
1 2 .... 6 3 3
999 × 6 + 24 54 54 54
7 2
15
= (1000 – 1) 6 + 3 = 5997
53. (c)
A
46. (d)
999
999 7 29 15 14
1000 – 1 1
7 7 7
7 × 999 = 7 × (1000 – 1) = 6993 7 8 1 49 16 24
2 2
2 7 5 14 53
6993 2
6993 24
1000
993 993 2 14 53 28 130 56
= 6993 + 6 = 6999 =
1000 1000 65 106 24 65 53 53
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54. (d)
1 1 1 1 5
B=
a 2 2 4 19 19
1– 3 3 3
1 1 5 5 5
1– 2
a 2 2
1
1– a
29 5
19 (A + B) 19 34
a a 19 19
1– 1–
1–
1– a 1 –1 a = 1–1 = 0 58. (d)
1– a a
1 1 13
55. (a) A=
1 9 48
3 3
4 13
29 9 1
– 9
7 4 1 53 14 1
7 8 1 28 65 24 13
2 2 then, 24A =
2 7 5 53 2
2
24
59. (b)
r
53 53 1 1 1
si
= =1
130 130 1 1 19
5– 5– 5–
1 4 53
56. (b)
5
an by 3–
5–
1
4
3–
19
n
A=
3
3 53 53
2
1–
ja
3
265 – 19 246
R s
60 (d)
Let,
a th
5
= A = 157 and B = 133
3 5 5
3
1 = 3 9 = 12 Given that,
3
A 2 B2 AB 1 1 1
ty a
= =
A 3 – B3 A – B 157 – 133 24
1 1 1
B= 61 (d)
di M
1 1 1
3– 3– 3– Let,
1 7 3
2– 2–
5 5 5 A = 0.09, B = 0.04, C = 0.16
7 Given that,
A2 B2 C2 2AB 2BC 2CA
1 1 3
ABC
5 4 4
3–
3 3 2
A B C
= A + B + C = 0.09 + 0.04 + 0.16 =0.29
5 3 ABC
A+B
12 4 62. (a)
5 9 14 7 4913 + 343
A
= = 289 + 49 – 119
12 12 6
3 3
57. (a) 17 7
= 2 2
1 1 1 17 7 17 7
A = 1+ 1 1
1 1 9
1 1 1 a 3 b3
1 10 10
1 = a 2 b2 ab
9 9
10 29 = (a + b)
1
19 19 = (17 + 7) = 24
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63. (c) 69. (d)
2
3 3 3
3.321 2.681 1.245 3 3.321 2.6811.245
2 2
(3x + 5)² + (3x – 5)²
3.321 2.681 1.245 3.321 2.681 2.6811.245 1.245 3.321
2[(3x)2 + (5)2]
This is the formula
= 2[9x2 + 25]
a 3 b3 c 3 – 3abc 70. (c)
a 2 b2 c 2 – ab – bc – ca = (a+b+c)
Put, y = 0
= (3.321 + 2.681 + 1.245) = 7.247
64. (b) 256x 4 1
80x 2 16x 2 5
80 80 80 70 70 70 50 50 50 840000
6400 4900 2500 5600 3500 4000 71. (a)
3 3
803 + 70 + 50 – 3×80×70×50 s 2 t 2 2st u 2
22 2 s 2 t 2 2 tu u 2
80 70 + 50 – 80×70 –70×50–50×80
= (80 + 70 + 50) = 200 s t 2 u2
65. (a) 2 2
s t u
r
4.23 0.008
si
4.23 0.84 0.04 s t us t – u
s t us t u
4.23 - 0.23
4.22 +4.2×0.2+ 0.22
an by =
stu
n
stu
= 4.2 – 0.2 = 4 72. (a)
66. (c)
ja
R s
1 1 1 1 1 1
143×143+143×139+139×139 x = y = z 1, P , q ,r
11 2 11 2 11 2
143×143×143-139×139×139
a th
Let a = 143, b = 139 1 1 1
2
a +ab+b 1 2 2 2p 2 2q 2 2r
=
a 3 –b3 a–b
ty a
1 1 1 3
1
1 1 2 1 2 1 2 1 3
= =
143 –139 4
di M
73. (d)
67. (d)
x y
6.35×6.35×6.35+3.65×3.65×3.65
x y
6.35×6.35+3.65×3.65–6.35×3.65
a 3 + b3 2 3 2 3
= a + b =
a + b2 – ab
2
2 3 2 3
a = 6.35
b = 3.65 5–2 6
= =2 6 –5
1 1000 –1
10000
635 + 365
10000
= 0.1
74. (b)
68. (d)
A
1801 × 1801 = (1801)2
2
2
a 2 – b2 – 2bc – c 2 a b c = (1800 + 1)2
2 2 2
2
a b 2ab – c a b – c2 = 18002 + 2 × 1800 × 1 + 12
= 3240000 + 3600 + 1
a – b c a b c
= 3243601
a b – c a b c
75. (a)
a b c
18 –1 17
a b c 0.18
90 90
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76. (b) 85. (c)
43213 – 43 43170 4317 Let number be x
0.43213
99900 99900 9990 Given, 3(2x + 9) = 75 x = 8
77. (d) 86. (a)
5.7676 Let the number be x and y
–2.3333 Given, x + y = 8 and xy = 15
= 3.43
3.4343 Divide both equation
78. (c) x y 8
xy 15
0.142857 0.285714
1 1 8
=
142857 285714 1
y x 15
999999 999999 2
79. (c) 87. (b)
r
Given, A + B = 3 (B + C) ........(i)
0.11 0.22 3 A + B + C = A + 30 ..........(ii)
si
11 22 33 B=5C ............(iii)
= O
an by × 33 = 1
99 99 99 Fom (ii)
80. (d) B + C = 30
n
39 13 5C + C = 30 (put B = 5C)
0. 39 0.39
ja
99 33 6C = 30 C = 5
R s
81. (b) B = 30 – 5 = 25
a th
From (i)
349 – 3
2.3 49 2 A + 25 = 3 × 30, A = 90 – 25 = Rs 65
990
88 (b)
346 1980 346 2326
ty a
2
990 990 990 252 252 2
Number of pants = 100
82. (d) 1 5
di M
2
2
0.537 – 0.335 0.234
537 5 335 3 234 2 141 4
Number of shirts = 80
990 990 990 7
532 332 232 432 89. (d)
=
990 990 990 990
Factor of 50 = 1, 2, 5, 10, 25, 50
83. (b)
Above person will be kept marbles
Let the number of oranges of Natu and Backku
Total marbles = 1 + 2 + 5 + 10 + 25 + 50 = 93
be x and y respectively
Given, x + 10 = 2 (y – 10) x – 2y = – 30........(i) 90 (d)
y + 10 = x – 10 x – y = 20 ..........(ii) Let the number be 10x + y
A
(ii) – (i) y = 50 Then sum of number and number obtained by
reversing the digits
Hence, x = 20 + 50 = 70
= 10x + y + 10y + x
84. (b)
Given, A + B = 120 = 11 (x + 9)
B + C = 130 If x + y = 11 then number will be perfect square
possible pairs are = (2, 9), (3, 8), (4, 7) and (5, 6)
C + A = 140
8 numbers are possible
On adding, 2(A + B + C) = 390 A + B + C =195
C = (A + B + C) – (A + B) = 195 – 120 = 75
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SURDS AND INDICES ( ?kkrkad vkSj dj.kh
)
(CLASSROOM SHEET)
6. 4489 – 472 + 392 = ?
TYPE-01
CRPF HCM 28/02/2023 (Shift- 01)
BASIC QUESTIONS (a) 2 (b) 6
1. If x = 64 + 121 – 169 , then find the value of x². (c) 1 (d) 4
7. Solve/gy djsa
;fn x = 64 + 121 – 169 gS] rksx² dk eku Kkr djsaA
r
SSC CGL TIER- II 07/03/2023 3
21 59 16 3 722 49 ?
si
(a) 16 (b) 25
(a) 4 (b) 5
(c) 36 (d) 49
a n by
(c) 6 (d) 7
2. Find the value of
n
625 625 625 8. If A = 8 3
(81²)³ and B = 6 3
(34 )³ 6 6 (3²)6 then
625 + + +
100 10000 1000000 what will the value A + B?
ja
R s
625 625 625
625 +
100
+
10000
+
1000000
dk eku Kkr ;fn A = 8 3 (81²)³ rFkk B = 6 3 (34 )³ 6 6 (3²)6 rks
a th
dhft,A A + B dk eku D;k gS\
SSC CHSL MAINS 10/01/2024 (Shift-01) (a) 9 (b) 4
(a) 27.025 (b) 27.775 (c) 6 (d) 3
ty a
(c) 27.075 (d) 27.75
3 3
3 3 3 3
3. Find the value of the given expression. 9. Find the value of 343 216
di M
fn, x, O;atd dk eku Kkr dhft,A
3 3
8 + 1681
3 3
343 3 3
216 dk eku Kkr dhft,A
SSC CGL TIER- II 02/03/2023 SSC CHSL MAINS 10/01/2024 (Shift-01)
(a) 5 (b) 6 (a) 42 3
(b) 56
(c) 4 (d) 7
(c) 42 (d) 3
42
4. Find the value of : (9 (36 (144 625))) (0.07)² (0.24)² 25
10. What is the value of
dk eku Kkr dhft,A dk eku D;k gS\
(a) 6 (b) 4 (a) 2.5 (b) 0.1
(c) 0.25 (d) 0.01
A
(c) 5 (d) 3
5. Simplify (solve) the following. 11. What is the value of 3
0.02 0.000049
fuEufyf[kr dks ljy (gy) dhft,A dk eku D;k gS\
(a) 3 (b) 0.003
10 + 25 + 108 + 154 + 225 (c) 0.03 (d) 0.3
16 +19.25 × 4² 12. Which of the following statement(s) is/are
TRUE?/fuEufyf[kr esa ls dkSu&lk@ls dFku lR;@gSgS
SSC CGL TIER- II 06/03/2023
I. 121 12321 1234321 = 1233
7 1
(a) (b) II.
18 9 0.64 64 36 0.36 15
2 5 (a) Only I (b) Only II
(c) (d) (c) Neither in or II (d) Both I and II
9 18
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13. 104.04 + 1.0404 + 0.010404 is equal to 0.0004 3 0.000008
; fn A = v kSjB =
104.04 + 1.0404 + 0.010404 d s cjkcj gS 16000 3 125000 4 810
4
(a) 0.306 (b) 0.0306
3
(c) 11.122 (d) 11.322 0.729 4 0.0016
gS] rks
A × B D;k gS\
14. If 0.16
5329 = 73 , then value of
SSC CGL MAINS (08/08/2022)
5329 + 53.29 + 0.5329 + 0.005329
(a) 5 × 10–8 (b) 7 × 10–7
+ 0.00005329 is
7
;f n 5329 = 73 ] rks
–8
(c) 10 (d) 6 × 10–8
4
5329 + 53.29 + 0.5329 + 0.005329 d k eku gS
0.324 0.081 4.624
+ 0.00005329 19. is
r
1.5625 0.0289 72.9 64
(a) 81.1003 (b) 81.0113
(a) 2.4 (b) 24
si
(c) 81.1103 (d) 81.1013
(c) 0.024 (d) 0.24
80 – 112
a n by
15. The value of is: 20. What is the value of
45 – 63
29.16 0.2916 0.0036
n
80 – 112 ?
dk eku gS% 1.1664 116.64 0.36
45 – 63
ja
R s
29.16 0.2916 0.0036
(a)
3
(b) 1
3 dk eku D;k gS\
1.1664 116.64 0.36
4 4
a th
1 7 SSC CGL MAINS (08/08/2022)
(c) 1 (d) 1
3 9 101 103
(a) (b)
20 20
ty a
72 363 175
16. The value of is. 26 27
32 147 252 (c) (d)
di M
5 5
dk eku Kkr djsaA
1.24 2.79
55 45 21. Find the value of
(a) (b) 2.64 5.94
42 56
1.24 2.79
45 55 d k eku Kkr dhft,A
(c) (d) 2.64 5.94
28 28 SSC CGL TIER II 26/10/2023
17. If 3 0.08 × 0.8 × p = 0.008 × 0.8 × 3
q then find 31 33
(a) (b)
p 44 64
the value of q . 31 33
(c) (d)
66 31
p
; fn gS] rksq
A
3
0.08 0.8 p = 0.008 × 0.8 × 3 q ,
TYPE-02
d k eku Kkr dhft,A
SSC CHSL MAINS 02/11/2023 (Shift- 01)
SMALLEST AND GREATEST VALUE
(a) 84 × 10–8 (b) 83 × 10–9 22. Which is the largest among the numbers,
(c) 83 × 10–8 (d) 84 × 10–9 3 4
5, 7, 13
3
0.0004 0.000008
18. If A = 4
16000 3 125000 4 810
and t ks la[;kvksa5, 3
7, 4
13 esa lcls cM+h gS
(a) 5 (b) 3 7
3
0.729 4 0.0016
B= , then what is A × B? (c) 4 13 (d) All are equal
0.16
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3
29. The greatest among the numbers
23. Which one among 6 , 2 5 and 6
12 is the
3
3
largest? 0.09, 0.064 , 0.5 and is
5
3
6 , 2 5 vkSj6 12 esa ls dkSu lcls cM+k gS\ 3
0.09, 3
0.064 ] 0-5 vkSj esa ls lcls cM+h la[;k gS
(a) 3
6 (b) 2
5 5
(c) 12 6 (d) All are equal 3
(a) 0.09 (b)
24. The greatest among the numbers 5
2
8, 4 13, 5 16, 10 41 is: (c) 0.5 (d)
0.064 3
30. Arranging the following in ascending order
2
8, 4 13, 5 16, 10 41 esa lcls cM+h la[;k gS% 334, 251, 717 we get
(a) 4
13 (b) 5
16 fuEufyf•r dks vkjksgh Øe esa O;ofLFkr 3djuk
34
, 251,
7 ge çkIr djrs gSa
17
r
(c) 10 41 (d) 2 8
(a) 334 > 251 > 717 (b) 717 > 251 > 334
25. Which shows the correct ascending order of the
si
(c) 3 > 7 > 2
34 17 51
(d) 251 > 334 > 717
given value?
31. The smallest among the numbers 2250, 3150, 5100
dkSu &lk fn, x, ekuksa dks lgh c<+rs gq, Øe esa lgh
a n by
and 4200
n'kkZrk gS\ 2250, 3150, 5100 vkSj4200 la[;kvksa esa lcls NksVh
n
DOE PRT 11/11/2019 (Shift- 03) (a) 4200 (b) 5100
150
(c) 3 (d) 2250
ja
(a) 5, 3 11, 26 3 (b) 5, 2 6 3, 3 11
R s
32. Which of the following is true?
(c) 3
11, 5, 2 3 6 6
(d) 2 3, 5, 11 3 fuEufyf•r esa ls dkSu lk lgh gS\
a th
26. Arranging the following in descending order, (a) (22.5)27 > (7.5)54 (b) (22.5)27 < (7.5)54
(c) (22.3)27 = (7.5)54 (d) (22.3)27 (7.5)54
3
we get 4, 2, 6 3, 4
5 33. Choose the incorrect reaction(s) from the
ty a
fuEufyf•r dks vojksgh Øe esa O;ofLFkr djus ij] gesa following:
fuEufyf•r esa ls xyr çfrfØ;k pqusa%
3
4, 2, 6 3, 4
5 çkIr gksrk gS
di M
(i) 6 2= 5 3
3 4 6
(a) 4 5 2 3 (ii) 6 2 5 3
4 3 6
(b) 5 4 3 2 (iii) 6 2 5 3
(a) (ii) and (iii) (b) (i)
(c) 2 63 34 45
(c) (i) and (ii) (d) (i) and (iii)
(d) 6
3 45 34 2 34. Which value among 11 + 5, 14 + 2,
27. The greatest of the number
8 + 8, is the largest?
2 4 5 10
8, 13 , 16, 41
11 + 5, 14 + 2, 8 + 8 esa ls dkSu lk eku
2
8, 4 13 , 5 16, 10 41 dh lcls cM+h la[;k lcls cM+k gS%
A
(a) 4
13 (b) 5
16 (a) 11 + 5 (b) 14 + 2
(c) 10 (d) 2 8 (c) 8+ 8 (d) All are equal
41
28. Which one correctly represents the given 35. Which one among 10 + 4, 11 + 3,
values in descending order?
7 + 7 is the smallest number?
dkSu lk fn, x, ekuksa dks ?kVrs gq, Øe esa lgh n'kkZrk gS\
10 + 4, 11 + 3, 7+ 7 esa ls lcls NksVh
DOE PRT 11/11/2019 (Shift- 02)
la[;k dkSu lh gS%
12 6 4 4 12 6
(a) 25, 10, 3 (b) 3, 25, 10 (a) 10 + 4 (b) 11 + 3
(c) 6
10, 12 25, 4
3 (d) 6
10, 4 3, 12
25 (c) 7+ 7 (d) All are equal
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36. Which of the following statements(s) is/are 42. Which is the greatest among ?
true?
fuEufyf•r esa ls dkSu lk@ls dFku lR; gS@gSa\ 6 4, 8 3, 12 2 and 24 1
I. 11 + 7 < 10 + 8 6 4, 8 3, 12 2 vkSj 24 1 esa
II. 17 + 11 > 15 + 13 lcls cM+k dkSu gS\
(a) Only I (b) Only II (a) 6 4 (b) 8 3
(c) Both I and II (d) Neither I nor II
(c) 12 2 (d) 24 1
37. Which is largest among 16 2, 19 + 1 and
43. Which is the smallest among ?
18 + 2 5 – 4, 10 – 2, 20 – 1
16 2, 19 + 1 rFkk 18 + 2 eas ls dkSu lcls cM+k 5 – 4, 10 – 2, 20 – 1 esa lcls NksVk dkSu gS\
gS\ (a) (b)
5– 4 10 – 2
r
(a) 19 + 1 (b) 16 2 (c) 20 – 1 (d) All are equal
si
(c) All are equal (d) 18 2 44. Which of the following is greatest?
fuEu esa ls dkSu lcls cM+k gSA
a n by
38. If x = 11 + 20 , y = 15 + 17 and
(a) 32 3 (b) 33 2
z = 14 + 18 . Which of the following holds
n
3 2
true? (c) 23 2 (d) 2
;fn x = 11 + 20 , y = 15 + 17 vkSj
ja
45. Which of the following is greatest?
R s
z = 14 + 18 . fuEufyf[kr esa ls dkSu lk lR; gS\ fuEu esa ls dkSu lcls cM+k gSA
a th
(a) x < y < z (b) y < z < x (a) 3333 (b) 333 3
33
(c) y < x < z (d) x < z < y (c) 3333 (d) 33
39. Arranging a = 6 – 5, b = 5 – 4, c 4– 3 TYPE-03
ty a
in ascending order, we get.
INFINITE SERIES
a = 6 – 5, b = 5 – 4, c 4– 3 dks vkjksgh
di M
46. Find the value of
Øe esa j•us ij] ge ikrs gSaA a a a.....
(a) c < b < a (b) b < a < c
a a a..... dk eku Kkr dhft,A
(c) a < c < b (d) a < b < c
3
40. The greatest among 7 – 5, 5 – 3, 9 – 7,
(a) a (b) a 2
11 – 9 is (c) a3 (d) 0
7 – 5, 5 – 3, 9 – 7, 11 – 9 esa lcls cM+k gS
47. 3 3 3..... is equal to.
(a) 7– 5 (b) 5– 3
3 3 3..... ds cjkcj gS%
(c) (d)
A
9– 7 11 – 9 (a) 3 (b) 3
(c) 23 (d) 33
41. Which is the greatest among 19 – 17 ,
48. If x = 4 4 4 4... , then what is the value of
13 – 11 , 7 – 5 and 5 – 3 ? x?
19 – 17 , 13 – 11 , 7 – 5 vk S j
;fn x = 4 4 4 4... gS] rks
x dk eku D;k gS\
5 – 3 esa lcls cM+k dkSu gS\ UPSC CDS 16/04/2023
(a) 19 – 17 (b) 13 – 11 (a) 2 (b) 4
(c) 7– 5 (d) 5– 3 (c) 8 (d) 16
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49. 7 7 7 7.... = 343y–1 then y = ? 12 12 12 12 12 12 =?
55. Find
7 7 7 7.... = 343y–1 rksy = ?
12 12 12 12 12 12 dk eku Kkr djsa
4 3 32 64
(a) (b)
3 2 (a) 12 31 (b) 12 63
31 63
5
(c) (d) 1 (c) 12 32 (d) 12 64
1
56. 12 + 12 + 12 + ..... is equal to.
50. The value of 2 4 2 4 2 3 4.... is
3 3
12 + 12 + 12 + ..... ds cjkcj gS%
(a) 3 (b) 4
3 3
2 4 2 4 2 3 4.... dk eku gS (c) 6 (d) 2
r
57. Find the value of
(a) 2 (b) 2²
si
(c) 2³ (d) 25
30 + 30 + 30 + 30 + .....
a n by
51. 27 ÷ 27 ÷ 27 ÷ ..... = ?
30 + 30 + 30 + 30 + .....
n
27 ÷ 27 ÷ 27 ÷ ..... = ? dk eku Kkr dhft,
(a) 3 (b) 27 (a) 3 (b) 4
27
ja
(c) 5 (d) 6
R s
(c) 2
27 (d) 4
27
58. Let x = 272 272 272 272 .... then x
a th
...
......
x equals
x
x 1
52. If x =
2
then x = ? ekuk x = 272 272 272 272 .... rks x
ty a
... cjkcj gS
...... (a) 16 (b) 413
x
di M
x (c) 17 (d) 4.35
x 1
;fn x = rksx = ? 59. Find the value of the given expression.
2 fn, x, O;atd dk eku Kkr dhft,A
1 1
(a) (b)
8 4 20 – 20 – 20 – 20 – ......
1 1 SSC CGL TIER- II 03/03/2023
(c) (d) (a) 4 (b) 6
16 32
(c) 5 (d) 2
53. Find the value of a a a a 60. What is the value of :
72 – 72 – 72 – .... ÷
a a a a dk eku Kkr dhft,&
A
7
15 20 – 20 – 20 – .... dk eku D;k gS\
(a) a 8 (b) a16
(c) a (d) a4 (a) 4 (b) 2
(c) 3.6 (d) 8
54. Find the value of 7 7 7 7 7 7
61. Let x = 42 – 42 – 42 – 42 – .... – then x
equals
7 7 7 7 7 7 dk eku Kkr dhft,&
ekuk x = 42 – 42 – 42 – 42 – .... – rks x
63 31
(a) 7 64 (b) 7 32
cjkcj gS
15 (a) 6 (b) 7
(c) 7 16 (d) 77 (c) Between 6 and 7 (d) Greater than 7
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62. 5 5 5 .... =? 7 2 7 – 2 7 2 7 – 2 7 .... =?
69.
21 1 21 – 1 (a) 51 (b) 4
(a) (b)
2 2
3 15
21 3 21 – 3 (c) (d) 3
2
(c) (d)
2 2
70. Let x = 6 – 6 6 – 6 .... to infinity ; then
63. 31 31 31 31 .... =? x equals
(a) 5 5 – 1.5 (b) 2.5 5 0.5 ekukx = 6 – 6 6 – 6 .... ls vuar rd_ rks x
5 5 –1 2 31 1 cjkcj gS
(c) (d)
2 2 (a) 3 (b) 21
r
64. 21 – 1 21 1
1 – 1 – 1 – .... = ? (c) (d)
2 2
si
5 1 5 –1
(a)
2
(b)
2
TYPE-04
(c)
53
a n by
(d)
5–3 IF SQUARE ROOT IS INSIDE SQUARE ROOT
n
2 2
71. If x 31 2 30 31 2 30 , then what is
ja
65. Find 19 – 19 – 19 – 19.... =? the value of x ?
R s
19 – 19 – 19 – 19.... Kkr djsa ;fn X 31 2 30 31 2 30 rks x dk eku
a th
D;k gS\
77 – 1 19 3
(a) (b) CRPF HCM 22/02/2023 (Shift - 02)
2 2
ty a
(a) 2 (b) 2 2
77 1
(c) (d) Between 4 and 5
2 (c) 2 15 (d) 4
di M
72. The square root of 14 + 65
66. Find 35 2 35 2 35 2 35 .... =?
14 + 65 dk oxZewy gS
(a) 2 + 5 (b) 3 + 5
35 2 35 2 35 2 35 .... Kkr djsa
(c) 5 + 3 (d) 3 + 25
(a) 6 (b) 7
(c) 5 (d) 6.4 73. What is the value of positive square root of 30
+ 105 ?
67. Find 154 – 3 154 – 3 154 – 3 154 – .... =? 30 + 105 dk /ukRed oxZewy dk eku D;k gS\
(a) 25 + 3 (b) 4 + 25
154 – 3 154 – 3 154 – 3 154 – .... Kkr djsa\ (c) 5 + 5 (d) 6 + 5
(a) 13 (b) 14 74. What is the value of square root 14 + 83 ?
A
(c) 11 (d) 9
14 + 83 ds oxZewy dk eku D;k gS\
68. If P = 11 3 11 3 11 3 11 – .... and Q = (a) 6 – 5 (b) 9 + 5
(c) 7 + 8 (d) 8 + 6
11 – 3 11 – 3 11 – 3 11 – .... then P + Q = ? 75. If the positive square root of 41 + 242 is A
and positive square root of 36 – 162 is B,
;fn P = 11 3 11 3 11 3 11 – .... vkSjQ = then what is the value of A – B?
;fn 41 + 242 dk /ukRed oxZewyA vkSj36 – 162
11 – 3 11 – 3 11 – 3 11 – .... rksP + Q = ? dk /ukRed oxZewy
B gS] rks
A – B dk eku D;k gS\
(a) 47 (b) 65 (a) 3 (b) 1
(c) 41 (d) 53 (c) 2 (d) 5
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76. Which of the following statement is correct?
1 a4
fuEufyf[kr esa ls dkSu lk dFku lgh gS\ ;fn a = 11 4 6 , gS] rks 2 dk eku D;k gS\
I. The square root of 5 + 26 is 2 + 3 a
5 + 26 dk oxZewy
2 + 3 gSA 286 96 6 186 95 6
(a) (b)
II. If 15 4 14 = x y , then x + y = 15 25 22
and xy = 50. 272 90 6 95 98 6
(c) (d)
;fn 15 4 14 = x y , gS] rks
x + y = 15 21 24
rFkkxy = 50 gSA
(a) Neither I nor II (b) Only I 84. If x = – 3 + 3 + 8 7 + 4 3 where x > 0, then
(c) Both I and II (d) Only II the value of x is equal to:
77. If 14 a 3 8 6 , then what is the value of a?
;fn x = – 3 + 3 + 8 7 + 4 3 tgk¡ x > 0] rksx
;fn 14 a 3 8 6 gS] rks
a dk eku D;k gS\
dk eku blds cjkcj gS%
r
(a) 4 (b) 6 (a) 3 (b) 4
(c) 5 (d) 8
si
(c) 1 (d) 2
78. If 52 – 30 3 = a + b3, then (a + b) is equal to:
85. Evaluate 10 + 2 6 + 2 10 + 2 15
a n by
;fn 52 – 30 3 = a + b3, rks(a + b) cjkcj gS%
ewY;kadu djs
10 + 2 6 + 2 10 + 2 15
(a) 4 (b) –4
n
(c) –2 (d) 2 (a) 2+ 3 (b) 3+ 5
ja
79. If 54 – 20 2 = a + b2, then (2a + 3b) is equal to: (c) 2+ 3+ 5 (d) 3+ 5+ 7
R s
;fn 86. What is the value
54 – 20 2 = a + b2, rks (2a + 3b) cjkcj gS%
a th
(a) 8 (b) 4 3
(26 15 3) 3 (26 – 15 3) = ?
(c) –3 (d) 11
80. If 43 – 24 3 = a + b3, then the value of
3
(26 15 3) 3 (26 – 15 3) dk eku D;k gS\
(a) 6 (b) 5
ty a
(3a + 5b) is equal to which of the following?
(c) 4 (d) 3
vxj 43 – 24 3 = a + b3, gks] rks (3a + 5b) dk eku
di M
fuEufyf[kr esa ls fdlds cjkcj gksxk\ TYPE-05
(a) –8 (b) 3
(c) 12 (d) –11 IF SQUARE ROOT IS IN THE DENOMINATOR
81. If a = 6 – 11 and b = 6 11 , then what is 3 2
the value of (b – a)? 87. The square root of is
3– 2
;fn a = 6 – 11 vkSjb = 6 11 gS] rks
(b – a)
dk eku D;k gS\ 3 2
dk oxZewy gS
(a) 6 (b) 2 3– 2
(c) 22 (d) 3
(a) 3 2 (b) 3– 2
5 3 – 48 – 4 2 50
A
82. If x = and y = 7 – 4 3
3 6 (c) 2 3 (d) 2– 3
then the value of (x – y) is closest to:
5 3 – 48 – 4 2 50 6 1 1
;fn x = vkSjy = 7 – 4 3 88. 2 equals to
3 6 3 2 3 3–2
rks(x – y) dk eku fudVre gS%
(a) 0.5 (b) 0.8 6 1 1
2
3 – 2 cjkcj gS
(c) 0.4 (d) 0.6 3 2 3
83. If a = 11 4 6 , then what is the value of
(a) (2 – 3) (b) – (2 3)
4
1 a
? (c) 1 (d) 2
a2
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3 2 2 6 2 3 1 1
89. – is equal to 95. Simplify – +
6 3 3 1 62 100 – 99 99 – 98
3 2 2 6 2 3 1 1 1
– ds cjkcj gS 98 – 97
–
97 – 96
+....+
2– 1
6 3 3 1 62
(a) 3 (b) 2 ljy dhft,A
(c) 0 (d) (a) 0 (b) 9
3 (c) 10 (d) 11
6 96. The value of
90. What is the simplified value of ?
5 4 3 1 1 1 1
6 – – +
4 – 15 15 – 14 14 – 13 13 – 12
dk ljyhÑr eku D;k gS\
5 4 3
1 1 1 1
24 3 – 30 13 3 – 15 – – is:
(a) (b) 12 – 11 11 – 10 10 – 3 3– 8
r
23 27
1 1 1 1
24 5 – 25 15 3 – 24 – –
si
+
(c) (d) 4 – 15 15 – 14 14 – 13 13 – 12
25 23
1 1 1 1
a n by
1 1 1 – – dk
91. The value of +....+ 12 – 11 11 – 10 10 – 3 3– 8
2 1 3 2 4 3
eku Kkr djsaA
n
1
is SSC CGL MAINS 29 Jan 2022
100 99
ja
(a) 2 2 2 (b) 2 – 2 2
R s
1 1 1 1
+....+ dk (c) 4 – 2 2 (d) 4 2 2
a th
2 1 3 2 4 3 100 99
eku gS 8 5 5 2 3 10 1
(a) 1 (b) 9 97. – – = a + b2
10 – 2 10 – 5 2– 5 2– 2
(c) (d)
ty a
99 99 – 1
then (15a – 2b) value is equal to
92. What is the value of
di M
1 1 1 1 8 5 5 2 3 10 1
3 4
4 5
5 6
6 7
? ;fn – – = a
10 – 2 10 – 5 2– 5 2– 2
1 1 1 1 + b2 gS] rks
(15a – 2b) dk eku crkb,aA
dk eku
3 4 4 5 5 6 6 7
(a) 3 (b) – 6
D;k gS\
(c) 6 (d) –3
(a) 3– 7 (b) 6– 5
(c) 7– 3 (d) 8– 5 7 5 14 10 10
98. What is the value of ÷ + ?
93. The value of 7– 5 14 – 10 5
1 1 1 1
.... is 7 5 14 10 10
1 2 2 3 3 4 8 9 ÷ + dk eku D;k gS\
A
7– 5 14 – 10 5
dk eku gS
(a) 1 (b) 0 SSC CGL MAINS (08/08/2022)
(c) 2 (d) 2 (a) 2 +1 (b) 2 2 +2
1 1 2+2
94. The value of + (c) (d) 2 2 +1
3.25 2.25 4.25 3.25
1 1 4
1 1 99. – –
+ + is 11 – 2 30 7 – 2 10 8 4 3
5.25 4.25 6.25 5.25
dk eku gS (a) 0 (b) 1
(a) 1.00 (b) 1.25
(c) 1.50 (d) 2.25 (c) 2 (d) 5
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100. The value of
38 – 5 3 a b 3
1 1 2 ;fn
23 ,b > 0 gS] rks
(b – a) dk
– – 26 7 3
10 84 eku fdruk gksxk\
12 – 140 8 – 60
SSC CGL MAINS 03 Feb 2022
1 1 2
– – (a) 7 (b) 18
10 84 dk eku gS
12 – 140 8 – 60 (c) 29 (d) 11
(a) 0 (b) 1 23
(c) 2 (d) 3 105. The value of 5 3 7 2 – 6 – is:
2 3 6
101. If x = 5 – 21 , then the value of
23
x 5 37 2– 6 – is: dk eku Kkr
2 3 6
is:
32 – 2x – 21 dhft,A
r
(a) 15 (b) 16
x
;fn x = 5 – 21 rks dk eku gS% (c) 12 (d) 10
si
32 – 2x – 21
1 1
106. in simplified
a n by
1 1 2 3– 5 2 – 3 – 5
(a) ( 3 – 7) (b) ( 7 – 3)
2 2 form equals
n
1 1 1 1
(c) ( 3 – 7) (d) (7 – 3) ljyhÑr :i ls
2 3 – 5
ds
2 2 2 – 3 – 5
ja
cjkcj gS
R s
52 6 – 5– 2 6
102. If = a 2 b 3 , then (a) 1 (b) 2
a th
(4 3 – 5 2) (3 2 4 3)
1
the value of (a – b) is: (c) (d) 0
2
52 6 – 5– 2 6
ty a
;fn = a 2b 3, rks(a 6 1 1
(4 3 – 5 2) (3 2 4 3) 107. If ×
= a
2 3 2 3– 5 2 – 3 – 5
dk eku gS%
di M
– b) + b6 where a and b are positive integers, then
1 1 what will be te value of (4a – 3b)?
(a) (b) –
3 15 6 1 1
;fn × = a
1 1 2 3 2 3– 5 2 – 3 – 5
(c)
15
(d)
3 + b6 tgk¡ a vkSjb ikftfVo iw.kkZadksa (baVhtj) gSa
(4a – 3b) dk ewY; D;k gksxk\
26 – 7 3 b a 3 (a) 15 (b) 15
103. If b > 0, then what is (c) 10 (d) 12
14 5 3 11
2 3 4
the value of (b – a ) ? 108. Solve ?
2 3 6 8 16
A
26 – 7 3 b a 3 2 3 4
; fn
11 b > 0 gks] rks dks gy dhft,\
14 5 3 2 3 6 8 16
(b – a ) dk eku fdruk gksxk\ (a) 2 (b) 2 –1
SSC CGL MAINS 03 Feb 2022
2
(a) 5 (b) 25 (c) (d) 3 – 2
3
(c) 12 (d) 9
109. The expression
38 – 5 3 a b 3
104.
26 7 3 23 , b > 0, then the value of 15 10 + 5 is equal to:
(b – a) is: 10 + 20 + 40 – 5 – 80
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4 4
15 10 + 5 113. Simplify: 3 6 59 3 6 59
O;atd ds cjkcj gS
10 + 20 + 40 – 5 – 80
4 4
3 6 59 3 6 59
(a) 10 3 + 2 5 (b) 5 + 2 2 ljy dj%s
(a) 5² (b) 54
(c) 5 3 + 2 2 (d) 5 – 2 5
(c) 58 (d) 512
1
TYPE - 06 9 2 2
44 22
SIMPLIFICATION OF POWER 8 –
114.
2 2 –2 is equal to
–5 5
3
5 x – 3 5 1
110. Simplified from of is 9 2
r
44 2 22
8 –
ds cjkcj gS&
si
2 2 –2
–5 5
3
5 x – 3 5
lss ljyhÑr gS
a n by
(a) 32 (b) 8
(c) 1 (d) 0
n
(a) x5 (b) x –5
32 0.13
× 320.07
1 115. Solve:
2 0.25
× 4 0.075 × 80.2
ja
(c) x (d)
R s
x
32 0.13
× 320.07
a th
1/2
4 9/4 2 × 22 gy djs
111. 8 – 2 0.25
× 4 0.075 × 80.2
2 2–2
1
ty a
(a) 2 (b)
1/2
3
4 9/4 2 × 22
8 –
di M
–2
1
2 2 (c) 1 (d)
2
(a) 32 (b) 8 116. The value of
(c) 1 (d) 0 (625)6.25 (25)2.6 (0.49)4 (0.343)4
× is:
1 –
2
3
4 6.75
(5 ) ( 5) 2.4
(0.2401)5
0.12 0.08
112. If A = (243) × (243) and B =
÷
216 (625)6.25 (25)2.6 (0.49)4 (0.343)4
4
4 6.75
(5 ) ( 5) 2.4 ×
(0.2401)5
dk eku gS%
1 – 3
B
then what is the value of ?
27 A TGT SST (Female) 14/11/2018 (Shift- 01)
2 (a) 25 (b) 0.35
A
1 – 3
;fn A = (243)0.12 × (243)0.08 vkSjB = ÷ (c) 175 (d) 0.25
216
117. On simplification
–4
1 3
B 1 2 4
rks dk eku D;k gS\ (49) 2 (512)0.25 (8)0.25 (729) 3 (64) 3
27 A 1
DOE PRT 13/11/2019 (Shift- 01) (343) 3 (256)–0.375 (81)0.5
3 4 reduces to /dk U;wured eku gS&
(a) (b)
4 27 PGT CS (Female) 26/07/2018 (Shift- 01)
4 8 (a) 2.25 (b) 1.25
(c) (d)
3 3 (c) 1.5 (d) 2.5
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118. If 5 5 × 53 ÷ 5–3/2 = 5a+2 , the value of a is: 126. Solve for x:
3x – 3x –1 = 486
;fn 5 5 × 53 ÷ 5–3/2 = 5a+2 ] a dk eku gS% x ds fy, gy djsa%
(a) 4 (b) 5
(c) 6 (d) 8 3x – 3x –1 = 486
a+3 a+6 a+1 (a) 7 (b) 9
3 × 4 × 25 4
119. If = , then the value (c) 6 (d) 5
27 a+1 × 8a – 2 × 125a+4 1526 127. If 2x –1 + 2x + 1 = 320, then the value of x is
of a + 9 is: ;fn 2x –1 + 2x + 1 = 320] rksx dk eku gS
3a+3 × 4 a+6 × 25a+1 4 (a) 0 (b) 9
;fn = gS rks a + 9 dk
27 a+1 × 8a – 2 × 125a+4 1526 (c) 10 (d) 7
eku gS% 128. If 8x + 1 – 8x – 1 = 63, find x
(a) 4 (b) 6 ;fn 8x + 1 – 8x – 1 = 63 gS] rksx Kkr dhft,
(c) 5 (d) 8 (a) 0 (b) 1
n
(c) –1 (d) 63
(243) 5 32n 1
r
120. The value of is: 129. If 2x + 3y = 17 and 2x + 2 – 3y + 1 = 5, then:
9n 3n–1 ;fn 2x + 3y = 17 rFkk2x + 2 – 3y + 1 = 5 gS] rks%
si
n
(243) 5 32n 1 SSC CGL TIER- II 07/03/2023
dk eku gS% (a) x = 1, y = 3 (b) x = 3, y = 3
a n by
9n 3n–1
(a) 3 (b) 9 (c) x = 3, y = 2 (d) x = 1, y = 2
(c) 6 (d) 12 130. If 5x – 3y = 13438 and 5x–1 + 3y+1 = 9686, then
n
121. The simplest value of the expression x+y = ?
1 ;fn 5x – 3y = 13438 vkSj5x–1 + 3y+1 = 9686, rks
p 14 p
ja
4 × 2 × 2p is: x+y = ?
R s
–p
(a) 9 (b) 11
2× 2
(c) 13 (d) 15
vfHkO;fDr dk lcls ljy eku gS%
a th
1 131. If x x x
(x x )x then, x equals
1
p p
4 × 2 × 2p
4
;fn x x x
gS] rks
x cjkcj gS
(x x )x
–p
2× 2
ty a
(a) 4 (b) 8 4 2
(a) (b)
(c) 4 p (d) 8 p 9 3
di M
122. If 272x–1 = (243)³ then value of x is
9 3
;fn 272x–1 = (243)³ rks x dk eku gSA (c) (d)
4 2
(a) 3 (b) 6
(c) 7 (d) 9 1
132. If 9 x – 2 – 22x–2 = 4x – 32x–3, then x is
123. If 3x+y = 81 and 81x–y = 3, then the value of x is
;fn 3x+y = 81 vkSj81x–y = 3] rksx dk eku gS; x–
1
;fn 9 2
– 22x–2 = 4x – 32x–3, rks x gS
15
(a) 42 (b) 3 2
8 (a) (b)
2 5
17
(c) (d) 39 3 4
8
(c) (d)
124. Find the value of x + y if 2x = 8y + 1 and 9y = 3x – 9 4 9
A
x + y dk eku Kkr dhft, ;fn 2x = 8y + 1 vkSj9y = 3x – 9 133. What are the values of x and y that satisfy
(a) 22 (b) 24 both the equations?
(c) 26 (d) 27
6
125. If 32 x – y = 3 x +y = 27, then the value of 3x–y will 20.7x . 3–1.25y = 8 and 40.3x . 90.2y = 8(81)1/5
27
be:
x vkSjy ds os dkSu ls eku gSa tks nkuksa lehdj.kksa
;fn 32 x – y = 3x +y = 27, rc 3x–y dk eku gksxk% larq"V djrs gSa\
1 6
(a) 3 (b)
3 20.7x . 3–1.25y = 8 vkSj40.3x . 90.2y = 8(81)1/5
27
1 (a) x = 2, y = 5 (b) x = 5, y = 2
(c) 3 (d) (c) x = 2.5, y = 6 (d) x = 3, y = 5
27
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ANSWER KEY
1.(c) 2.(b) 3.(d) 4.(b) 5.(c) 6.(d) 7.(b) 8.(d) 9.(c) 10.(d)
11.(d) 12.(d) 13.(d) 14.(c) 15.(c) 16.(d) 17.(d) 18.(d) 19.(c) 20.(b)
21.(c) 22.(a) 23.(b) 24.(d) 25.(c) 26.(a) 27.(d) 28.(d) 29.(d) 30.(a)
31.(b) 32.(b) 33.(c) 34.(c) 35.(b) 36.(a) 37.(d) 38.(d) 39.(d) 40.(b)
41.(d) 42.(d) 43.(a) 44.(b) 45.(d) 46.(a) 47.(b) 48.(b) 49.(a) 50.(a)
r
si
51.(a) 52.(c) 53.(b) 54.(a) 55.(d) 56.(b) 57.(d) 58.(c) 59.(a) 60.(b)
a n by
61.(a) 62.(a) 63.(b) 64.(b) 65.(a) 66.(b) 67.(c) 68.(d) 69.(d) 70.(c)
n
71.(a) 72.(b) 73.(c) 74.(d) 75.(d) 76.(b) 77.(d) 78.(c) 79.(d) 80.(b)
ja
R s
81.(b) 82.(a) 83.(a) 84.(d) 85.(c) 86.(c) 87.(a) 88.(d) 89.(c) 90.(a)
a th
91.(b) 92.(c) 93.(c) 94.(a) 95.(d) 96.(c) 97.(b) 98.(a) 99.(a) 100.(a)
101.(b) 102.(b) 103.(a) 104.(c) 105.(c) 106.(c) 107.(a) 108.(b) 109.(c) 110.(c)
ty a
di M
111.(d) 112.(b) 113.(b) 114.(d) 115.(c) 116.(a) 117.(a) 118.(a) 119.(c) 120.(b)
121.(b) 122.(a) 123.(c) 124.(d) 125.(c) 126.(c) 127.(d) 128.(b) 129.(c) 130.(c)
131.(c) 132.(a) 133.(b)
A
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Surds & Indices/?kkrkad vkSj dj.kh
( Practice Sheet With Solution)
Level-01 7. Which one of the following is the biggest?
3
4, 4 6, 6 15 and 12
245
1. The value of is
7 7 7..... fuEufyf•r esa ls dkSu lk lcls cM+k gS\
7 7 7..... dk eku gS 3
4, 4 6, 6 15 and 12
245
(a) 5 (b) 6 (a) 3
4 (b) 4
6
(c) 7 (d) None
r
(c) 156 (d) 12 245
8. If cube root of 175616 is 56 , then the value of
si
2. The value of 12 12 12 ..... is
3
175.616 3 0.175616 3 0.000175616 is
d k eku gS
a n by
12 12 12 ..... equal to
(a) 4 (b) 3 ; fn 175616 dk ?kuewy56 gS] rks3 175.616 ]
n
(c) 2 (d) 12 3
0.175616 ] 3 0.000175616 dk eku cjkcj gS
ja
(a) 0.168 (b) 62.16
3. The value of 30 – 30 – 30........ is:
R s
(c) 6.216 (d) 6.116
a th
30 – 30 – 30........ d k eku gS 9.
1
Which among 2 , 3 , 4 4 , 6 6 and 12 2
2
1
3
1 1 1
is the
(a) 5 (b) 6 largest?
(c) 7 (d) 30 1 1 1 1
ty a
1
22 , 33 , 4 4 , 66 vkSj12 2 esa ls dkSu lk lcls cM+k gS\
x 1 x 3
a b
4. If then the value of x is:
di M
1 1
b a (a) 2 2 (b) 3 3
x 1 x 3
a b 1 1
; fn b
a rc x dk eku gS% (c) 4 4 (d) 6 6
10. The greatest of the numbers (2.89)0.5, 2 – (0.5)2,
3
(a) (b) 3 0.5
2
1 1
2 1– , 3 is
(c) 2 (d) 2
3
5. The value of (256)0.16 × (256)0.09 is:
0.5
(256)0.16 × (256)0.09 d k eku D;k gS\ 1 1
l cls cM+h la[;k
(2.89)0.5, 2 – (0.5)2] 1– , 3 gS
A
(a) 256.25 (b) 64 2
(c) 4 (d) 16
6. Which one of the following is the least? (a) (2.89)0.5 (b) 2 – (0.5)2
3
3, 3 2, 2 and 4? (c) 2 3 (d)
fuEufyf•r esa ls dkSu lk lcls NksVk gS\ 11. The greatest number among 260, 348, 436, and
524 is :
3, 3 2, 2 vkS
j 34
buesa lcls cM+h la[;k dkSu lh260gS
, 348, 436] 524 \
(a) 2 (b) 3
2 (a) 260 (b) 348
(c) 3 (d) 3
4 (c) 436 (d) 524
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12. The largest numbers among 0.16, 0.16 , (0.16)2, 1 1
0.04 is:
(a) 14 3 (b) 2512
fuEu0.16, 0.16 ,(0.16)2, 0.04 esa lcls cM+h la[;k (c) 16 6
1
(d) 12 2
1
dkSu&lh gS\
20. Which of the following is largest among
(a) 0.16 (b) 0.16 fuEufyf[kr esa lcls cM+k dkSu gS\
(c) (0.16)2 (d) 0.04
1 1 1 1
13. Which is the greatest among 19 – 17 , (125) 6 , (11) 3 , (12) 6 , (5) 4 ?
13 – 11 , 7 – 5 and 5 – 3 ?
1
SSC CHSL 21/03/2023 (Shift-03)
1
(a) (12) (b) (11) 3
19 – 17 , 13 – 11 , 7 – 5 5 – 3
6
lcls cM+h la[;k dkSu lh gS\ (c) (125) 6
1
(d) (5) 4
1
(a) 19 – 7 (b) 13 – 11
The value of/dk eku Kkr djsaA
r
21.
(c) 7– 5 (d) 5– 3
si
3 3
3
–2744 × –216
9n × 35 × 27
14. If = 27 , then the value of n is: 64
a n by
4 3
3 × 81 729
9n × 35 × 27
3 SSC CPO 09/11/2022 (Shift-01)
n
;fn 4
= 27 ] rksn dk eku gS% (a) 164 (b) 512
3 × 81
(c) 189 (d) 156
ja
(a) 0 (b) 2
R s
(c) 3 (d) 4 1
22. The value of is closest to:/dk eku buesa
a th
15. The value of 5x × 25x –1 ÷ (5x –1 × 25x –1) 7–4 3
5x × 25x –1 ÷ (5x –1 × 25x –1) dk eku gS ls fdlds fudVre gS\
(a) 2 (b) 5
SSC CPO 09/12/2019 (Shift-02)
ty a
(c) 7 (d) 9
(a) 4.1 (b) 4.2
16. If m and n are whole numbers such that mn
= 121, then (m – 1)n+1 = ? (c) 1.2 (d) 3.7
di M
;fn m vkSj n iw.kZ la[;k,¡ gSa tSlsmfd
n
= 121] rks 23. The value of 11 2 18 is closest to/dk eku fdlds
n+1
(m – 1) = ? fudVre gS\
(a) 100 (b) 1000
SSC CPO 11/12/2019 (Shift-02)
(c) 10000 (d) 10
(a) 4.8 (b) 4.4
17. The value of/dk eku gS 5 35 26 448 7 (c) 3.8 (d) 4.1
ICAR Mains, 10/07/2023 (Shift-1) 24. Find the value of/dk eku Kkr djsaA
(a) 4 (b) 7
(c) 6 (d) 5 (3 (27 (73 (64)))
18. The value of/dk eku fdruk gksxk
3
50 3 216 3 512 UP Constable 28/01/2019 (Shift-01)
A
ICAR Mains, 10/07/2023 (Shift-2) (a) 1 (b) 2
(a) 8 (b) 6 (c) 3 (d) 4
(c) 4 (d) 5 25. Simplify/ljy dhft,
19. Which of the following is the smallest among
337
1 1 1 1
14 3 , 12 2 , 16 6 and 2512 ? 694 191 484
fuEufyf[kr esa lcls NksVk dkSu gS\
1 1 1 1 RPF Constable 04/02/2019 (Shift-01)
14 3 , 12 2 , 16 6 vkSj2512 ? (a) 29 (b) 25
SSC CHSL 10/03/2023 (Shift-03) (c) 24 (d) 26
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26. Evaluate/dk eku Kkr djsaA (a) 21.12 (b) 17.05
(c) 15.15 (d) 19.32
93 32 274 225 33. Solve/gy djsa%
RRB ALP & TECH 31/08/2018 (Shift-03)
1031 61 138 36
(a) 9 (b) 11
(c) 12 (d) 10 UPSI 30/11/2021 (Shift-01)
27. Solve the following/fuEufyf[kr dks gy dhft,A (a) 35 (b) 65
(c) 32 (d) 48
51 134 5 42 16 9 34. Solve/gy djsa%
2 2
NTPC CBT-01 11/01/2021 (Shift-02) 4 3 5 = x – 5 + 191
(a) 197 (b) 520 UPSI 30/11/2021 (Shift-01)
(c) 8 (d) 64
(a) –127 (b) –110
28. Find the value of X
r
(c) –117 (d) –137
X dk eku Kkr djsaA
35. Solve/gy djsa%
si
X 587 121 2316 180 1296 26.01 2.89 46.24 2.89
a n by
UPSI 12/11/2021 (Shift-02) + 26.01 2.89 46.24 2.89
n
(a) 2 (b) 1 UPSI 2/12/2021 (Shift-03)
(c) 0.5 (d) 4 (a) 36 (b) 12
ja
29. What value should come in place of the
R s (c) 48 (d) 24
following question mark?
36. Find the value of x
fuEufyf[kr iz'u fpUg ds LFkku ij D;k eku vkuk pkfg,\
a th
x dk eku Kkr djsaA
12.96 7.84 3.24 31.36 ?
(140 x ) 13 144
UPSI 12/11/2021 (Shift-02)
ty a
UPSI 24/11/2021 (Shift-03)
(a) 56.58 (b) 26.35
(a) 125 (b) 115
(c) 78.45 (d) 31.36
di M
(c) 110 (d) 120
30. What value should come in place of the
following question mark? 37. Solve/gy dhft,%
fuEufyf[kr iz'u fpUg ds LFkku ij D;k eku vkuk pkfg,\ 38.44 9.61 86.49 9.61
2
6 (5 5)2 ? ( 5)2 195 UPSI 20/11/2021 (Shift-1)
UPSI 15/11/2021 (Shift-02) (a) 24 (b) 36
(a) –57 (b) –63 (c) 48 (d) 42
(c) –61 (d) –59 38. Simplify/ljy dhft,
31. Find the value of X
3 5 125
X dk eku Kkr djsaA =?
80 6 5
A
155 – X 135 81 NTPC CBT-01 13/01/2021 (Shift-01)
UPSI 16/11/2021 (Shift-02) 5 4
(a) (b)
(a) 17 (b) 15 4 5
(c) 11 (d) 12 7 9
(c) (d)
32. Solve/gy dhft,% 9 5
2
1 3
–
19.36 4.84 43.56 4.84 = ? 39. Find the value of/dk eku Kkr djsa\ –
729
UPSI 16/11/2021 (Shift-03) UPSI 15/12/2017 (Shift-03)
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45. If a = b, b = c, c = a and b2 = ac, then y equals:
x y z
1
(a)
81
(b) – 81 ;fn ax = b, by = c, cz = a vkSjb2 = ac gS] rks
y cjkcj gS%
1 xy xz
(c) (d) 81 (a) (b) 2 x – z
81 x +z
40. Simplify/ljy dhft,% xz 2xz
(c) 2 z – x (d) x + z
2 2
542 321 – 542 – 321
542 321 x 1
46. If x = 5 + 2 6 , then is equal to:
NTPC CBT-01 09/01/2021 (Shift-02) x
642 x 1
(a)
271
(b) 1 ;fn x = 5 + 2 6 gS] rks cjkcj gS%
x
1 (a) 2 (b) 2 2
(c) 4 (d)
r
271
(c) 3 (d) 2 3
Simplify/ljy dhft,%
si
41.
2430.13 2430.07
5.53 – 4 3
a n by
47. The value of
30.25 22 16
7 490.075 3430.2
0.25
n
NTPC CBT-02 16/06/2022 (Shift-02) 2430.13 2430.07
(a) 9.5 (b) 0.75 dk eku gSA
7 490.075 3430.2
0.25
ja
(c) 1.5 (d) 14.25
R s
42. Find the value of/dk eku Kkr dhft,A 3 5
(a) (b)
a th
7 7
13 3
– 43 ÷ 8 – {2 + 6 × 9} (c)
7
(d)
7
13 – 8 2 3 5
Simplify/ljy djsa :
ty a
NTPC CBT-01 21/01/2021 (Shift-01) 48.
1 1 1 1
217 211 6.252 × 0.0144 2 + 0.027 3 × 81 4
di M
(a) – (b) –
8 8
(a) 0.14 (b) 1.4
685 685
(c) (d) – (c) 1 (d) 1.2
8 8
49. If 32x + 3 – 244 × 3x = – 9, then which of the
Level-02 following statements is true?
;fn 32x + 3 – 244 × 3x = – 9, rks fuEufyf•r esa ls dkSu
43. The value of (xb+c)b–c (xa+c)c–a (xb+a)a–b x 0 is
lk dFku lR; gS\
(xb+c)b–c (xa+c)c–a (xb+a)a–b x 0 dk eku gS (a) 'X' is a positive number
(a) 1 (b) 2 (b) 'X' is a negative number
(c) 'X' can be either a positive number or a
(c) –1 (d) 0
A
negative number
44. If 6 x = 3 y = 2 Z, then what is the value of (d) None of the above
1 1 1 50. If x = –0.5, then which of the following has the
?
y z x smallest value?
;fn x = –0.5 gS] rks fuEufyf•r esa ls fdldk eku lcls
1 1 1
;fn 6 = 3 = 2 gS] rksy z x dk eku D;k gS\
x y Z NksVk gS\
1 1
SSC CHSL 20/03/2023 (Shift-02) (a) 2 x (b)
x
(a) 1 (b) 0 1
(c) 3 (d) 6 (c) (d) 2 x
x2
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0.48 0.70 z 2
51. Given that 10 = x, 10 = y and x = y , then
the value of z is close to: eku yhft, x 6 – 6 6 – 6 ........ rks x
;g ns•rs gq, fd 100.48 = x, 100.70 = y vkSjxz = y2] cjkcj gS
rksz dk eku fdlds djhc gS%
21 – 1 21 1
(a) 45 (b) 88 (a) (b)
2 2
(c) 2.9 (d) 3.7
(c) 2 (d) 12
b 1 c 2
52. If 3
7a 35 20 is a whole number
then, which one of statements below is 57. The value of 5 5 5 5 is
consistent with it?
5 5 5 5 dk eku gS
;fn 3
7a 35
b 1
20
c 2
,d iw.kZ la[;k gS] rks
uhps fn, x, dFkuksa esa ls dkSu lk blds vuq:i gS\ (a) 5 (b) 516
15
(a) a = 3, b = 2, c = 1
r
17
(b) a = 3, b = 1, c = 1 (c) 10 (d) 516
si
(c) a = 2, b = 1, c = 2
58. The value of/dk eku D;k gksxk\
(d) a = 1, b = 2, c = 1
a n by
53. If a, b, c are non-zero and 14a = 36b = 84c, then 0.12 0.012 0.0092
is:
0.012 0.0012 0.00092
n
1 1
6b – is equal to
c a
(a) 102 (b) 10
ja
;fn a, b, c 'kwU;srj gSa vkSj
14a = 36b = 84c gS] rks
R s (c) 0.1 (d) 0.01
1 1 59. The value of/dk eku D;k gksxk\
a th
6b –
c a
cjkcj gS
0.032 0.212 0.0652
(a) 3 (b) 2 is
0.0032 0.0212 0.00652
ty a
(c) 1 (d) 5
(a) 0.1 (b) 10
54. Simplify/dks ljyhd`r dhft,&
di M
(c) 102 (d) 103
2 3 2 3 3 1
a
2 3 2 3 3 1 60. If 0.05 0.5 a = 0.5 × 0.05 × b , then b is
equal to
(a) 2 – 3 (b) 2 3
a
(c) 16 – 3 (d) 4 – 3 ;fn 0.05 0.5 a = 0.5 × 0.05 × b rks cjkcj gS
b
(a) 0.0025 (b) 0.025
55. Let x 4 4 – 4 4 – ... then
(c) 0.25 (d) 0.00025
x equals
61. Simplify:/ljy djsa\
eku yhft, x
A
4 4 – 4 4 – . .. 12.12 – 8.12 0.252 0.2519.95
rksx cjkcj gS (a) 1 (b) 2
(a) 3 (b) 13 (c) 3 (d) 4
62. The square root of/dk oxZewy Kkr djsa\
3 –1 13 1
(c) (d)
2 2 0.753
+[0.75 + (0.75)2 + 1]?
1 0.75
56. Let x 6 – 6 6 – 6 .. . . .. .. . then (a) 4 (b) 3
x equals (c) 2 (d) 1
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63. The value of /dk eku Kkr dhft,\
69. The value of 6 – 17 – 2 72 is closest to/dk eku
1 1
+ + fdlds fudVre gS\
3.25 2.25 4.25 3.25
SSC CPO 12/12/2019 (Shift-01)
1 1 (a) 2.1 (b) 2.4
+
5.25 4.25 6.25 5.25 (c) 2.7 (d) 1.7
(a) 1.00 (b) 1.25
(c) 1.50 (d) 2.25 70. The value of 9 – 2 11 – 6 2 is closest to:/dk eku
64. The value of /dk eku Kkr djsa\ fdlds fudVre gS\
1 1 2 SSC CPO 13/12/2019 (Shift-01)
– – (a) 2.7 (b) 2.9
12 – 140 8 – 60 10 84 (c) 2.4 (d) 2.1
(a) 0 (b) 1
1
r
(c) 2 (d) 3 71. The value of is closest to/dk eku
17 12 2
si
65. Given 10 24 40 60 p q r fdlds fudVre gS\
a n by
then, find (p + q + r) SSC CPO 13/12/2019 (Shift-02)
(a) 1.2 (b) 0.14
fn;k x;k 10 24 40 60 p q r
n
(c) 1.4 (d) 0.17
rc (p + q + r) dk eku gSA
72. Simplify/ljy dhft,
ja
(a) 9 (b) 10
R s
(c) 8 (d) 11 73 5 7–3 5
a th
66. The simplified value of/dk ljyhd`r eku D;k gksXkk 3 5 3– 5
1 1 NTPC CBT-02 14/06/2022 (Shift-02)
?
2 3– 5 2– 3– 5 (a) 14 (b) 6
ty a
(a) 0 (b) 1 (c) 3 (d) 4
1 73. Find the value of/dk eku Kkr dhft,A
di M
(c) 2 (d)
2
3 –1 3 1 2– 3 2 3
67. If 3x = 9y = 27z and
3 1 3 –1 2 3 2– 3
1 1 1 32
+ + = , find z.
3x 6y 9z 3 NTPC CBT-01 13/01/2021 (Shift-02)
(a) 18 (b) 14
1 1 1 32
;fn 3x = 9y = 27z vkSj 3x + 6y + 9z = 3 gS] rks (c) 16 (d) 20
74. Find 'x'/Kkr djsA
z Kkr dhft,A
(x)2 + (146)2 = (232)2 – (52)2 – 5468
CRPF HCM 01/03/2023 (Shift - 01)
UPSI 13/12/2017 (Shift-03)
A
1 5
(a) (b) (a) 158 (b) 183
5 8
(c) 156 (d) 162
1 8
(c) (d) 75. Find the value of x in the following equation.
32 5
68. Solve the equation 32x + 1 – 3x = 3x + 3 – 32. fuEu lehdj.k esax dk eku Kkr djsaA
lehdj.k 3 2x + 1
–3 =3 x x+3
–3 2
dks gy dhft,A 82.3 × 40.8 × 160.4 = 2x
CRPF HCM 01/03/2023 (Shift - 02) UPSI 13/12/2017 (Shift-02)
(a) 2, – 2 (b) 2, 3 (a) 11.4 (b) 10.2
(c) –1, 2 (d) –1, 1 (c) 7.7 (d) 10.1
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76. If 952.4 × 951.3 ÷ 95.9989 = 95n find the value of n? 1 1
p –1q 2 3 p6 q –3 3
;fn 952.4 × 951.3 ÷ 95.9989 = 95n rksn dk eku Kkr djsa\ ;fn] a b
p3 q –2 ÷ p –2 q 3 = p q gS rksa + b dk eku
UPSI 15/12/2017 (Shift-02)
gS] tgk¡P vkSjq vyx&vyx /ukRed vHkkT; la[;k,¡ gSa
(a) 2.7011 (b) 0.027011
2
(c) 0.27011 (d) 27.011 (a) –1 (b) –
3
77. Find the value of/dk eku Kkr djsa\
4
(c) (d) 2
1024 0.13
1024 0.07
3
70.25 490.075 3430.2 82. If x =
UPSI 20/12/2017 (Shift-02) 4 + 10 + 2 5 + 4 – 10 + 2 5 , then the
value of x lies between:
3 3
(a) (b) ;fn x = 4 + 10 + 2 5 ] rksx dk eku
49 16 + 4 – 10 + 2 5
r
buds chp fLFkr gS%
3 4
si
(c) (d) CRPF HCM 28/02/2023 (Shift - 03)
52 7
(a) 3.8 and 4.2 (b) 3.4 and 4.8
Find the value of the following./fuEu dk eku Kkr djsA
a n by
78.
(c) 3 and 3.4 (d) 2.6 and 3
83. If (5.55)x = (0.555)y = 1000, then the value of
812 3
n
8 3
1 1
UPSI 20/12/2017 (Shift-02) – is
x y
ja
(a) 5
R s (b) 9
1 1
(c) 11 (d) 3 ;fn (5.55)x = (0.555)y = 1000] rks x – y dk eku gS
a th
Level-03 1
(a) 1 (b)
1 3
79. Given that x2018 y2017 = and x2016 y2019 = 8, the
ty a
2 2
value of x2 + y3 is (c) (d) 3
3
di M
1
;g ns•rs gq, fd x2018 y2017 = vkSjx2016 y2019 = 8, If x = 4096 7 4 3
84. , then which of the followng
2
equals 64?
x2 + y3 dk eku gSA
37 31 ;fn x = 40967 4 3 gS] rks fuEufyf•r esa ls dkSu lk 64
(a) (b)
4 4 ds cjkcj gS\
35 33 7
(c) (d) x2 x7
4 4 (a) (b)
x4 3 x4 3
80. Given A = 265 and B = (264 + 263 + 262 + ...+20),
which of the following is true? 7
A = 265vkSjB = (264 + 263 + 262 + ...+20) fn;k x;k x2 x7
(c) (d)
A
gS] fuEufyf•r esa ls dkSu lk lR; gS\ x2 3
x2 3
(a) B is 264 larger than A 1
85. If 9x – 2 – 22 x – 2 4 x – 32 x – 3 , then x is
(b) A and B are equal
(c) B is larger than A by 1 1
;fn 9x – 2 – 22 x – 2 4 x – 32 x – 3 ] rksx gS
(d) A is larger than B by 1
1 1 3 2
p–1q 2 3 p6 q –3 3 (a) (b)
81. If 3 –2 ÷ –2 3 = pa q b , then the value of a 2 5
p q p q
+ b, where P and q are different positive 3 4
(c) (d)
primes, is 4 9
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2x + 4 x + 1
86. If 2 – 17 × 2 = –4, then which of the
–2
following is true? n n
9 n 32 3 – – 27
2
;fn 2 2x + 4
– 17 × 2 x+1
= –4, rks fuEufyf•r esa ls dkSu91. If
1 , then m – n = ?
33m 23 729
lk lR; gS\
(a) x is a positive value –2
n n
(b) x is a negative value 9 n 32 3 – – 27
;fn 2
1 rksm – n = ?
(c) x can be either a positive value or a 33m 23 729
negative value
(d) None of these (a) 3 (b) 1
87. If 4 = 5, 5 = 6, 6 = 7, .......127
x1 x2 x3 x124
= 128 then (c) 2 (d) –2
find x1x2x3.....x124? 92. If 2 x + y – 2z
=8 8z – 5 – y
;5 4y – 6z
= 25y + z ;34x – 3z =9x + z
;fn 4 x 1
= 5, 5x 2 = 6, 6 x3 = 7, .......127 x124 = 128 gS] rks then the value of 2x + 3y + 5z is:
x1x2x3.....x124 Kkr dhft;s\ ;fn 2x + y – 2z = 88z – 5 – y ; 54y – 6z = 25y + z ; 34x – 3z =9x + z
gS rks
2x + 3y + 5z dk eku crkb,A
r
5
(a) 2 (b)
2
si
(a) 56 (b) 44
7 (c) 32 (d) 28
(c) 3 (d)
a n by
2
93. If ax = b , by = 3
c, cz a then xyz = ?
88. Evaluate/dk ewY;kadu djsa
n
48
;fn ax = y
b, b =
3
c, cz a rksxyz
4
+ 1 ?
ja
5 +1 4 8
5 +1 5 +1 16
5 +1
1 1
R s (a) (b)
6 9
(a) 25 (b) 125
a th
(c) 5 (d) 625 1 1
(c) (d)
3a+3 × 4a+6 × 25a+1 4 24 12
89. If = then the value of
27a –1 × 8a – 2 ×125a+4 1526 94. If (x – 2a)(x – 5a) (x – 8a)(x –11a) + ka4 is a
ty a
a + 9 is: perfect square then k =?
;fn (x – 2a)(x – 5a) (x – 8a)(x –11a) + ka4 ,d iw.kZ
di M
3a+3 × 4a+6 × 25a+1 4
;fn = gS] rksa + 9 dk eku
27a –1 × 8a – 2 ×125a+4 1526 oxZ gS rks
k =?
Kkr dhft,A (a) 49
(a) 4 (b) 6
(b) 81
(c) 5 (d) 8
(c) 64
–6
–3
(3 x +7)
3 2 x +3 4 2
5 (d) 72
2 2 3
=
90. If 3 , then the 95. If x(x – 1) (x – 2)(x – 3) + 1 = k2 , then which
3
one of the following is a possible expression
value of is: for k?
2 – 42x
;fn x(x – 1)(x – 2)(x – 3) + 1 = k2 rks fuEufyf[krk
A
–6
–3
2 x +3 2 (3 x +7) 5
3
2
4
2 3 esa ls ds fy, dkSu lk lHkao gksxk\
;fn 3
=
3 gS] rks
(a) x2 – 3x + 1
2 – 42x dk eku Kkr dhft,A (b) x2 – 3x – 1
(a) 5 (b) 6 (c) x2 + 3x – 1
(c) 3 (d) 4 (d) x2 – 2x – 1
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ANSWER KEY
1.(c) 2.(a) 3.(a) 4.(c) 5.(c) 6.(b) 7.(a) 8.(c) 9.(b) 10.(c)
11.(b) 12.(b) 13.(d) 14.(c) 15.(b) 16.(b) 17.(a) 18.(c) 19.(b) 20.(c)
21.(c) 22.(d) 23.(b) 24.(c) 25.(d) 26.(d) 27.(c) 28.(c) 29.(d) 30.(d)
31.(c) 32.(a) 33.(c) 34.(d) 35.(d) 36.(b) 37.(b) 38.(b) 39.(d) 40.(c)
41.(c) 42.(b) 43.(a) 44.(b) 45.(d) 46.(b) 47.(a) 48.(d) 49.(c) 50.(b)
51.(c) 52.(a) 53.(a) 54.(c) 55.(d) 56.(a) 57.(b) 58.(b) 59.(b) 60.(b)
r
si
61.(d) 62.(c) 63.(a) 64.(a) 65.(d) 66.(d) 67.(c) 68.(c) 69.(b) 70.(c)
a n by
71.(d) 72.(c) 73.(a) 74.(c) 75.(d) 76.(a) 77.(d) 78.(d) 79.(d) 80.(d)
n
81.(b) 82.(b) 83.(b) 84.(c) 85.(a) 86.(c) 87.(d) 88.(c) 89.(c) 90.(a)
ja
91.(c) 92.(b)
R s93.(d) 94.(b) 95.(a)
a th
ty a
di M
A
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SOLUTIONS
1. (c) LCM of 2, 3 = 6
Let, 1
6
1
6
1
6
1
6
x=
3 2 , 2 3 , 2 2 , 4 3
7 7.......
33, 22, 23, 42
Squaring both sides 27, 4, 8, 16
x2 = 7 × 7 7....... Least value = 4 3
2
2
x = 7x 7. (a)
x=7 1 1 1 1
2. (a) 4 3 , 6 4 , 15 6 , 24512
In this type of series we devide the number in LCM of 3, 4, 6, 12 = 12
r
two factors whose HCF are 1
1 1 1 1
4 312 , 6 4 12 , 15 612 , 2451212
si
12 12 12 .....
= (4)4, (6)3, (15)2, (245)1
a n by
12
= 256 , 216, 225, 245
Largest value = 256 = 3
4
n
3
In positive series correct answer is always 8. (c)
larger digit
ja
3
175.615 3 0.175616 3 0.000175616
So, answer is 4
R s
5.6 + 0.56 + 0.056
3. (a)
a th
6.216
In this type of series we devide the number in
9. (b)
two factors whose HCF are 1
1 1 1 1 1
30 – 30 – 30....... 2 2 , 3 3 , 4 4 , 6 6 ,1212
ty a
LCM of 2, 3,4,6 = 12
30
26, 34,43, 62, 121
di M
1
5 Largest = 34 3 3
In negative series correct answer is always
10. (c)
smaller digit
So, answer is 5
1
4. (c) 2 0.5
(2.89) , 2 – 0.25, 1 , 3
ATQ, 1
1 –
a x –1 a 3– x 2
b b (1.7) , 1.75, 2, 1.732
x – 1 = 3 – x Largest value = 2
2x = 4 11. (b)
A
x=2 260 , 348 , 436, 524
5. (c) HCF of 60, 48, 36, 24 = 12
So, 25 , 34, 43 , 52
(256)0.16 × (256)0.09
32 , 81, 64 , 25
(256)0.25
Greatest = 81 348
1
256 4 = 4 12. (b)
2
0.16, 0.16 , (0.16) , 0.04
6. (b)
0.16, 0.4, 0.0256, 0.04
1 1 1 1
32 , 2 3 , 2 2 , 4 3 Largest = 0.4 = 0.16
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13. (d) 2 4 2 3
125 , 11 , 12 , 5
19 – 17 , 13 – 11 , 7– 5 , 5– 3 15625, 14641, 144, 125
1
2 2 2 2
, , , Largest = 1256
19 17 13 11 7 5 5 3
21. (c)
2 Given,
= Largest =
5 3
3
–2744 3 –216
= 5– 3
3
64
14. (c)
729
3
9 n 3 5 27
27 3
3 81
4 –14 –14 –14
3
–6 –6 –6
3 2 n 5 9
r
3
3 444
317 3
9 9 9
si
= 32n + 14 – 17 = 33
= 2n – 3 = 3
–14 –6
a n by
2n = 6
4
n=3
9
n
15. (b)
–14 –6 9
189
5x 52 x –2 53 x –2 4
ja
3 x –3
x –1
5 5 2 x –2
5
R s 22. (d)
5–2 Given that,
a th
= –3 5
5 1
16. (b) 7–4 3
m = 11, n = 2
ty a
1
(m – 1)n+1 = (11 –1)2+1 = 1000
17. (a) 7 – 2 12
di M
5 35 – 26 448 7 1 2 3
2– 3 2 3
5 38 = 4
2+ 3 = 3.7
18. (c)
23. (b)
3
50 3 216 3 512
11 2 18 = 9 2
3 50 6 8 = 4 = 3 + 1.4 = 4.4
19. (b) 24. (c)
1 1 1 1
14 12 16 25
3 2 6 12 3 27
73 64
A
LCM of 3, 2, 6, 12, = 12
144, 126, 162, (25)1 3 27 73 8
1
Smallest value = (25)1 = 25 12
3 27 81
20. (c)
1 1 1 1
125 , 11 , 12 , 5
6 3 6 4 , largest = ? 3 27 9
LCM of 6, 3, 6, 4 = 12
3 36 = 3 6
12 12 12 12
125 6 , 11 3 , 12 6 , 5 4 9 =3
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25. (d) 30. (d)
6 (5 5)2 ? ( 5)2 195
(694 – (337 – 191 – 484) ))
6 + 125 = ? – 5 + 195
= (694 – (337 – 191 – 22)) 131 = ? + 190
? = – 59
= (694 – (337 – 169)) 31. (c)
Squaring both sides
= (694 – (337 – 13)) 155 – X = 135 + 9
X = 155 – 144 = 11
= (694 – (324) 32. (a)
= (694 – 18) = 676 = 26 19.36 4.84 43.56 4.84
= (4.4 + 2.2 × 6.6 + 2.2) = 21.12
26. (d)
33. (c)
r
93 32 274 225
si
1031 61 138 36
= =
a n by
93 32 274 15 1031 61 138 6
= = 1031 61 144
n
93 32 289
= 1031 61 12
= 93 32 17 = 93 7 10
ja
27 (c)
R s = 1031 49
=
a th
1031 7
51 134 5 42 16 9
= 1024 = 32
= 34. (d)
51 134 5 49
ty a
2 2
4 3 5 = x – 5 + 191
= 51 134 35
4 + 45 = x – 5 + 191
di M
= 51 169 51 13 64 = 8 49 = x + 186
28. (c) – x = – 49 + 186
x = – 137
X= 587 – 121 2316 – 180 – 1296
35. (d)
X 587 – 11 2316 – 180 – 36 5.1 ÷ 1.7 × 6.8 ÷ 1.7 + 5.1 ÷ 1.7 × 6.8 ÷ 1.7
12 + 12 = 24
X= 576 2316 – 12 36. (b)
X = 24 ÷ 48
1 140 – x 13 144
X= 0.5
2
140 – x 13 12
A
29. (d)
On squaring both the sides,
12.96 7.84 3.24 31.36
140 – x = 25
1296 784 324 3136
= x = (140 – 25) = 115
100 100 100 100
37. (b)
= 3.6 × 2.8 ÷ 1.8 × 5.6
( 38.44 9.61 86.49 9.61)2
3.6 2.8
= 5.6 (6.2 3.1 9.3 3.1)2
1.8
= 5.6 × 5.6 = 31.36 (2 3)2 (6)2 36
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38. (b) 45. (d)
3 5 125 ATQ,
80 6 5 ax = by = cz = k (Let)
1 1 1
3 55 5 8 5 4 a kx , b ky , c kz
4 56 5 10 5 5
39. (d) b2 = ac
–2 2 1 1
–1 3 2
–729 3 y x z
729
= (– 9)2 = 81 2 xz
40. (c)
y xz
We know that, 2xz
y
2
(a b) – (a – b) 2 x+ z
r
=4 46. (b)
ab
si
2 2
(542 321) – (542 – 321) x 5 2 6
=4
542 321
a n by
41. (c) x 3 2
We know that,
n
a³ – b³ = (a – b) (a² + b² + ab) x –1 42 6 2 3 =
3 3
2 2 2 2
5.5 4 x 3 2 2 3
ja
30.25 22 16
R s
= (5.5 – 4) = 1.5 47. (a)
a th
42. (b)
2430.13 × 2430.07
13 3
–4 3
÷ 8 – {2 + 6 × 9} 7 × 490.075 × 3430.2
0.25
13 – 8 2
ty a
350.13 350.07
133 – 4 3 0.25 20.075 0.6
= ÷ 8 – 56 7 7 7
di M
13 – 4
= (132 + 42 + 13 × 4) ÷ 8 – 56
30.65 0.35 3
= (185 + 52) ÷ 8 – 56
70.25 0.15 0.6 7
–211 48. (d)
= 237 ÷ 8 – 56 =
8
1 1 1 1
43. (a)
6.25 2 0.0144 2 0.027 3 81 4
b2 c2 2
a2 2
b2
x x c x a
= 2.5 × 0.12 + 0.3 × 3
x b2 c 2 c2 a 2 a 2 b2 = 0.3 + 0.9 = 1.2
x0 = 1 49. (c)
44. (b) ATQ,
A
Let, 6x = 2y = 3z = k 3x (3x × 27 – 244) = – 9
1 1 1 Let, 3x = y
x y
6 = k , 2 = k , 3 = kz y (27y – 244) = –9
1 1 1 27y2 – 244y + 9 = 0
ky kz = kx 1
y 9, y
1 1 1 27
y z x 1
3x =
1 1 1 27
+ – =0
y z x 3x = 3–3
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x = –3 54. (c)
3x = 9 2 3 2 – 3 3 – 1
3x = 32 2 – 3 2 3 3 1
x=2
2 4 3 3 –1
x is either negative or positive
1 3 1
50. (b)
x = – 0.5 3 1 3 1
14
1 1 3 1 3 1
– 1
2x 2 0.5
2 –2 4–2 3
4 14 16 – 3
2
1 1
4 55. (d)
x 2 0.25 ATQ,
1 1
2 – 0 .5
2 0 .5 2 x 4 4 – 4 ......
r
1 1
–2 44 – 3 1
si
x –0.5
2
1
a n by
So, smallest value = 13 1
x
2
n
51. (c)
56. (a)
xz = y2 (100.48)z = (100.70)2
ja
10(0.48)z = 101.40 x 6 – 6 6.......
R s
0.48z = 1.40
a th
46 – 3 – 1
140
Z = 2.9 2
48
21 –1
52. (a) =
ty a
2
a b+1 c+2 Whole no. 57. (b)
3
7 × 35 × 20
di M
By option A
5 5 5 5
3
73 353 203 2 –1
4
15
5 4
= 516
7 × 35 × 20 = 4900 (Whole no) 2
So option (a) is correct 58. (b)
53. (a) 2 2 2
14a = 36b = 84c 14a = 62b = 84c = k (Let)
0.1 0.01 0.009
2 2 2
1
0.01 0.001 0.0009
a
14 k 0.01 0.0001 0.000081
= 10
1 0.0001 0.000001 0.00000081
2b
6k 59. (b)
A
1
84 k c 0.032 + 0.212 + 0.0652
= 10
We know that 14 × 6 = 84 0.0032 + 0.0212 + 0.00652
1 1 1 60. (b)
a 2b c 0.05 0.5 a 0.5 0.05 b
1 1 1 Squaring both sides
–
c a 2b 0.05 × 0.5 × a = 0.5 × 0.5 × 0.05 × 0.05 × b
1 a
6b 3 0.025
2b b
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61. (d) 66. (d)
2 2 2
– 8.1 0.25 0.2519.95 1 1
12.1
2– 5 3 2– 5 3
20.2 4 0.25 0.25 19.95
2– 5 – 3
2 – 5 3
20.2 4 2
0.25 20.2 = 4 2 – 5 – 3
62. (c) 2 2 – 5 2 2 – 5 1
0.753 2 4 2 10 2 2 2 – 5 2
+ 0.75 0.75 1
1 0.75
67. (c)
27 3x = 9y = 27z 3x = 32y = 33z
64 3 9 x = 2y = 3z
1
r
1 4 16
1 1 1 32
4
si
3x 6y 9z 3
27 3 9
a n by
1 1 1 1 32
16 4 16
9z 9z 9z 3
n
64
=2 1 32
16
3z 3
ja
63. (a)
R s
1
1 1 1 z=
a th
32
3.25 2.25 4.25 3.25 5.25 4.25
68. (c)
1 By option
ty a
6.25 5.25 Option (c) satisfy that equation
69. (b)
3.25 – 2.25 4.25 – 3.25 5.25
di M
– 4.25 6.25 – 5.25 6 – 17 – 2 72
= 6.25 – 2.25
= (2.5 – 1.5) = 1.00
=
6– 3–2 2
64. (a) = 3 + 2 2 = 2.4
ATQ,
70. (c)
1 1 2
– –
7– 5 5– 3 7 3 9 – 2 11 – 6 2
7 5 5 3 7– 3 2
– – = 9 – 2 32 2 –23 2
A
2 2 2
7 5 5 3 7 3
0
= 9–2 3– 2
2
65. (d) = 9–62 2
= 32 2
10 24 40 60 p q r
2
2 3 5 p q r = 2 12 2 2 = 2 1
p q r 10 = (1.4 + 1) = 2.4
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71. (d) 77. (d)
1 1 10240.13 1024 0.07
= 70.25 490.075 3430.2
17 12 2 17 2 72
0.13 0.07
1 =
1024
0.25 20.075
= 7 7 730.2
9 8 2
1024 0.2 102410
1 9– 8 3–2 2 0.25 0.15 0.6
= 7 7
9 8 9– 8 1 1
= (3 – 2 × 1.414) = 0.17 = 1024 5
4
72. (c) 7 7
78. (d)
73 5 7–3 5 3
+ 8 3 2
3 5 3– 5 81
r
1
1 8
7 3 53 – 5 7 – 3 5 3 5
=
3 3 1
si
812
812 8
4
a n by
1
21 9 5 – 7 5 –15 21 – 9 5 7 5 –15
= 81 4 3
4
n
79. (d)
12
= =3 1
ja
4 x2018 y2017 = .......(i)
R s 2
73. (a)
x2016 y 2019 = 8 ..........(ii)
a th
3 –1 3 1 2 3
2– 3
(i) ÷ (ii)
3 1 3 – 1 2 3 2 – 3
x2 1
y 2 16
ty a
( 3 – 1)² ( 3 1)² (2 – 3)² (2 3)²
(3 – 1) (4 – 3) y = 4x Put in (i)
di M
1
(4 – 2 3 4 2 3) (7 – 4 3 7 4 3) x2018 × 42017 × x2017 =
= 2
2 1
= 4 + 14 = 18 1 4035
x 4035
74. (c) 2
x2 = (232)2 – (52)2 – 5468 – (146)2
1
= 53824 – 2704 – 5468 – 21316 x
2
x2 = 24336
y=2
x = 156
75. (d) 1 33
x2 + y3 = 8
82.3 × 40.8 × 160.4 = 2x 4 4
A
(23)2.3 × (22)0.8 × (24)0.4 = 2x
80. (d)
26.9 × 21.6 × 21.6 = 2x A = 265
26.9 + 1.6 + 1.6 = 2x B = 264 + 263 + 262 + .......+ 20
210.1 = 2x 20 + 21 + 22 + ..........+ 264
x = 10.1
1 265 – 1
76. (a) =
95 2.4
× 951.3
÷ 95 0.9989
= 95 n 2 – 1
952.4 + 1.3 – 0.9989 = 95n = (265 –1)
n = 2.7011 So, A is greater than B by 1
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81. (b) 84. (c)
1 1
7 4 3
p–1q2 6 –3 x 4096
3 –2 p–2q 3 pa q b
3 3
p q p q
7– 4 3
1 1
(4096) = x
q4 p8 3 3
4 6 (64)2 = x
7 – 4 3
p
q
4 7–4 3
q3 64 = x 2
4 10
3
p 3
pa q b q pa qb
7
8 x2
p 3 p4 64 =
6
x2 3
3
q 85. (a)
10 1
x–
q p –4 p a q b
r
3 9 2
– 22 x – 2 4 x – 32 x – 3
si
2 x –1 2 x – 2 2 x – 3
10 3 – 2 22 x – 3
a = – 4 b=
3
32 x –1 32 x – 3 22 x 22 x – 2
a n by
10 2 1 1 1
a+b=–4+ –
= 32x 22 x 1
n
3 3
3 27 4
82. (b)
10 5
ja
32 x 22 x
x
R s
4 10 2 5 4 – 10 2 5 27 4
5 2.2
a th
= 3 × 8 = 27 × 22x
2x
x2 4 10 2 5 4 – 10 2 5 2 16–10– 2 5 3 2 x 3 3 3
x =
2 2 2
x2 8 2 6 – 2 5
ty a
86. (c)
x 2 8 2 6 – 4.4
x2 = 8 + 2 × 1.6 = 11.6 22x × 16 – 17 × 2x × 2 = – 4
di M
x = more than 3.4 Let, 2x = y
83. (b) 16y2 – 34y + 4 = 0
Given that, 32
(5.55)x = 103 y 2
16
(0.555)y = 103
2x = 21
ATQ,
x=1
3
= 5.55 = 10 x ......(1) 2x = 2–3
3
x = –3
& 0.555 = 10 .......(2)
y So, x is either positive or negative
By dividing (1) by (2) 87. (d)
A
3 4 x1 5, 5x 2 6, 6 x 3 7....127 x124 128(27 )
x
10 5.55 4x1x2x3 = 7
3
=
0.555 4x1x2x2....x124 = 27
y
10
7
3 3 27 = 4 2
–
x y
10 = 10
1
3 3 27 = 27 4 2 4
– =1
x y
7
1 1 1 7
– = 42
x y 3 2
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88. (c) 92. (b)
48 2 x + y – 2z = 88z– 5 – y; 54y – 6z = 25y + z ; 34x – 3z = 9x + z
4 x + y – 2z = 24z – 15 – 3y, 52y + 2z, 2x = 5z
1
5 1 4
5 1 8
5 1 16
5 1 x + 4y – 26z – 15, y = 4z
5z 5z
48 x= , y = 4 × 2 = 8, z = – 10z = – 15
4
16
5 –1
1
2 2
4 5×2 –15z
x= =5,z= –15 = 2
2 2
48
5
16
= 53 = 125 2x + 3y + 5z
20 + 24 + 10 = 44
89. (c) 93. (d)
3a 3 4 a 6 25a 1 4 1
ax = b 2 .......(i)
27 a –1 8a – 2 125a 4 1526
1
r
22a 12 by = c 3 .......(ii)
= 22
23a – 6
si
1
cz = a 2 .......(iii)
2– a + 18 = 22
a2x = b
a n by
a = 16
Put value of b in eqn. (ii)
a 9 16 9 5
a6xy = c
n
90. (a) Put value of c in eqn. (iii)
ja
–3 1
(2 x 3)
4 a6xyz = a 2
3
R s
2
3 1
a th
xyz =
12
(3 x 7)
–6 94. (b)
2
5
2 3 (x – 2a) (x – 11a) (x – 5a) (x – 8a) + ka4 = 0
ty a
= 3 (x2 – 11ax – 2ax + 22a2) (x2 – 8ax – 5ax + 40a2) + ka4
(x2 – 13ax + 22a2) (x2 – 13ax + 40a2) + ka4
di M
–3 2 –6 Let x2 – 13ax = g
3 × (2x + 3) × = × (3x + 7) × = (g + 22a2) (g + 40a2) + ka4
4 3 5
= g2 + 40a2g + 22a2g + 880a4 + ka4
90x + 135 = 48x + 112
= g2 + 62a2g + 880a4 + ka4 [a2 + b2 + 2ab = (a + b)2]
42x – 23 k = 81 will make it a perfect square
2 – 42x (g + 31a2)2 = 0
95. (a)
2 23 = 5 x (x – 1)(x – 2)(x – 3) + 1 = k2
91. (c) Value putting method
Take value of x in negative
–2
n n Put x = – 1
9 n 32 3 – – 27
A
2 1 – 1 (–1 –1) (– 1 – 2)(–1 –3) + 1
3 23
3m
729 –1 (– 2)(–3) (–4) + 1
24 + 1 = 25
33n 32 – 33n 25 = k2 = k = 5
33m 8 Now put the value of x in options to check
whether the option satisfied the value of k
33n 32 – 1 1
= (a) x2 – 3x +1
33m – 8 36
(–1)2 –3(–1) + 1
1 1 1+3+1=5k
m – n = 2r
33m – 3n 36 option A is correct
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CLASSIFICATION OF NUMBERS /la[;kvksa dk oxhZdj.k
(CLASSROOM SHEET)
Classification of Numbers (la[;kvksa dk oxhZadj.k)
Number
la[;k
Real Numbers Imaginary Numbers
okLrfod la[;k dkYifud la[;k
r
Complex Numbers
si
Rational Number Irrational Numbers
ifjes; la[;k vifjes; la[;k lfeJ la[;k
an by
Integers Fractions (Decimals)
n
iw.kkZad fHkUu
(n'keyo)
ja
R s
Positive Integers Zero Negative Integers
a th
/ukRed iw.kkZad 'kwU; udkjkRediw.kkZad
ty a
Non-Negative Non-Positive
xSjudkjkRed xSj/ukRed
di M
Whole Numbers Natural Numbers
iw.kZ
la[;k izkÑfrdla[;k
Prime Number Unity Composite Numbers
vHkkT;la[;k ,dkad HkkT;la[;k
Ex. –3, –5
Real Number/okLrfod la[;k
Note: (–1) i , i² = – 1, i³ = –i, i4 = 1
All those numbers which can be represented on
A
number line are called as Real Numbers Rational Numbers/ifjes; la[;k&
(os lHkh la[;k,¡ tks la[;k js•k ij çLrqr dh tk ldsa] okLrfod
p
la[;k,¡ dgykrh gSaA) All those numbers which can be expressed in
q
form, where p & q both are integers & q 0 are
– –5 –4 –3 –2 –1 0 1 2 3 4 5
Ex: 0, 1, –1, 16.83, –22.87, 2, all are Real Numbers rational numbers.
Imaginary Numbers./dkYifud la[;k, p
(lHkh la[;k,¡ tksq ds :i esa fy•h tk ldsa] tgk¡ p vkSjq
Cannot be denoted on number line.
la[;k js[kk ij fu:fir ugha fd;k tk ldrk gSA nksuksa iw.kkZadq gSa
0 vkSj
ifjes; la[;k,¡ gSaA)
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2 4 –8 0 22 1. The largest 3-digit prime number is:
Ex: , , , , , 2 & 0.2 all are Rational Numbers.
3 7 7 5 7 3 vadksa dh lcls cM+h vHkkT; la[;k gS%
Irrational Numbers/vifjes; la[;k,¡ & DP CONSTABLE 22/11/2023 (Shift- 03)
(a) 983 (b) 991
p
Those numbers which can't be expressed in (c) 987 (d) 997
q
form are called as Irrational Numbers. 2. The sum of the smallest three-digit prime
p number and the largest three-digit prime
(os lHkh la[;k,¡ tksq ds :i esa u fy•h tk ldsa] vifjes; number is:
la[;k,¡ dgykrh gSaA) lcls NksVs rhu&vadh; vHkkT; la[;k vkSj lcls cM+s rhu
Ex: 2, 3, 5 & are called as Irrational Numbers. vadksa dh la[;k dk ;ksx gS%
DP CONSTABLE 03/12/2023 (Shift- 01)
Integers (iw.kkZad)
(a) 1104 (b) 1098
Integers are a collection of all positive & negative
(c) 1100 (d) 1093
natural numbers & zero.
r
Note: Each prime number can be written in (6p ± 1)
(çkÑfrd la[;kvksa ds /ukRed vkSj Í.kkRed la[;kvksa
form, But every (6p ± 1) form may not be
si
vkSj 'kwU; ds lewg dks iw.kkZad dgrs gSaA) necessarily prime no.
i.e., – , ....... – 5, – 4, ....... –1, 0, 1, 2, 3, ......
an by izR;sd vHkkT; la[;k dks
(6p ± 1) :i esa fy[kk tk ldrk gS]
Natural Numbers (Positive Integers) ysfdu gj (6p ± 1) :i vko';d :i ls vHkkT; la[;k ugha
gks ldrk gSA
n
(izkÑfrd la[;k,a)
Counting Numbers are called as natural Ex. 13 6 × 2 + 1 (prime)
numbers ja 25 6 × 4 + 1 (not a prime number)
R s
(fxuus ;ksX; la[;k,a izkÑfrd la[;k,a dgykrh gSa)
a th
Ex: 1, 2, 3, ........... Between Nu mber of prime no.
1-50 15
Whole Numbers (Non-Negative Integers)
1-100 25
(laiw.kZ iw.kkZad)
ty a
1-200 46
If we add zero (0) to the collection of Natural 1-500 95
Numbers then that collection is called as
di M
1-1000 168
Whole Numbers.
(;fn ge lHkh izkÑfrd la[;kvksa ds lewg esa 'kwU; tksM+
3. nsa
If m is the number of prime numbers between 0 and
rks og lewg lEiw.kZ la[;k dgykrk gSaA) 50; and n is the number of prime numbers between
Ex: 0, 1, 2, ............. 50 and 100, then what is (m – n) equal to ?
;fn 0 vkSj 50 ds chp vHkkT; la[;kvksa dh la[;k
m gS_
Prime numbers/vHkkT; la[;k,a
vkSj 50 vkSj 100 ds chp vHkkT; la[;kvksa dh la[;k
n
It has only two factor- 1 & itself.
gS] rks
(m – n) fdlds cjkcj gS\
buds dsoy nks xq.ku•aM gksrs gSa& 1 vkSj Lo;aA
Prime number between 1 to 100 [CDS 2020 (I)]
1 ls 100 ds chp dh vHkkT; la[;k,a (a) 4
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, (b) 5
A
47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (c) 6
Even and smallest prime no/le vkSj lcls NksVh (d) 7
vHkkT; la[;k 4. How many prime numbers are there between 40
Smallest 3 digit prime number/lcls NksVh 3 vadksa and 50?
dh vHkkT; la[;k 101 40 vkSj 50 ds chp fdruh vHkkT; la[;k,¡ gSa\
largest 3 digit prime number/lcls cM+h 3 vadksa dh SSC GD 10/01/2023 (Shift-01)
vHkkT; la[;k 997 (a) 4 (b) 5
Only pair of consecutive odd prime no/ yxkrkj fo"ke
(c) 3 (d) 2
vHkkT; la[;k dk dsoy ;qXe 3, 5, 7
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How to check the given number is prime rhu la[;k,¡ tks ,d&nwljs dh lg&vHkkT; gSa] ,slh gSa fd
igys nks dk xq.kuiQy 391 gS vkSj vafre nks dk xq.kuiQy
or not?
943 gSA eè; la[;k Kkr dhft,A
fdlh la[;k ds vHkkT; gksus dh tkap dSls djsaA
DP CONSTABLE 16/11/2023 (Shift- 01)
To check whether a number is prime number or
not, first take the square root of the number. (a) 19 (b) 41
Round of the square root to the immediately (c) 17 (d) 23
lower integer. Then check divisibility of number
by all prime below it. If number is not divisible 9. Let a and b be two numbers such that a and
by any prime number then number is prime b – a are co-primes and b and b + a are co-primes,
number. respectively. Which of the followings is true?
dksbZ la[;k vHkkT; gS ;k ugha] ;g Kkr djus ds fy,] lcls eku yhft, a vkSj b nks la[;k,¡ gSa tSls fd a vkSj
b – a lg&vHkkT;
igys la[;k dk oxZ ewYk ysa] oxZ ewy dks mlls NksVs iw.kkZad ds gSa vkSj
b vkSjb + a Øe'k% lg&vHkkT;
:i esa fy[k ysaA mlds ckn la[;k dh foHkkT;rk dh tkap mlls gSaA fuEufyf•r esa ls dkSu lk lR; gS\
NksVh lHkh vHkkT; la[;kvksa }kjk djsaA ;fn fdlh Hkh vHkkT; DP CONSTABLE 16/11/2023 (Shift- 03)
la[;k ls foHkkT; ugha gksrh rks ;g vHkkT; la[;k gSA
r
(a) b is prime
Ex. 137 is prime number or not?
(b) a and b are co-prime
si
137 vHkkT; la[;k gS ;k ugha\
(c) a is prime
137 11 prime number less than or equal
an by (d) a and b are not co-prime
to 11 are 2, 3, 5, 7 and 11, 137 is not divisible
10. The sum of 3 prime numbers is 100. One
by any of there. Hence it is prime numbers.
n
number is greater than another number by 36.
11 ls NksVh ;k cjkcj vHkkT; la[;k,a
2, 3, 5, 7 rFkk11 gS] Find the largest number.
137 bueas ls fdlh ds Hkh }kjk foHkkT; ugha gS blfy, ;g ,d
ja fdUgha rhu vHkkT; la[;kvksa dk ;ksx 100 gSA ;fn ,d
R s
vHkkT; la[;k gSA
la[;k nwljh la[;k ls 36 vf/d gS rks rks lcls cM+h la[;k
a th
5. Consider the following numbers:
Kkr dhft,A
fuEufyf[kr la[;kvksa ij fopkj dhft,%
(a) 61 (b) 67
I. 437 II. 797 III. 1073
(c) 59 (d) 71
ty a
How many of the above numbers are prime?
11. x, y and z are distinct prime numbers where
mi;qZDr la[;kvksa esa ls fdruh la[;k,¡ vHkkT; gSa\ x < y < z. If x + y + z = 70, then what is the value of z?
di M
[CDS 2023 (I)] x, y rFkk z fof'k"V vHkkT; la[;k,a gSa] xtgk¡
< y < z gSA
(a) Only one (b) Only two ;fn x + y + z = 70 gS] rksz dk eku D;k gS\
(c) All three (d) None SSC CGL MAINS (08/08/2022)
6. Which of the following numbers is NOT a prime (a) 31 (b) 29
number? (c) 43 (d) 37
fuEufyf•r esa ls dkSu lh la[;k vHkkT; la[;k ugha
\ gS 12. x, y and z are prime numbers such that x + y + z = 38.
DP CONSTABLE 20/11/2023 (Shift- 02) What is the maximum value of x?
(a) 1271 (b) 1171 x, y vkSjz vHkkT; la[;k,a bl rjg gSa fdx + y + z = 38]
(c) 1471 (d) 1571 rksx dk vf/dre eku D;k gS\
7. If (k2 + 3) and (k3 + 5) are prime numbers then
(a) 19 (b) 23
what will be the value of k?
A
(c) 31 (d) 29
;fn (k2 + 3) vkSj(k3 + 5) vHkkT; la[;k,¡ gSa krks
dk
eku D;k gksxk\ 13. Four Prime numbers are taken in ascending
order. The product of first three is 385 and of
DP CONSTABLE 14/11/2023 (Shift- 02)
last three is 1001. Find the smallest prime no.
(a) 2 (b) 4
pkj vHkkT; la[;k,a c<+rs Øe esa yh xbZ gSA buesa ls çF
(c) 7 (d) 3
rhu dk xq.kuiQy 385 rFkk vfUre rhu dk xq.kuiQy 1001
8. Three numbers which are co-prime to each
other are such that the product of the first two gSaA buesa ls çFke vHkkT; la[;k dkSu& lh gS\
is 391 and that of the last two is 943. Find (a) 5 (b) 7
the middle number. (c) 11 (d) 17
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Co-prime numbers/lg&vHkkT; (a) 2
(c) 1
(b) 3
(d) 4
If the HCF of two numbers is 1. 19. Three numbers which are coprime to one
;fn nks la[;kvksa dk e-l-i- 1 gSA another are such that the product of the first
Ex. (2, 3), (11, 13), (16, 9), (25, 19) etc. two is 551 and that of the last two is 1073. The
sum of the three numbers is?
Twin-prime numbers/;qXe vHkkT; la[;k,a
rhu la[;k,a tks fd ,d nwljs ds fy, lgvHkkT; la[;k,a
When two consecutive prime numbers are with
gSA igyh nks la[;kvksa dk xq.kuiQy 551 rFkk vafre n
an interval of 2, then they are called twin
prime numbers. la[;kvksa dk xq.kuiQy 1073 gS] rks rhuksa la[;kvksa dk
tc nks yxkrkj vHkkT; la[;k 2 ds varjky ds lkFk gks rks mls Kkr djsA
;qXe vaHkkT; la[;k dgrs gSaA (a) 75 (b) 81
Ex. (3, 5)(5, 7)(11, 13) (c) 85 (d) 89
14. Which of the given pairs form a co-prime
numbers pair?
Composite Numbers/la;qDr ;k HkkT; la[;k
It has more than two factors.
fn, x, ;qXeksa esa ls dkSu&lk ,d lg&vHkkT; la[;k ;qXe
r
buds nks ls vf/d xq.ku[akM gksrs gSaA
cukrk gS\
Ex. 4, 6, 8, 9,10,12,14,15 etc.
si
SSC GD 12/01/2023 (Shift-01)
Smallest composite no. (lcls NksVh HkkT; la[;k
)4
(a) (21, 42) (b) (9, 63)
(c) (36, 15)
an by (d) (11, 21)
15. Which of the following is a pair of co-primes?
Smallest odd composite no. (lcls NksVh fo"ke HkkT;)la[;k
9
n
fuEufyf•r esa ls dkSu lk lg&vHkkT; ;qXe gS\ Note:
(i) 1 is neither prime nor composite number.
ja
DP CONSTABLE 14/11/2023 (Shift- 01)
1 u rks vHkkT; vkSj u gh HkkT; la[;k gSA
R s
(a) (198, 175) (b) (7, 35)
(ii) If a and b are any two odd primes then a2 + b2
(c) (217, 651) (d) (32, 62)
a th
and a2 – b2 is composite numbers.
16. Which of the following pairs of numbers are
relatively prime to each other? ;fn a rFkkb nks fo"ke vHkkT; la[;k,a gSa
a2 +rks
b2 rFkka2
– b HkkT; la[;k,a gksxhaA
2
fuEufyf•r esa ls la[;kvksa dk dkSu lk ;qXe ,d nwljs ls
ty a
vis{kkÑr vHkkT; gS\ Perfect Numbers/lEiw.kZ la[;k
DP CONSTABLE 23/11/2023 (Shift- 01) If the sum of all the factors of a number (except
di M
(a) (103, 113) (b) (51, 119) that number) is equal to the given number, then
(c) (27, 51) (d) (98, 567) that number is called a perfect number.
17. Which type of numbers are NOT co-prime? ;fn la[;k ds lHkh xq.ku•aMks dk ;ksx(ml la[;k dks NksM+dj
fdl çdkj dh la[;k,¡ lg&vHkkT; ugha gSa\ ] nh xbZ la[;k ds cjkcj gks] rks ml la[;k dks lEiw.kZ la[;k
SSC GD 31/01/2023 (Shift-03) dgrs gSaA
(a) Any two prime numbers
Ex. 6, 28, 496, 8128 etc.
dksbZ nks vHkkT; la[;k,¡
(b) Any two consecutive odd numbers Factor of 28 1, 2, 4, 7, 14
dksbZ Hkh nks yxkrkj fo"ke la[;k,¡ 1 + 2 + 4 + 7 + 14 = 28
(c) Any two consecutive numbers Thus, 28 is a perfect no.
dksbZ Hkh nks yxkrkj la[;k,¡
A
Note: 6 is a smallest perfect no.
(d) Any two consecutive even numbers (6 lcls NksVh lEiw.kZ la[;k
gS)
dksbZ Hkh nks yxkrkj le la[;k,
Even and Odd Numbers (le ,oa fo"ke la[;k,a)
18. Twin prime numbers are the prime numbers
whose difference is always equal to 2. The Even Numbers (le la[;k,a)
number of twin primes between 35 and 100 is: Any integer which on division by 2 gives zero
tqM+oka vHkkT; la[;k,¡ os vHkkT; la[;k,¡ gSa ftudk varj
(0) as the remainder is called as an even
ges'kk 2 ds cjkcj gksrk gSA 35 vkSj 100 ds chp tqM+oka
number.
vHkkT; la[;kvksa dh la[;k gS% (,d ,slk iw.kkZad ftldks 2 ls foHkkftr djus ij 'ks"kiQy
DP CONSTABLE 16/11/2023 (Shift- 01) 'kwU; vkrk gSA mls le la[;k,¡ dgrs gSaA)
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Ex: 0, 2, 4, 6, 8,………… Difference between a Rational & an
Even no can be expressed as 2n Irrational Number.
le la[;kvksa dks2n ds :i esa O;Dr fd;k tk ldrk gSA (ifjes; rFkk vifjes; la[;kvksa esa varj)
Odd Numbers (fo"ke la[;k,a) The decimal expansion of a Rational number
Any integer which on division by 2 gives 1 as is either terminating or non-terminating
the remainder is called as an odd number . (repeating) whereas the decimal expansion of
an irrational number is non-terminating non-
(,d ,slk iw.kkZad ftldks 2 ls foHkkftr djus ij 'ks"kiQy
repeating.
1 vkrk gS mls fo"ke la[;k,a dgrs gSaA) ,d ifjes; la[;k dk n'keyo çlkj ;k rks lkar vkorhZ
Ex: 1, 3, 5, 7,…………..
Odd no can be expressed as 2n ± 1
gksrk gS] ogha vifjes; la[;k dk n'keyo çlkj vlkar
fo"ke la[;kvksa dks
2n ± 1 ds :i esa O;Dr fd;k tk vukorhZ gksrk gSA
ldrk gSA
Rational Number (ifjes; la[;k, a
)
r
si
Terminating (lkar) Non Terminating
an by (Repeating)
(vuolkuh vkorhZ)
n
1 1
0.5 0.3333.... 0.3
2 3
ja 1 1
R s
0.25 0.1666666.... 0.16
4 6
a th
1 1
0.2 0.1111111.... 0.1
5 9
1 1
ty a
0.125 0.090909.... 0.09
8 11
1 1
di M
0.1 0.142857142857.... 0.142857
10 7
1
0.05
20
Irrational numbers are non terminating & non Then do the prime factorization of the
repeating
denominator and check denominator is made
Ex: 3.14159..... up of which primes
2 1.414.....
blds ckn gj dk vHkkT; xq.ku[kaM djsa vkSj ns[ksa fd g
3 1.732.....
How to check whether a Rational number is
fdu vHkkT; la[;kvksa ls cuk gSA
A
Terminating or Non-terminating ? Case 1 : If its made by using 2 or 5 only then
(,d ifjes; la[;k dk lkar ;k vlkar vkorhZ gksus dk ijh{k.k its terminating
dSls djrs gSa\)
;fn ;g 2 vFkok 5 ls gh cuk gS rks ;g lkar vkorhZ gSA
First check whether the rational number is in
its simplest form or not . If it's not in its simplest Case 2 : If any prime is used other than 2 or
form convert it into its simplest form . 5 then its non - terminating .
lcls igys ;g tkaps fd nh xbZ ifjes; la[;k vius ljyre ;fn ;g 2 vFkok 5 ls ugha cuk gS rks ;g vlkar vukorhZ gSA
:i gS vFkok ugha] ;fn ;g ljyre :i esa ugha gS rks
cuk,aA
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20. Which of the following numbers are 22. If the points P and Q represent real numbers
terminating decimals? 0.73 and 0.56 on the number line, then what
fuEufyf[kr esa ls dkSu lh la[;k n'keyo dks lekIr dj is the distance between P and Q ?
jgh gS\ ;fn fcUnqP vkSj Q la[;k js[kk ij okLrfod la[;kvksa
0.73 vkSj 0.56 dks fu:fir djrs gSa] rksP vkSjQ ds
23 37
(a)
8
(b)
15 chp dh nwjh D;k gS\
[CDS 2020 (I)]
47 69
(c) (d) 1 1
25 40 (a) (b)
6 5
Conversion of Recurring decimal in
16 11
p p (c) (d)
q
form./ q
:i esa ,d vkorhZ n'keyo 45 90
23. Which of the following number is a
dk :ikarj.kA terminating decimal?
buesa ls dkSu lh la[;k] lkar n'keyo la[;k gS
r
Ex: X = 0.2353535………..
10X = 2.353535……….. (1)
si
6937 5896
100X = 23.53535……….. (a) (b)
42 75
1000X = an by
235.3535………..
Subtract (1) from (2)
(2)
(c)
5873
128
(d)
6917
42
n
990X = 233
24. If radius of a sphere is rational, then which of
233 the following is/are correct?
X = ja ;fn ,d xksys dh f=kT;k ifjes; gS] rks fuEufyf[kr esa ls
R s
990
Now, we discuss the shortcut for this: dkSu&lk lgh gS \
a th
1. Its surface area is rational.
vc] ge blds fy, 'kkWVZdV ij ppkZ djrs gSaA bldk i`"Bh; {ks=kiQy ifjes; gSA
Ex: X = 0.235 2. Its volume is rational.
ty a
bldk vk;ru ifjes; gSA
235 – 2 Select the correct answer using the code given
X=
di M
990 below:
uhps fn, x, dwV dk iz;ksx djds lgh mÙkj pqfu,%
233
X= d1
990
[CDS 2020 (II)]
Ex: 123 1 122 (a) 1 Only (b) 2 Only
A 0.123
990 990 (c) Both 1 and 2 (d) Neither 1 nor 2
123 12 111 25. Which one of the following is not correct?
B 0.123
900 900 fuEufyf[kr esa ls dkSu lk lgh ugha gS\
123 [CDS 2019 (II)]
C 0.123
999 (a) 1 is neither prime nor composite
47123 471 46652 1 u rks vHkkT; gS vkSj u gh HkkT;
A
D 0.47123
99000 99000 (b) 0 is neither positive nor negative.
21. By what smallest natural number X = 0 u rks /ukRed gS vkSj u gh ½.kkRed
0.349999.............. be multiplied so that it (c) If p × q is even, then p and q are always
becomes an integer? even
og lcls NksVh çkÑfrd la[;k ftlls X = 0.349999..... ;fn p × q le gS] rksp vkSjq ges'kk le gksaxs
dks xq.kk djus ij ,d iw.kZ vad çkIr gksA
(d) 2 is an irrational number
(a) 20 (b) 100
(c) 300 (d) 900 2 ,d vifjes; la[;k gS
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ANSWER KEY
1.(d) 2.(b) 3.(b) 4.(c) 5.(a) 6.(a) 7.(a) 8.(d) 9.(b) 10.(b)
11.(d) 12.(c) 13.(a) 14.(d) 15.(a) 16.(a) 17.(d) 18.(b) 19.(c) 20.(b)
21.(a) 22.(a) 23.(c) 24.(d) 25.(c)
r
si
an by
n
ja
R s
a th
ty a
di M
A
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Classification of Number
( Practice Sheet With Solution)
1. Find out which of the following sets form co- 7. Which of the following pairs is not a pair of twin
prime numbers. primes?
fuEufyf[kr esa ls dkSu&lk ;qXe tqM+ok vHkkT;ksa dk ;qXe u
fuEufyf•r leqPp;ksa esa ls dkSu&lk lg vHkkT; la[;k,¡ cukrk gSA
ALP & Tec. 21/08/2018 (Shift-02)
NTPC CBT-02 18/01/2017 (Shift-01)
(a) 11, 13 (b) 71, 73
(a) (12, 7) (b) (21, 42) (c) 131, 133 (d) 191, 193
(c) (43, 129) (d) (3,9) 8. Which from the following is not a rational
number ?
2. Which of the following is an odd composite fuEufyf[kr esa ls dkSu ,d ifjes; la[;k ugha gS\
number?
RPF SI 05/01/2019 (Shift-01)
fuEu esa ls dkSu&lh ,d fo"ke HkkT; la[;k gSA
(a) 3
1728 (b)
NTPC CBT-02 18/01/2017 (Shift-02)
(c) 2.487627287 (d) 8.36712846781
(a) 13 (b) 17 9. If a and b are odd numbers, then which of the
(c) 12 (d) 15 following will be an even number?
3. ______is an irrational number. ;fn a vkSjb fo"ke la[;k,¡ gSa] rks fuEufyf[kr esa ls dkSu&lh
le la[;k gksxh\
______,d vifjes; la[;k gSA
Hajipur Group-D 06/05/2012 (Shift-01)
NTPC CBT-02 19/01/2017 (Shift-02)
(a) a + b (b) a + b + 1
11 (c) ab (d) ab + 2
(a) (2 × 3) (b) 10. If A is integer and B is natural number than B
8
–A=?
2 ;fn A iw.kkaZd BgS]
çkd`r la[;k gS] rksB – A = ?
(c) (d) 9 A. Whole number B. Null number
3
C. Natural number D. Even number
4. All irrational number are numbers.
Group-D 21/02/2016
lHkh vifjes; la[;k,¡ ____la[;k,¡ gksrh gSaA
(a) B (b) D
NTPC CBT-02 17/01/2017 (Shift-03) (c) A (d) C
(a) Real (b) Integers 11. Which among the following is not an irrational
number?
(c) Whole (d) Imaginary
fuEufyf•r esa ls dkSu lh ,d vifjes; la[;k ugha gS\
5. All rational numbers are_____ numbers
Group-D CBT 18/09/2018 (Shift-03)
lHkh ifjes; la[;k,¡------------ la[;k,¡ gSaA
(a) (b) 7652
NTPC CBT-02 17/01/2017 (Shift-01)
(c) 5428 (d) 6084
(a) Whole (b) Integer 12. Which of the following statements is true
(c) Real (d) Irrational regarding the group of numbers 4, 121, 16 and
256?
6. In a prime number……..
la[;k 4] 121] 16 vkSj 256 ds lewg ds lacaèk esa fuEufyf[kr
,d vHkkT; la[;k esa ------------
esa ls dkSu lk dFku lR; gS\
NTPC CBT-1 30/03/2016 (Shift-02)
ASM Bhopal (Non-Tech.) 11/10/1998
(a) There are more than two divisors. (a) All of them are odd number
(b) Divided by itself and 1. (b) All of them are prime numbers
(c) It has no divisor. (c) All of them are square numbers
(d) Is not a positive integer (d) All of them are divisible by 7
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13. All natural numbers and 0 are called the 17. The sum of twice the largest two-digit prime
____________ numbers. number and three times the largest three-digit
lHkh çkd`r la[;k,¡ vkSj 0 -------------- la[;k,¡ dgykrh gSaA prime number will be equal to the sum.
nks vadksa dh lcls cM+h vHkkT; la[;k dk nksxquk vkSj rhu vadk
Kolkata Ticket Collector (TC) 03/12/2006
lcls cM+h vHkkT; la[;k dk rhu xquk dk ;ksx ;ksx ds cjkcj gksxkA
(a) Whole (b) Prime NTPC CBT-1 22/08/2022 (Shift-02)
(c) Integer (d) Rational
(a) 3185 (b) 3029
14. The number which is neither prime nor (c) 2195 (d) 6523
composite is:
18. Which of the following pairs is co-prime?
og la[;k tks u rks vHkkT; gS vkSj u gh HkkT; gS%
fuEufyf[kr ;qXeksa eas ls dkSu&lk lg&vHkkT; gS\
Bhubaneswar ASM 30/08/2009
UPSI 14/12/2017 (Shift-02)
(a) 0 (b) 1
(a) (156, 126) (b) (23, 92)
(c) 3 (d) 2
(c) (24, 351) (d) (18, 35)
15. Which of the following pairs are co-primes?
19. 3 16 is ________
fuEufyf[kr esa ls dkSu ls tksM+s lg&vHkkT; gSa\
3 16 ,d ________ gSA
Group D 20/09/2018 (Shift-02)
(a) 348, 296 (b) 114, 213 UPSI 12/12/2017 (Shift-01)
(c) 59, 97 (d) 3025, 4920 (a) Irrational number
16. The decimal ex pansion of rational number (b) Imaginary number
37 (c) Integer
2 will terminate after.
2 ×5 (d) Prime number
20. Let P, Q and R be specific integers. R is a positive
37
ifjes; la[;k 22 × 5 dk n'keyo çlkj ds ckn lekIr gks even integer while P and Q are positive odd
integers. Which of the following expression cannot
tk,xkA be true?
DMRC Customer Relations Assistant 22/07/2012 eku ysaP, Q vkSj R fof'k"V iw.kk±d RgSaA
,d /ukRed le
iw.kk±d gSa tcfd
P vkSjQ /ukRed fo"ke iw.kk±d gSaA fuEufyf[kr
(a) One decimal place
O;atd esa ls dkSu&lk lR; ugha gks ldrk\
(b) Two decimal places
UPSI 12/12/2017 (Shift-01)
(c) Three decimal places
(a) Q(P – R) is odd (b) (P – R)2 is even
(d) Four decimal places
(c) Q2(P – R) is odd (d) R(P – Q)2 is even
ANSWER KEY
1.(a) 2.(d) 3.(a) 4.(a) 5.(c) 6.(b) 7.(c) 8.(b) 9.(a) 10.(a)
11.(d) 12.(c) 13.(a) 14.(b) 15.(c) 16.(b) 17.(a) 18.(d) 19.(c) 20.(b)
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SOLUTIONS
1. (a) 12. (c)
Co-prime:- Pair whose HCF is 1 4. is a square of 2
HCF of (12,7) = 1 121 is a square of 11
16. is a square of 4
= This is a co-prime number.
256 is a square of 16
2. (d)
13. (a)
Odd composite numbers are odd numbers with more
Natural Number = 1 , 2 , 3 ,--------
than 2 factors.
Whole Number = 0 , 1 , 2 , 3 ,--------
15 = 3 × 5
14. (b)
3. (a) 1 is neither prime nor composite
(2 3) 6 is an irrational number. 15. (c)
Co-prime numbers means pairs of numbers which have
4. (a) no common factors other than 1.
All irrational numbers are real numbers. 59, 97 is a co-prime pair.
5. (b) 16. (b)
All rational numbers are real. 37 37
= 22 5 = = 1.85
6. (b) 20
The prime number is divisible by 1 and itself. So, the decimal expansion of a rational number ends
after two decimal places.
7. (c) 17. (a)
Factors of 133 = 1, 7, 19 Largest 2 digit prime number = 97 and Largest 3 digit
133 It is not a prime number. prime number = 997
Required sum
8. (a)
= (2 × 97 + 3 × 997) = (194 + 2991) = 3185
Since the decimal expansion of is turbulent and
non-recurring. Therefore is an irrational number. 18. (d)
Among the options, since 23 is a prime number, it is
9. (a)
co-prime with all other numbers.
The sum of any two odd numbers makes an even
[A pair of numbers is said to be co-prime when their
number.
common factor is at most 1]
(a + b) = even number HCF (18 and 35) = 1
10. (c) 19. (c)
Whole Number = Integer – Natural Number 1
2
3 16 3 42 = 3 (4 )2 = 12
2 is an integer as well as a Natural Number.
This is an integer.
=2–2=0
20. (b)
(b – a) is a whole number. Let P = 3, Q = 5, R = 2, the values of P, Q and R
11. (d) Putting in option
(a) Q(P – R) = 5 (3 – 2) = 5 odd
An irrational number is a real number which cannot
be expressed as a ratio of integers, all options except (b) (P – R)² = (3 – 2)² = 1 even
6084 are irrational numbers. Option b cannot be true.
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UNIT DIGIT (bdkbZ vad
)
(CLASSROOM SHEET)
1. The digit in unit's place of the product 8. Find the last digit of 2136?
fuEufyf[kr xq.kuiQy dk bdkbZ vad Kkr djsaA 2136 dk vafre vad Kkr dhft,\
49237 × 3995 × 738 × 83 × 9 is : Group D 05/12/2018 (Shift-02)
(a) 6 (b) 3
(a) 0 (b) 7
(c) 7 (d) 9
(c) 6 (d) 8
9. The units digit of the expression
2. The unit digit in 3 × 38 × 537 × 1256 is :
ifj.kkeh la[;k esa bdkbZ vad Kkr djsaA
3 × 38 × 537 × 1256 dk bdkbZ vad Kkr djsaA
256251 + 36528 + 7354 is :
r
(a) 4 (b) 2
(a) 6 (b) 5
(c) 6 (d) 8
si
(c) 4 (d) 0
3. Find the unit digit of the product of all the odd
10. What is the unit digit in the Expansion of
prime number. an by
lHkh fo"ke vHkkT; la[;kvksa ds xq.kuiQy dk badkbZ vad
6732?
6732 ds O;atd esa bdkbZ vad D;k gS\
n
Kkr dhft,A
[CDS - 2021 (I)]
(a) 0 (b) 1
ja (a) 1 (b) 3
R s
(c) 2 (d) 5
(c) 7 (d) 9
4. Find the unit digit of the product of all the
11. What is the digit in the unit place of 399 ?
a th
prime number.
399 ds bdkbZ LFkku ij dkSu&lk vad gksxk\
lHkh vHkkT; la[;kvksa ds xq.kuiQy dk badkbZ vad Kkr
[CDS - 2020 (II)]
dhft,A
ty a
(a) 1 (b) 3
(a) 0 (b) 1
(c) 7 (d) 9
(c) 2 (d) 5
di M
12. What is the unit place digit in the expansion
5. For any natural number n, what is always the
of 773?
last digit of the result 6n – 5n?
fdlh Hkh izkd`r la[;k
n ds fy,] 6n – 5n ds ifj.kke dk 773 ds izlkj esa bdkbZ dk vad D;k gS\
vafre vad ges'kk D;k gksrk gS\ [CDS - 2019 (I)]
NTPC CBT-1, 28/12/2020 (Shift-02) (a) 1 (b) 3
(a) 1 (b) 3 (c) 7 (d) 9
(c) 7 (d) 5 13. Consider the following statements:
6. Find the unit digit in given factor of fuEufyf[kr dFkuksa ij fopkj dhft, %
(3451)51 ×(531)43.
1. Unit digit in 17174 is 7
(3451)51 ×(531)43 ds fn, x, xq.ku[kaM esa bdkbZ vad
17174 esa bdkbZ vad 7 gSA
Kkr dhft,A
A
2. Difference of the squares of any two odd
RPF SI – 11/01/2019 (Shift-01)
numbers is always divisible by 8
(a) 6 (b) 4
fdUgha Hkh nks fo"ke la[;kvksa ds oxksZa dk varj
(c) 1 (d) 9
7. The digit in unit place of the number
8 ls foHkkT; gSA
3. Adding 1 to the product of two consecutive
ifj.kkeh la[;k esa bdkbZ vad Kkr djsaA
odd numbers makes it a perfect square.
(1570)2 + (1571)2 + (1572)2 + (1573)2 is :
nks ozQekxr fo"ke la[;kvksa ds xq.kuiQy esa 1 tksM+
(a) 1 (b) 2
,d iw.kZ oxZ izkIr gksrk gSA
(c) 3 (d) 4
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Which of the above statements are correct? 20. Let x = (433)24 – (377)38 + (166)54. What is the
mi;ZqDr esa ls dkSu ls dFku lgh gSa\ units digit of x?
[CDS - 2019 (II)]
eku ys fd x = (433)24 – (377)38 + (166)54 gS] rks
x
dk bdkbZ dk vad D;k gS\
(a) 1, 2 and 3 (b) 1 and 2 only
SSC CGL MAINS 29/01/2022
(c) 2 and 3 only (d) 1 and 3 only
(a) 8 (b) 7
14. What is the last digit of the sum S = 997 + 279 ?
(c) 6 (d) 9
;ksxiQyS = 997 + 279 dk vafre vad D;k gS\
21. The digit in the unit's place of
UPSC CDS 16/04/2023
fuEu O;atd dk bdkbZ vad Kkr dhft,A
(a) 3 (b) 6
[(251)98 + (21)29 – (106)100 + (705)35 – 164 + 259] is
(c) 7 (d) 9
(a) 1 (b) 4
15. Find the units digit of 4343 – 2828. (c) 5 (d) 6
4343 – 2828 dk bdkbZ vad Kkr djsaA 22. The unit digit in the product
r
(a) 1 (b) 2 ifj.kkeh la[;k esa bdkbZ vad Kkr djsaA
(c) 3 (d) 4
si
771 × 663 × 365 is :
16. What is the digit in the unit's place of the (a) 1 (b) 2
398 – 389 ls
an by
number represented by 398 – 389?
fu:fir la[;k ds bdkbZ LFkku ij dkSu lk23.
(c) 3
What is the unit digit of /dk
(d) 4
bdkbZ vad D;k gS\
n
vad gS\ [4523 1632
× 2224 1632
× 3225 1632
]
[CDS - 2019 (II)]
ja NTPC 18/01/2017 (Shift-03)
R s
(a) 3 (b) 6 (a) 1 (b) 0
(c) 4 (d) 5
a th
(c) 7 (d) 9
24. The last digit (1001)2008 + 1002 is :
17. Find the unit digit of the sum:
(1001)2008 + 1002 dk vafre vad gS %
fuEufyf[kr ;ksxiQy dk bdkbZ vad Kkr djsaA
(a) 0 (b) 3
ty a
1255 848 (c) 4 (d) 6
11
3 1618
di M
25. Find the last digit of the expression
(a) 0 (b) 2 fuEufyf[kr O;atd dk vafre vad Kkr dhft,A
(c) 4 (d) 6 12 + 22 + ........... + 322 + 332.
23 38 54
18. Let x = (633) – (277) + (266) what is the (a) 0 (b) 5
unit digit of x? (c) 7 (d) 9
;fn x = (633)23 – (277)38 + (266)54 gS] rks
x dk bdkbZ 26. Find the unit digit of/bdkbZ vad Kkr djsaA
vad D;k gS\ 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 + 1010.
SSC CGL Tier II 11/09/2019 (a) 0 (b) 5
(a) 7 (b) 6 (c) 7 (d) 9
(c) 4 (d) 8 27. Find the unit digit of (123456789)123456789.
A
19. If x = (164)169 + (333)337 – (727)726 (123456789)123456789 dk bZdkbZ vad Kkr djsaA
what is the unit digit of x? (a) 2 (b) 6
;fn x = (164) 169 337
+ (333) 726
– (727) gS] (c) 1 (d) 9
rksx dk bdkbZ vad D;k?gS 28. Find the unit digit of 373941
SSC CGL Tier II, 11/09/2019 41
dk bdkbZ vad Kkr djsaA
3739
(a) 5 (b) 7
(a) 7 (b) 9
(c) 8 (d) 9
(c) 3 (d) 1
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29. Find the unit digit of 92332 (a) 4 (b) 8
(c) 9 (d) 6
923 dk bdkbZ vad Kkr djsaA
32
13 47
39. The unit digit of (137 ) is :
(a) 9 (b) 1
(c) 3 (d) 4 (13713)47 dk bZdkbZ vad D;k gS\
30. Find the unit digit of 4 2930 – 4 3029 (a) 1 (b) 3
(c) 5 (d) 7
30
4 29 – 4 30
29
dk bdkbZ vad Kkr djsaA
40. If the unit digit of 433 * 456* 43N is (N + 2),
(a) 2 (b) 8 then what is the value of N?
(c) 4 (d) 6
;fn 433 * 456* 43N dk bdkbZ dk vad(N + 2) gS]rks
31. Find the unit digit of 22324 – 92423
N dk eku D;k gS\
223 – 924 dk bdkbZ vad Kkr djsaA
24 23
(a) 1 (b) 3
(a) 8 (b) 2 (c) 8 (d) 6
(c) 4 (d) 6 41. Find the unit digit of the expression :
32. Find the unit digit of (123456789)123456789!.
r
fuEufyf[kr O;atd dk bdkbZ vad Kkr djsaA
(123456789)123456789! dk bZdkbZ vad Kkr djsaA
si
4 5 6 7 8 9 10
(a) 2 (b) 6 23 × 34 × 45 × 56 × 67 × 78 × 89 ?
(c) 1 (d) 9
an by (a) 2 (b) 1
33. Find the unit digit of 973234! × 234973!.
(c) 0 (d) 3
973234! × 234973! dk bZdkbZ vad Kkr djsaA
n
42. The last digit of the expression
(a) 2 (b) 6
(c) 7 (d) 9 4 + 9² + 4³ + 45 + 96 + ...... 499 + 9100
ja dk vafre vad gS%
R s
34. Find the unit digit of the expression :
fuEufyf[kr O;atd dk bdkbZ vad Kkr djsaA (a) 4 (b) 6
a th
2222 4444 8888 9999
2222 + (4444) + (8888) + (9999) (c) 5 (d) 0
(a) 0 (b) 2 43. The unit digit of 1³ + 2³ + 33 + 4³ + 5³ + .... +
(c) 5 (d) 9 101³
ty a
35. Find the unit digit of the expression :
1³ + 2³ + 33 + 4³ + 5³ + .... + 101³ dk bdkbZ vad
fuEufyf[kr O;atd dk bdkbZ vad Kkr djsaA
D;k gS\
di M
1! + 2! + 3! + 4! +.................+100!.
(a) 1 (b) 0 (a) 0 (b) 5
(c) 3 (d) 5 (c) 6 (d) 1
36. The last digit of the following expression is : 44. The unit digit of 1 + 24 + 34 + 44 + 54 + .... +
4
fuEufyf[kr O;atd dk vafre vad Kkr djsaA 754
11 + 22 + 33 + 44 + ........... 1010 14 + 24 + 34 + 44 + 54 + .... + 754 dk bdkbZ vad D;k
(a) 0 (b) 5 gS\
(c) 7 (d) 9 (a) 0 (b) 5
37. The right most non-zero digits of the number
302720 is : (c) 2 (d) 1
A
la[;k 302720 esa lcls nk,a v'kwU; la[;k gS % 45. What is the unit digit of 15 + 25 + 35 + 45 + 55 +
(a) 1 (b) 3 .... + 955?
(c) 7 (d) 9 15 + 25 + 35 + 45 + 55 + .... + 955 dk bdkbZ vad D;k
38. Which of the following can't be the unit's digit
gS\
of a perfect square?
fuEufyf[kr esa ls dkSu&lk iw.kZ oxZ dk bdkbZ vad ugha(a)gks
0 (b) 5
ldrk gS\ (c) 2 (d) 1
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ANSWER KEY
1.(a) 2.(d) 3.(d) 4.(a) 5.(a) 6.(c) 7.(d) 8.(d) 9.(d) 10.(a)
11.(c) 12.(c) 13.(c) 14.(b) 15.(a) 16.(b) 17.(a) 18.(c) 19.(c) 20.(a)
21.(b) 22.(d) 23.(b) 24.(b) 25.(d) 26.(c) 27.(d) 28.(c) 29.(a) 30.(b)
31.(a) 32.(c) 33.(b) 34.(c) 35.(c) 36.(c) 37.(a) 38.(b) 39.(b) 40.(d)
41.(c) 42.(d) 43.(d) 44.(a) 45.(d)
r
si
an by
n
ja
R s
a th
ty a
di M
A
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Unit Digit/ bdkbZ vad
( Practice Sheet With Solution)
Level-01 9. What is the units digit of 1! + 2! + 3! + .....+99!
+ 100! is?
1. Find the unit digit in (6736)32567
1! + 2! + 3! + .....+99! + 100! dk bdkbZ vad D;k
(6736) 32567 esa bdkbZ dk vad Kkr dhft,A gksrk gS\
(a) 3 (b) 6 (a) 3 (b) 1
(c) 4 (d) 2 (c) 5 (d) 6
2. Find the unit digit of 795 – 358 10. What is the unit of 973234! × 234973!?
795 – 358 dk bdkbZ vad Kkr dhft,A 973234! × 234973! dk bdkbZ vad D;k gS\
r
(a) 4 (b) 7 (a) 2 (b) 6
si
(c) 3 (d) 0 (c) 7 (d) 9
3. Find the unit digit of (17)1999 + (11)1999 –71999 11. Find the unit digit of/dk bdkbZ vad D;k gS\
(17) 1999
an by
+ (11) 1999
–7 1999
dk bdkbZ vad Kkr dhft,A
32 37 38
n
(a) 7 (b) 1
(c) 9 (d) 5 (a) 4 (b) 3
4. ja
Find the unit digit in the square root of 15876 (c) 9 (d) 5
R s
12. Find the unit digit of/bdkbZ vad D;k gksxk\
15876 ds oxZewy esa bdkbZ dk vad Kkr dhft,A
2942! + 4421!
a th
(a) 8 (b) 6
(a) 6 (b) 7
(c) 4 (d) 2
(c) 9 (d) 3
5. Find the unit digit of 111!
13. Find the unit digit of/bdkbZ vad D;k gksxk\
ty a
111! dk bdkbZ vad Kkr dhft;sA 55552345 + 66665678
(a) 8 (b) 5 (a) 1 (b) 3
di M
(c) 1 (d) 0 (c) 5 (d) 7
169 337
6. If x = (164) + (333) – (727)726, then what 14. The unit digit in the sum of (124)372 + (124)373 is.
is the unit digit of x?
(124)372 + (124)373 ds ;ksx esa bdkbZ vad gSA
;fn x = (164) + (333)
169 337
– (727) 726
rksx dk bdkbZ (a) 5 (b) 4
vad Kkr djsaA (c) 20 (d) 0
(a) 5 (b) 9 15. What is the unit digit in the product of
(c) 8 (d) 7 (658 × 539 × 476 × 312)?
7. What is the unit digit of the sum of first 150 (658 × 539 × 476 × 312) ds xq.kuiQy esa bdkbZ
whole numbers? vad D;k gS\
çFke 150 iw.kZ la[;kvksa ds ;ksx dk bdkbZ vad D;k gS\ (a) 4 (b) 2
A
(a) 9 (b) 5 (c) 8 (d) None of these
(c) 0 (d) 1 16. Find the last digit of 2136?
8. If in a two digit number, the digit at units 2136 dk vafre vad Kkr dhft,\
place is z and the digit at tens place is 8, then (a) 6 (b) 3
the number is (c) 7 (d) 9
;fn ,d nks vadksa dh la[;k esa] bdkbZ ds LFkku ij
zzvad
17. What is the unit digit of/dk bdkbZ vad D;k gS\
gS vkSj ngkbZ ds LFkku ij vad 8 gS] rks la[;k gS [45231632×22241632×32251632]
(a) 80z + z (b) 80 + z (a) 1 (b) 0
(c) 8z + 8 (d) 80 z + 1 (c) 4 (d) 5
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18. The unit digit of/dk bdkbZ vad D;k gS\ 28. If x is a positive integer, what is the unit digit of
(13713)47 (24)2x+1 × (33)x+1 × (17)x+2 × (9)2x
(a) 1 (b) 3
(c) 5 (d) 7 ;fn x ,d /ukRed iw.kkZad gS](24)rks 2x+1 × (33)x+1 ×
19. What is the unit digit of/dk bdkbZ vad D;k gS\ (17)x+2 × (9)2x dk bdkbZ vad D;k gS\
234! 973! (a) 2 (b) 4
973 × 234
(a) 2 (b) 6 (c) 6 (d) 8
(c) 9 (d) 7 29. What is the ten's digit in 7400?
20. What is the unit digit in the product of ? 7400 esa ngkbZ dk vad D;k gS\
(3547)153 × (251)72 (a) 1 (b) 0
(3547)153 × (251)72 ds xq.kuiQy esa
bdkbZ vadD;k gSA (c) 2 (d) 9
(a) 5 (b) 6
30. Find unit digit of/dk bdkbZ dk vad D;k gS\
(c) 7 (d) 1
100! 99! 98! 1!
1! + 2! + 3! + ...+100!
Level-02
(a) 6 (b) 4
21. Find the unit digit of (23)21× (24)22 × (26)23 ×
r
(27)24 × (25)25 (c) 9 (d) 7
si
(23)21× (24)22 × (26)23 × (27)24 × (25)25 dk bdkbZ 31. What is the unit digit of/bdkbZ vad Kkr dhft,%
vad Kkr dhft, 1.(1!)1! + 2.(2!)2!+ 3.(3!)3!+.......+ 101.(101!)101!
(a) 5
(c) 0
an by (b) 2
(d) 3
(c) 6 (d) 2
n
(c) 0 (d) 1
22. Find the unit digit of (235) 215 + (314) 326 +
(6736)213 + (3167)112. 32. Find the unit digit of/dk bdkbZ vad Kkr dhft,A
ja dk bdkbZ 4 198
+6 12345
+ 348 66
+ 24 11
+ 1.
R s
215 326 213 112
(235) + (314) + (6736) + (3167)
vad Kkr dhft, (a) 3 (b) 2
a th
(a) 6 (b) 5 (c) 1 (d) 0
(c) 0 (d) 8 33. What is the right most integer of the
23. Find the unit digit of/dk bdkbZ vad Kkr dhft,A expression 65776759 + 54697467 + 65776759 + 54697467
ty a
1! + 2! + 3! + 4! + 5! + .......... + 3333! O;atd dk lcls nkfguk iw.kkZad D;k gSA
(a) 1 (b) 9 65776759 + 54697467 + 65776759 + 54697467
di M
(c) 3 (d) 4
(a) 4 (b) 6
24. The unit digit of [(2543 × 5642) + 45625 + 2342 +
(c) 9 (d) 0
7623] is-
[(2543 × 5642) + 45625 + 2342 + 7623] dk bdkbZ vad gS&34. What is the unit digit of/dk bdkbZvad D;k gSA
413 547 624 812
(a) 1 (b) 2 217 ×819 ×414 × 342 .
(c) 3 (d) 4 (a) 2 (b) 4
25. The unit digit in the square root of 66049 is (c) 6 (d) 8
66049 ds oxZewy esa bdkbZ vad gS 35. How many total tens digit in the calculation
(a) 3 (b) 7 from series 1 to 99?
(c) 8 (d) 2 J`a[kyk 1 ls 99 rd dh x.kuk esa dqy fdrus ngkbZ vad gSa\
26. The unit digit in the expression :/O;atd esa bdkbZ (a) 98 (b) 90
A
vad gSA (c) 99 (d) 100
(36234) (33512) (39180) – (5429) (25123) (31512)
36. What is the unit digit of
(a) 8 (b) 0
167 × 2183 × 497 × 839 × 235 × 111 × 1039
(c) 6 (d) 5
× 251 × 563?
27. Find the unit digit of the expression/O;atd esa
bdkbZ vad D;k gksxk\ 167 × 2183 × 497 × 839 × 235 × 111 × 1039
312 + 322 + 332 + 342 + 352 + 362 + 372 + 382 + 392 × 251 × 563 dk bdkbZ vad D;k gS\
(a) 1 (b) 4 (a) 0 (b) 5
(c) 5 (d) 9 (c) 1 (d) 7
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37. The unit digit of/dk bdkbZ vad D;k gS\ 44. The digit in the unit place of 15 + 25 + ... + 995
13 + 23 + 33 + 43 + ....+1013 15 + 25 + ... + 995 ds bdkbZ LFkku esa vad
(a) 1 (b) 3
(a) 0 (b) 5 (c) 0 (d) 2
(c) 6 (d) 1
45. The unit digit of/dk bdkbZ vad D;k gksxk\
38. Find the unit digit of/dk bdkbZ vad D;k gksxk\
184 + 284 + 384 + 484 + 584 +...........7584
33
22566 (a) 0 (b) 5
(a) 0 (b) 3 (c) 2 (d) 1
(c) 4 (d) 5 46. What is the ten's digit in the expression/O;atd
39. Find the unit digit of/dk bdkbZ vad D;k gksxk\ esa ngkbZ dk vad D;k gS\
64 64 64 943268 × 147347 × 9164 × 3285 × 1139
(a) 3 (b) 4 (a) 5 (b) 0
(c) 5 (d) 6 (c) 6 (d) 8
If a and b are positive integers and x = 4a and
40. Find the unit digit of/dk bdkbZ vad D;k gksxk\ 47.
y = 96, which of the following is a possible unit
r
5251
54 53 digit of xy?
;fn a vkSj b èkukRed iw.kk±d gSax vkSj = 4a vkSj
si
(a) 3 (b) 4
(c) 5 (d) 6 y = 96] rks fuEufyf[kr esa ls dkSuxy lk dk laHkkfor
41. Find the unit digit of
an by 45
67
89
10
is?
bdkbZ vad gS\
(a) 1 (b) 4
n
910 (c) 8 (d) 7
56
78
dk bdkbZ vad Kkr dhft;s\
4 48. What is the unit digit of/ dk bdkbZ vad D;k gksxk\
(a) 4 ja
(b) 6
R s
4 5 6 7 8 9
(c) 0 (d) 2 23 34 45 56 67 78
Find the unit digit of/dk bdkbZ vad D;k gksxk\ (a) 3 (b) 0
a th
42.
(c) 6 (d) 6
(217)413 × (519)547 × (414)624 × (342)812
87
(a) 2 (b) 4 49. Find the unit digit of 5773941 is?
(c) 6 (d) 8
ty a
57739 dk bdkbZ vad Kkr dhft;s\
4187
Level-03
(a) 4 (b) 6
di M
43. Find the unit digit of (888)9235! + (222)9235! + (c) 0 (d) 3
(666)2359! + (9999)9999!
50. What is the unit digit of/dk bdkbZ vad D;k gS\
(888)9235! + (222)9235! + (666)2359! + (9999)9999! dk
bdkbZ vad Kkr dhft,A 2 ×334 45
×4 56
×567
×6 78 89
×7 ×8 9 10
?
(a) 6 (b) 0 (a) 2 (b) 1
(c) 9 (d) 4 (c) 0 (d) 3
ANSWER KEY
1.(b) 2.(a) 3.(b) 4.(b) 5.(d) 6.(c) 7.(b) 8.(b) 9.(a) 10.(b)
A
11.(a) 12.(b) 13.(a) 14.(d) 15.(a) 16.(d) 17.(b) 18.(b) 19.(b) 20.(c)
21.(c) 22.(d) 23.(c) 24.(a) 25.(b) 26.(c) 27.(c) 28.(d) 29.(b) 30.(c)
31.(d) 32.(c) 33.(a) 34.(d) 35.(b) 36.(b) 37.(d) 38.(d) 39.(d) 40.(b)
41.(a) 42.(d) 43.(c) 44.(c) 45.(a) 46.(c) 47.(b) 48.(b) 49.(d) 50.(c)
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SOLUTIONS
1 (b) 8. (b)
unit digit of (0,5,6,1)n is same Unit place = z
unit digit of (6736)32567 is 6. then number 10x + z and ten's digit is 8
2. (a) x=8
795 – 358 it is of the form 80 + z
= 73 – 32 9. (a)
=3–9=4 All numbers starting with 5! will end in zero.
3. (b) unit digit of 1! + 2! + 3! + ... + 100! is same
171999 + 111999 – 71999 as unit digit of 1!+2!+3!+4! = 1 + 2 + 6 + 24 = 33.
= 73 + 1 – 73 unit digit = 3
=3+1–3=1 10. (b)
4. (b) 973234! × 234973!
r
Square root of 15876 is 126 234 !
si
unit digit 6 = Remainder 0
4
'OR'
We know,
42 = 16
an by 973!
4
= Remainder 0
n
62 = 36 unit digit = 34 × 44
1×6=6
4 ja
R s
15876 11. (a)
6 We know, (even)even/odd = even.
a th
158 > 122 3738
32 even
124 or 126
We know, 1252 = 15625 < 15876 even unit digit
ty a
from options only option (a) is even.
15876 = 126
12. (b)
unit digit is 6.
di M
2942! + 4421!
5. (d)
94 + 44 = 1 + 6 = 7
111!
13. (a)
Since, any factorial number greater than or
equal to 5! has zero at unit's place. 55552345 + 66665678
'O' is the required answer. 5 + 6 = 11 unit digit = 1
6. (c) 14. (d)
x = (164)169 + (333)337 – (727)726 124372 + 124373
4169 + 3337 –7726 124372 [1 + 124]
41 + 31 –72 124372 [125]
4 + 3 – 9 = 7 – 9 17 – 9 = 8 Even no. × multiple of 5 = unit digit '0'
A
7. (b) '0' is the unit digit.
n n 1 15. (a)
Sum = ATQ,
2
Multiply each unit digit in the series by the
whole numbers = 0, 1, 2, 3, ..., 149
unit digit
150 = 8 × 9 × 6 × 2 = 72 × 12
Sum = 149 × = 11175
2 Now,
unit digit = 5 unit digit = 2 × 2 = 4
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16. (d) 25. (b)
unit digit of 2136 3
660 49
2132 32
7
unit digit = 9 2
660 > 25
17. (b)
ATQ, 66049 is either 253 or 257
Consider
45231632 × 22241632 × 32251632
2552 = 65025 < 66049
34 × 44 × 54 number is 257
unit digit = 0 unit digit = 7
18. (b) 26. (c)
ATQ, (36)234 (33512) (39180) – (54)29 (25123) (31512)
(13713)47 = (7)13 × 47 6 × 30 × 1 – 4 × 5 × 1
Now, 6–0=6
r
Power divided by 4 27. (c)
(7)1×3 = (7)3 = 343
si
312 + 322 + 332 + 342 + 352 + 362 + 372 + 382 +
unit digit = 3 392
19. (b)
We know
an by =1+4+9+6+5+6+9+4+1
= 45
n
234! & 973! both are divisible by 4 unit digit = 5
then, 28. (d)
(973)4n × (234)4nja
R s
(24)2x+1 × (33)x+1 × (17)x+2 × (9)2x
=1×6=6 = 41 × 3x+1 × 7x+2×1 = 4 × (3 × 7)x+1.7×1
a th
Unit digit = 6 = 4×(21)x+1 × 7
20. (c) = 4 × 1 × 7 = 28
(3547)153 + (251)72 Unit digit = 8
ty a
Unit digit = 71 × 14 29. (b)
=7×1=7 Divide 7400 by 100
di M
21. (c) 100 100
(23)21 × (24)22 × (26)23 × (27)24 × (25)25 7400 74 2401 01
= 31 × 42 × 6 × 74 × 5 100 100 100 100
=3×6×6×1×5=0 tens digit = '0'.
22. (d) 30. (c)
(235)215 + (314)326 + (6736)213 + (3167)112 (1!)100! + (2!)99! + (3!)98! + ........... (99!)1!
= 5 + 42 + 6 + 74 1100! + 299! + 698! + 2497! + (120)96! + .......
=5+6+6+1=8
23. (c)
1! + 2! + 3! + 4! + 5! + ........... + 3333!
A
unit digit = 0
5! onwards unit digit will zero. unit digit = 1 + 2 + 6 + 44 + 0
4 4
unit digit = 1 + 2 + 6 + 4 + 0 + 0 ...... =1+6+6+6=9
=3 31. (d)
24. (a) 1.(1!) 1! + 2.(2!) 2! + 3.(3!) 3! + ........... +
(2543 × 5642) + (456)25 + 2342 + 7623 101.(101!)101!
= (5 × 6) + 6 + 9 + 6 5! onwards unit digit will zeoro.
= 30 + 6 + 9 + 6 = 51 1×1 + 2×(2)2 + 3×(6)6 + 4 ×(24)24
unit digit = 1 unit digit = (1 + 8 + 8 + 4) = 1
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32. (c) 39. (d)
ATQ, We know
(4)198 + (6)12345 + (348)66 + (24)11 + 1 (4)odd = unit digit = 4
(4)2 + (6)1 + (8)2 + (4)3 + 1 (4)even = unit digit = 6
64 64
6 + 6 + 4 + 4 + 1 = 21 then, 64 64 = 4 64 = unit digit = 6
unit digit = 1 40. (b)
33. (a)
5251 5251
65776759 + 54697467 + 65776759 + 54697467 54 53 = 4 53 = (4)odd = unit digit = 4
6759 41. (a)
= 3 Remainder 10
4 Unit digit of 67
89
45
7467 When the power of the base of a number is odd,
= 3 Remainder
4 then even if it is raised to an even or odd
Unit digit = 7³ + 9³ + 7³ + 9³ power, the result will remain odd.
then,
r
=3+9+3+9=4
unit digit = (4)odd = 4
34. (d)
si
42. (d)
217413 × 819547 × 414624 × 342812
ATQ,
413
4
= 1,
547
2
an by
= 1,
624
2
= 0,
812
4
=0
All powers divided by 4
n
413 547 624 812
Unit's digits are:- 7 4 9 4 4 4 2 4
= 71 × 93 × 44 × 24
7×9×6×6 ja
R s
unit digit = 8 =7×9×6×6
= unit digit = 8
a th
35. (b)
43. (c)
1 to 10 tenth's digit = 1
(888)9235! + (222)9235! + (666)2359! + (9999)9999!
11 to 99 tenth's digit = 89
= 84 + 24 + 64 + 94 [factorials are multiple of 4]
ty a
Total (1+ 89) = 90
=6+6+6+1=9
36. (b)
44. (c)
di M
We know
15 + 25 + ... + 995
When, we multiply by 5 to any odd number we = (1 + 2 + 3 + ... + 9) + (1 + 2 + 3 + ... + 9)+ ..
get unit digit 5 10 times
37. (d)
91
We know sum of cube of a natural number 9× × 10 = 450
series 2
unit digit = 0
n n +1 2 45. (a)
=
, here n = 101
2 Given that,
184 + 284 + 384 + 584 + .......7584
101102 2 All powers multiple of 4
A
= (101 × 51)2
2 14×n + 24×n + 34×n + 44×n + 54×n + 64×n + 74×n + 84×n
+ 94×n + 104×n....... 754×n
(5151)2 = unit digit = 1 14 + 24 + 34 + 44 + 54 + 64 + 74 + 84 + 94 + 104
38. (d) 1+6+1+6+5+6+1+6+1+0
We know Unit digit for 1 to 10 = 3
(5)odd/even = unit digit = 5 Unit digit for 1 to 70 = 3 × 7 = 1
then, Unit digit for 1 to 75
33
6633
= 1 + 14 + 24 + 34 + 44 + 54
22566 = 5 = unit digit = 5
= (1 + 1 + 6 + 1 + 6 + 5) = 0
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46. (c) 48. (b)
For ten's digit divide given expression by 100, ATQ,
the remainder will be ten's digit
23 3 3 3 × (3)multiple of 4 (5)even × (6)odd × (7)multiple of 4
943268 147347 9164 3285 1139
=2×1×5×6×1
100
= unit digit = 0
943268 147347 2291 657 1139 49. (d)
=
5
Given that,
3 2 1 2 4 3 20 60 4187
= 57739
5 5 20 100
ATQ,
60 is remainder ten's digit = 6
41 devided by 4
47. (b)
39
4187 187
x = 4a , y = 9b 57739 739 74
r
unit digit of x = 4 or 6
= 73 = 343
si
unit digit of y = 9 or 1
unit digit = 3
possibilities are:-
an by 50. (c)
4×1=4
ATQ,
4×9=6
n
4 5 6 7 8 9 10
6×9=4 23 34 4 5 56 67 78 89
6×1=6 ja (2)81 × (3)4n × (4)odd × (5)even × (6)odd × (7)4n × (8)1
R s
The unit digits of xy will be 4 or 6 =2×1×4×5×6×1×8
a th
by options, (B) is correct. = unit digit = 0
ty a
di M
A
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DIVISIBILITY (foHkkT;rk
)
(CLASSROOM SHEET)
1. 210102 can be divided exactly by 7. What is the least value of x so that number
8x5215 becomes divisible by 9?
210102 dks iw.kZr% fdlds }kjk foHkkftr fd;k tk ldrk gS\
SSC CPO 16/03/2019 (Shift- 02)
x dk U;wure eku D;k gksxk rkfd la[;k
x5215]
8 9 ls
(a) 7 (b) 3
foHkkT; gks\
SSC CPO 09/11/2022 (Shift-01)
(c) 4 (d) 8
2. Number 30744 is divisible by which one digit (a) 3 (b) 1
number? (c) 5 (d) 6
la[;k 30744 ,d vad okyh fdl la[;k ls foHkkT; gS\ 8. A 9-digit number 846523X7Y is divisible by 9, and
r
SSC CPO 16/03/2019 (Shift- 03) Y – X = 6. Find the value of 2X 4Y .
si
(a) All the other numbers except 5 and 7 ,d 9-vadh; la[;k gS846523X7Y tks 9 ls foHkkT; gS]
(b) only 2, 3 and 6 vkSjY – X = 6 gSA rc] 2X 4Y dk eku Kkr dhft,A
an by
(c) only 2, 3, 6 and 9
(d) All the other numbers except 5
SSC CGL 26/07/2023 (Shift-02)
n
(a) 4 (b) 2
3. Number 106974 is divisible by which one digit
(c) 6 (d) 8
number?
ja 9. If the 4-digit number x67y is exactly divisible
R s
la[;k 106974 ,d vad okyh fdl la[;k ls foHkkT; gS\ by 9, then the least value of (x + y) is _____.
SSC CPO 15/03/2019 (Shift- 03) ;fn x67y ,d ,slh 4&vadh; la[;k gS tks 9 ls iw.kZr%
a th
(a) only 2, 3, 4 and 7 (b) only 2 and 3 foHkkT; gS] (xrks+y) dk U;wure eku_________gSA
(c) only 2, 3 and 4 (d) only 2, 3 and 7 SSC CGL 02/12/2022 (Shift-01)
4. Which of the following is NOT divisible by 6. (a) 9 (b) 0
ty a
fuEufyf[kr esa ls dkSu lk 6 ls foHkkT; ugha gSA (c) 5 (d) 3
(i) 1,97,232 (ii) 9,72,132 10. The number 1254216 is divisible by which of
di M
the following numbers?
(iii)8,00,552 (iv) 17,90,184
SSC CHSL 02/08/2023 Shift-02
la[;k 1254216 fuEufyf[kr esa ls fdl la[;k ls foHkkT; gS\
SSC CGL MAINS 06/03/2023
(a) (i) (b) (iii)
(c) (iv) (d) (ii) (a) 16 (b) 5
5. Find the value of k such that the number (c) 8 (d) 11
k53206k is divisible by 6. 11. If the 8-digit number 123456xy is divisible by
k dk eku Kkr dhft,] ftlls k53206k, 6 ls foHkkT; 8, then the total possible pairs of (x,y) are:
gksA ;fn 8 vadksa dh la[;k123456xy, 8 ls foHkkT; gS]
SSC CGL 19/04/2022 (Shift- 03) rks(x, y) ds dqy fdrus laHkkfor ;qXe gksaxs\
(a) 4 (b) 1 SSC CGL 03/12/2022 (Shift-04)
A
(c) 2 (d) 7 (a) 8 (b) 13
6. If the number 87m6203m is divisible by 6, then (c) 10 (d) 11
find the sum of all possible values of ‘m’. 12. If the four-digit number 463y is divisible by 7,
then what is the value of y?
;fn la[;k 87m6203m, 6 ls foHkkT; gks] 'm'
rks ds ;fn pkj vadksa dh la[;k463y, 7 ls foHkkT; gS] rks
y
lHkh laHkkfor ekuksa dk ;ksxiQy Kkr djsaA dk eku D;k gksxk\
SSC CHSL 05/08/2021 (Shift- 3) SSC CGL MAINS 07/03/2023
(a) 10 (b) 15 (a) 4 (b) 6
(c) 16 (d) 20 (c) 3 (d) 5
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13. Which number among 24963, 24973, 24983 20. Find the greatest possible value of (a + b) for
and 24993 is divisible by 7? which the 8-digit number 143b203a is divisible
24963, 24973, 24983 vkSj24993 esa ls dkSu&lh by 15.
la[;k 7 ls foHkkT; gS\
(a + b)dk vf/dre laHko eku Kkr djsa] ftlds fy, 8-
SSC CGL 18/07/2023 (Shift-02)
(a) 24973 (b) 24983
vadh; la[;k 143b203a, 15 ls foHkkT; gS
(c) 24963 (d) 24993 SSC CHSL 09/06/2022 (Shift- 02)
14. Which of the following numbers is divisible by (a) 15 (b) 17
11? (c) 16 (d) 14
fuEufyf•r esa ls dkSu lh la[;k 11 ls foHkkT; gS\ 21. What are the values of R and M, respectively,
SSC CGL MAINS 07/03/2023 if the given number is perfectly divisible by
16 and 11?
(a) 5214341 (b) 5648741
(c) 6598321 (d) 2378965
;fn nh xbZ la[;k 16 vkSj 11 ls iw.kZr% foHkkT; gS] r
R vkSjM ds eku Øe'k% D;k gSa\
15. Which of the following numbers is NOT divisible
by 11? 34R05030M6
r
fuEufyf•r esa ls dkSu lh la[;k 11 ls foHkkT; ugha gS\ SSC CPO 10/11/2022 (Shift-01)
(a) 4 and 6 (b) 7 and 5
si
SSC CHSL 02/08/2023 Shift-01
(c) 5 and 5 (d) 5 and 7
(a) 1735624 (b) 752563
16.
(c) 1661308 an by (d) 1904529
Find the smallest value of a so that 42a48b (a
22. An 11-digit number 7823326867X is divisible
by 18. What is the value of X?
,d 11&vadh; la[;k 7823326867X, 18 ls foHkkT; gSA
n
X
> b) is divisible by 11.
a dk U;wure eku Kkr djsa ftlesa
42a48b (a > b) la[;k
dk eku D;k gS\
ja
11 ls foHkkT; gksA SSC CGL 19/07/2023 (Shift-01)
R s
SSC CGL 17/08/2021 (Shift 02) (a) 6 (b) 4
a th
(a) 4 (b) 5 (c) 8 (d) 2
(c) 0 (d) 9 23. A six-digit number 11p9q4 is divisible by 24.
Then the greatest possible value for p and q is:
17. Find the greatest value of b so that 30a68b (a
;fn Ng vadksa dh la[;k11p9q4, 24 ls foHkkT; gS]
ty a
> b) is divisible by 11.
b dk vf/dre eku Kkr djsa] ftlls 30a68b (a > b) la[;k rksp vkSjq vfèkdre laHkkfor eku Kkr dhft,A
di M
11 ls foHkkT; gksA SSC CGL MAINS 26/10/2023
SSC CGL 13/08/2021 (Shift-03) (a) 42 (b) 32
(a) 4 (b) 9 (c) 56 (d) 68
(c) 3 (d) 6 24. The 6-digit number 439xy5 is divisible by 125.
How many such 6-digit numbers are there?
18. If a 4-digit number 273x is divisible by 12 and a
7-digit number y854z 06 is divisible by 11 , then 16&vadh; la[;k439xy5 125 ls foHkkT; gSA ,slh fdruh
what is the value of (x + y + z) ? 6&vadh; la[;k,¡ gSa\
;fn 4& vadksa dh ,d la[;k273x,12 ls foHkkT; gS vkSj SSC PHASE IX 2022
7- vadks dh ,d la[;k y854z06, 11 ls foHkkT; gS] rks (a) 4 (b) 2
(x + y + z) dk eku D;k gksxk\ (c) 5 (d) 3
A
SSC PHASE IX 2022
25. The six-digit number 537xy5 is divisible by
(a) 18 (b) 12 125. How many such six-digit numbers are
(c) 6 (d) 20 there?
19. Which of the following numbers is divisible by N% vad okyh la[;k
537xy5, 125 ls foHkkT; gSA ,sls N%
15? vadksa okyh fdruh la[;k,a gSa\
fuEufyf[kr esa ls dkSu&lh la[;k 15 ls foHkkT; gS\ SSC CHSL 19/04/2021 (Shift- 1)
SSC CHSL 01/06/2022 (Shift- 2)
(a) 4 (b) 2
(a) 1,65,485 (b) 3,06,045
(c) 3 (d) 5
(c) 2,12,695 (d) 2,95,145
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26. Which of the following numbers is divisible (a) 4 (b) 2
by both 7 and 11? (c) 3 (d) 6
fuEufyf[kr esa ls dkSu&lh la[;k 7 vkSj 11 nksuksa ls\ foHkkT;
32. IfgSa
the number 5X8146Y is divisible by 88, then
CHSL 2019 19/03/2020 (Shift- 01) what is the value of X?
(a) 16,324 (b) 10,098 ;fn la[;k 5X8146Y, 88 ls foHkkT; gS] X
rksdk eku
(c) 10,108 (d) 10,087 D;k gS\
27. Which of the following numbers is divisible SSC CHSL 30/05/2022 (Shift- 3)
by both 9 and 11? (a) 3 (b) 1
fuEufyf[kr esa ls dkSu&lh la[;k 9 vkSj 11 nksuksa ls foHkkT;
(c) 2 (d) 4
gSa\ 33. If 4M37094267N is divisible by both 8 and
CHSL 2019 19/03/2020 (Shift- 03) 11, where M and N are single digit integers,
then the valued M and N is :
(a) 10,089 (b) 10,098
(c) 10,108 (d) 10,087
;fn la[;k 4M37094267N, 8 vkSj11 nksuksa ls foHkkT;
gS] tgk¡M vkSjN ,dy vad iw.kk±d gSa]M
rksvkSjN dk
r
28. If a 10-digit number 75462A97B6 is divisible
eku gS &
si
by 72, then the value of 8 A – 4B is:
CHSL 2019 20/10/2020 (Shift- 01)
75462A97B6 ,d ,slh 10&vadh; la[;k gS tks 72 ls
an by (a) M = 5, N = 6 (b) M = 2, N = 5
foHkkT; gS] rks
8 A – 4B dk eku Kkr djsaA (c) M = 5, N = 2 (d) M = 5, N = 4
n
34. What is the least value of x+y, if 10 digit
SSC CGL MAINS 29/01/ 2022
number 780x533y24 is divisible by 88?
(a) ja (b) ;fn 10 vadksa dh la[;k780x533y24, 88 ls foHkkT; gS]
R s
28 21
rksx+y dk U;wure eku D;k gS\
a th
(c) 30 (d) 27 SSC CHSL 03/08/2023 Shift-04
29. If a nine-digit number 789x6378y is divisible (a) 4 (b) 3
by 72, then the value of xy is: (c) 1 (d) 2
ty a
;fn ukS & vadh; ,d la[;k 789x6378y la[;k 72 ls 35. If the nine-digit number 7p5964q28 is
foHkkT; gS] xy
rksdk eku fdruk gksxk\ completely divisible by 88, what is the value
di M
SSC CGL MAINS 03/02/ 2022 of (p² – q), for the largest value of q, where p
and q are natural numbers ?
(a) 10 (b) 12
;fn ukS vad okyh la[;k7p5964q28, 88 ls iw.kZr%
(c) 08 (d) 15
foHkkT; gS]qrks
ds vf/dre eku ds fy, (p² – q) dk eku
30. If the 7-digit number 612x97y is divisible by Kkr djsa] tgk¡
p vkSjq izkÑfrd la[;k,a gSaA
72 , and the 6 - digit number 34z178 is divisible
SSC CGL 16/08/2021 (Shift 02)
by 11 , then the value of (x- 2y + 3z) is:
(a) 72 (b) 9
;fn 7 vadksa dh la[;k
612x97y, 72 ls foHkkT; gS] vkSj
(c) 0 (d) 81
6 vadksa dh la[;k
34z178,11 gS] rks
(x- 2y + 3z) dk
36. If the nine-digit number 9m2365n48 is
eku gS%
completely divisible by 88, what is the value
A
SSC PHASE IX 2022 of (m2 × n2) for the smallest value of n, where
(a) 6 (b) 5 m and n are natural numbers?
(c) 2 (d) 7 9m2365n48 ,d ,slh ukS vadksa dh la[;k gS] tks 88 ls
31. If the number 6336633P is divisible by 132, then iw.kZr% foHkkT; ngS]
ds rks
lcls NksVs eku ds fy,
(m2 ×
the value of P is: n2) dk eku D;k gksxk] tgk¡
m vkSjn izkÑr la[;k,¡ gSa\
;fn la[;k 6336633P, 132 }kjk foHkkT; gS]Prks
dk SSC CGL 13/04/2022 (Shift- 03)
eku D;k gksxk\ (a) 32 (b) 64
SSC CPO 10/11/2022 (Shift-02) (c) 20 (d) 36
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37. If the number 55p1067q9 is exactly divisible 43. If the 6 – digit number 57zxy8 is divisible by
by 99, then pq is equal to: each of 7,11 and 13 , then (x – 2y + z) is:
;f n la[;k 55p1067q9 la[;k 99 ls iwjh rjg foHkkT; ;f n 6 vadksa dh la[;k57zxy8 7,11 vkSj13 eas ls
gS] rkspq dk eku fdruk gksxk\ izR;sd ls foHkkT; gS](xrks
– 2y + z) dk eku gSA
SSC CPO 11/11/2022 (Shift-03) SSC PHASE IX 2022
(a) 35 (b) 28 (a) –1 (b) 2
(c) 36 (d) 42 (c) 1 (d) –2
38. If the 9-digit number 4x92y6588 is exactly 44. Which of the following is the least 6-digit
number that is divisible by 93?
divisible by 99(x + y < 10), then what is the
value of 2(y – x) ? fuEufyf[kr esa ls dkSu&lh 6 vadksa dh og lcls NksVh la[;k
gS] tks 93 ls foHkkT; gS\
;f n 9& vadksa dh la[;k
4x92y6588 99(x + y < 10)
SSC CHSL 10/06/2022 (Shift- 3)
99 ls iw.kZr% foHkkT; gS]
2(y rks
– x) dk eku D;k gS\
(a) 100068 (b) 100070
SSC PHASE IX 2022
(c) 100075 (d) 100065
(a) –2 (b) 1
r
45. Which of the following numbers is NOT divisible
(c) 2 (d) –1 by 150?
si
39. Which of the following numbers is divisible by fuEufyf[kr esa ls dkSu&lh la[;k 150 ls foHkkT; ugha gS
55? SSC CHSL 07/06/2022 (Shift 01)
an by
fuEUkfy[kr esa ls dkSu&lh la[;k 55 ls foHkkT; gS\ (a) 320550 (b) 333300
n
SSC CHSL 06/06/2022 (Shift- 03) (c) 453750 (d) 201300
(a) 178765 (b) 185625 46. The largest six-digit number exactly divisible
(c) 171125 ja (d) 164485
by 243 is:
R s
6 vadksa dh og cM+h ls cM+h la[;k dkSu&lh gS tks 243
40. If the 5-digit number 535ab is divisible by 3,
iw.kZr% foHkkT; gS\
a th
7 and 11, then what is the value of
SSC CHSL 11/08/2021 (Shift- 3)
(a2 – b2 + ab) ?
(a) 999947 (b) 999949
;f n 5-vadh; la[;k 535ab, 3, 7 vkSj11 ls foHkkT; gS] (c) 999943 (d) 999945
ty a
rks(a2 – b2 + ab) dk eku D;k gS\ 47. What is the greatest five-digit number that
SSC CGL MAINS 15/11/ 2020 is completely divisible by 8, 15, 16, 21 and
di M
(a) 95 (b) 83 5?
(c) 89 (d) 77 8] 15] 16] 21 vkSj 5 ls iwjh rjg ls foHkkT; gksus okyh lcls
41. If the 5-digit number 235xy is divisible by 3,
cM+h ikap&vadh; la[;k fuEu eas ls dkSu&lh gS\
SSC CHSL 04/08/2021 (Shift- 1)
7 and 11, then what is the value of
(3x – 4y) ? (a) 98320 (b) 99120
(c) 95760 (d) 92680
;f n 5-vadh; la[;k 235xy, 3, 7 vkSj11 ls foHkkT; gS]
48. 225 + 226 + 227 is divisible by
rks(3x – 4y) dk eku D;k gS\
225 + 226 + 227 fd lls foHkkT; gS\
SSC CGL MAINS 16/11/ 2020
CHSL 2019 19/10/2020 (Shift- 3)
(a) 8 (b) 9 (a) 6 (b) 7
A
(c) 5 (d) 10 (c) 5 (d) 9
42. The number 823p2q is exactly divisible by 7, 49. Which of the following numbers will completely
11 and 13. What is the value of (p–q)? divide 7 81 + 7 82 + 7 83 ?
la[;k 823p2q, 7, 11 vkSj13 ls iw.kZr% foHkkT;
(p–gSA fuEufyf[kr esa ls dkSu lh la[;k
7 81 + 7 82 + 7 83 dks iwjh
q) dk eku Kkr djsaA rjg ls foHkkftr djsxh\
SSC CGL 20/08/2021 (Shift 03) CHSL 2019 17/03/2020 (Shift- 01)
(a) 399 (b) 389
(a) 8 (b) 3
(c) 387 (d) 397
(c) 5 (d) 11
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50. 571 + 572 + 573 + 574 + 575 is divisible by which of 56. The sum of 3-digit numbers abc, bca and cab
the following number? is always divisible by:
571 + 572 + 573 + 574 + 575 fn, x, fodYiksa esa ls fdl 3 vadksa dh la[;k
abc, bca vkSjcab dk ;ksx ges'kk --
la[;k ls foHkkT; gS\ ------ ls foHkkT; gksrk gS
SSC CGL MAINS 03/02/2022 SSC PHASE IX 2022
(a) 71 (b) 69 (a) 35 (b) 41
(c) 89 (d) 73 (c) 37 (d) 31
51. If a positive integer 'n' is divisible by 3, 5 and 57. A 3-digit number ABC, where A is at the
7, then what is the next larger integer divisible hundreds place, B is at the tens place and C is
by all these numbersd? at the unit's place, is so-written as ABCABC and
is divided by the LCM of 7, 11 and 13. What
;fn ,d ldkjkRed iw.kk±dn' 3] 5 vkSj 7 ls foHkkT; gS]
will be the result?
rks lHkh la[;kvksa ds }kjk vxys dkSu&lk cM+k iw.kk±d
foHkkT; gksxk\ 3 vadksa dh ,d la[;kABC gS ftlesaA lSadM+s ds LFkku ij]
CHSL 2019 19/10/2020 (Shift- 01) B ngkbZ ds LFkku ij rFkk
C bdkbZ ds LFkku ij gSA bl la[;k
r
(a) n + 21 (b) n + 35 dksABCABC :i esa iqu% fy•k x;k rFkk bls 7] 11 vkSj 13
si
(c) n + 105 (d) n + 110
ds y?kqÙke lekioR;Z ls foHkkftr fd;k x;kA ifj.kke D;k
52. The greatest whole number by which the
an by
expression n4 + 6n3 + 11n2 + 6n + 24 is divisible gksxk \
for every natural number n, is :
n
UP Constable 28/01/2019 (Shift-02)
lcls cM+h iw.kZ la[;k ftlls O;atd
n4 + 6n3 + 11n2 + (a) ABC (b) CBA
6n + 24, n ds izR;sd eku ds fy, iw.kZr% foHkkT; gS\ (c) BCA
ja (d) AAB
R s
(a) 6 (b) 24 58. Consider a 6-digit number of the form XYXYXY.
a th
(c) 132 (d) 48 The number is divisible by:
53. If n is a whole number greater than 1, then XYXYXY tSlh 6&vadh la[;k ds ckjs esa fopkj djsaA ;g
n2(n2 – 1) is always divisible by : la[;k HkkT; gS%
;fn n ,d 1 ls cM+h iw.kZ la[;k gksn2rks
(n2 – 1) ges'kk
ty a
[CDS - 2023 (I)]
foHkkT; gksxk % (a) 3 and 7 only (b) 7 and 13 only
di M
(a) 16 (b) 2 (c) 3, 13 and 37 only (d) 3, 7, 13 and 37
(c) 10 (d) 8 59. What will be the greatest number 32a78b,
54. A six digit number is formed by repeating a which is divisible by 3 but NOT divisible by 9?
three digit number, for example 256256 or (Where a and b are single tigit number).
678678 etc. Any number of this form is always lcls cM+h la[;k32a78b dkSu lh gksxh] tks 3 ls foHkkT;
exactly divisible by :
gS ysfdu 9 ls foHkkT; ugha gS\ a(tgk¡
vkSjb ,dy 'kh"kZd
rhu vadksa dh la[;k dh iqujko`fÙk ls ,d N%vadh; la[;k la[;k gSa)A
curh gS] tSls & 256256 vFkok 678678 vkfnAbl izdkj SSC CHSL 09/08/2023 Shift-02
dh dksbZ la[;k ges'kk iw.kZr% foHkkftr gksxh % (a) 324781 (b) 329787
(a) 7 only (b) 11 only (c) 326787 (d) 329784
A
(c) 13 only (d) 1001 60. What is the value of x in the number 3426x if
55. The sum of 3-digit numbers abc, cab and bca the number is divisible by 6 but not divisible
is not divisible by: by 5?
3- vad okyh la[;kvksaabc, cab vkSjbca dk ;ksxiQy - la[;k 3426x esax dk eku Kkr dhft,] ;fn la[;k 6 ls
-------- ls foHkkT; ugha gS\ foHkkT; gS ysfdu 5 ls foHkkT; ugha gSA
SSC CGL 24/08/2021 (Shift 01) SSC CHSL, 10/08/2023 (Shift-4)
(a) a + b + c (b) 37 (a) 3 (b) 4
(c) 31 (d) 3 (c) 6 (d) 8
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61. What is the number of possible pairs of 68. Find the number of three digit natural numbers
(P, Q), if the number 357P25Q is divisible by divisible by 8, 12 and 15.
both 3 and 5?
8] 12 vkSj 15 ls foHkkftr gksus okyh rhu vadksa dh
ds laHkkfor ;qXeksa dh la[;k D;k gS] ;fn la[;k izkÑfrd la[;kvksa dh la[;k Kkr djsaA
(P, Q)
357P25Q, 3 vkSj 5 nksuksa ls foHkkT; gS\
(a) 3 (b) 5
(a) 7 (b) 6
(c) 8 (d) 9
(c) 5 (d) None of these
62. How many numbers between 3 and 200 are 69. How many numbers are there between 1 and
divisible by 7 ? 900 which are not divisible by 2, 3
or 5?
3 vkSj 200 ds chp fdruh la[;k 7 ls foHkkT; gSa\
(a) 27 (b) 28
1 ls 900 ds chp ,slh fdruh la[;k,¡ gSa tks 2] 3 vFkok
(c) 29 (d) 36
5 ls foHkkftr ugha gS\
63. How many numbers between 100 and 1000 are (a) 240 (b) 245
divisible by 17?
r
(c) 250 (d) None of these
100 vkSj 1000 ds chp fdruh la[;k,¡ 17 ls foHkkT; gSa\
70. How many numbers are there from 1 to 100
si
SSC CHSL 01/06/2022 (Shift- 1) which are neither divisible by 3 nor by 5?
(a) 51
(c) 52
an by (b) 53
(d) 54
1 ls 100 rd ,slh fdruh la[;k,¡ gSa tks uk rks 3 ls vkSj
uk gh 5 ls foHkkftr gS\
n
64. How many positive two-digit numbers can be (a) 53 (b) 54
divided by 6?
ja (c) 55 (d) None of these
R s
nks vadksa dh fdruh /ukRed la[;kvksa dks 6 ls fOkHkkftr
71. How many numbers are there from 700 to 950
fd;k tk ldrk gS\
a th
which are neither divisible by 3 nor by 7 ?
SSC CHSL 06/06/2022 (Shift 01)
(a) 12 (b) 15 700 ls 950 rd ,slh fdruh la[;k,¡ gSa tks uk rks 3 ls vkSj
(c) 14 (d) 13
uk gh 7 ls foHkkftr gS\
ty a
65. How many numbers between 800 and 2000 are SSC CGL 04/03/ 2020 (Shift- 03)
di M
divisible by 13? (a) 144 (b) 143
800 vkSj 2000 ds chp fdruh la[;k,¡ 13 ls foHkkT; gS\ (c) 146 (d) None of these
CHSL 2019 19/10/2020 (Shift- 01) 72. How many numbers are there from 200 to 800
(a) 90 (b) 92 which are neither divisible by 5 nor by 7 ?
(c) 91 (d) 93
200 ls 800 rd ,slh fdruh la[;k,¡ gSa tks uk rks 5 ls vkSj
66. How many numbers between 1000 and 5000
uk gh 7 ls foHkkftr gS\
are exactly divisible by 225?
SSC CGL 04/03/2020 (Shift- 03)
1000 vkSj 5000 ds chp fdruh la[;k,¡ gS tks 225 ls
iw.kZr% foHkkftr gS\ (a) 407 (b) 410
(a) 16 (b) 18 (c) 413 (d) 411
A
(c) 19 (d) 12 73. How many numbers are there from 500 to 650
67. How many numbers between 300 and 700 are (including both) which are neither divisible by
divisible by 5, 6 and 8? 3 nor by 7?
500 vkSj 700 ds chp fdruh la[;k 5] 6 vkSj 8 ls foHkkT; 500 ls 650 rd (nksuksa dks lfEefyr djrs gq,) ,slh fdruh
gSa\ la[;k,a gSa tks 3 vkSj 7 nkuksa ls foHkkT; ugha gSa\
CPO 2019 25/11/2020 (Shift- 02) SSC CGL 11/04/2022 (Shift- 02)
(a) 20 (b) 2 (a) 21 (b) 121
(c) 5 (d) 3
(c) 87 (d) 99
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74. Let m and n be natural numbers. What is the 75. How many five-digit numbers of the form
minimum value of (m + n) such that 33m + 22n XXYXX is/are divisible by 33?
is divisible by 121? XXYXX ds :i dh ik¡p vadks dh fdruh la[;k,¡ 33 ls
eku yhft, m vkSjn /u iw.kZ la[;k,¡ gSaA
(m + n) U;wure foHkkftr gksrh gSa\
eku D;k gS ftlls fd 33m + 22n, 121 ls HkkT; gks\ [CDS - 2018 (II)]
[CDS - 2023 (I)] (a) 1 (b) 3
(a) 3 (b) 4 (c) 5 (d) Infinite
(c) 5 (d) 10
ANSWER KEY
1.(b) 2.(d) 3.(d) 4.(b) 5.(a) 6.(a) 7.(d) 8.(c) 9.(c) 10.(c)
r
11.(b) 12.(a) 13.(b) 14.(a) 15.(b) 16.(b) 17.(c) 18.(a) 19.(b) 20.(d)
si
21.(c) 22.(d) 23.(c) 24.(a) 25.(a) 26.(a) 27.(b) 28.(a) 29.(c) 30.(c)
31.(d) 32.(a)an by
33.(c) 34.(d) 35.(b) 36.(b) 37.(a) 38.(c) 39.(b) 40.(d)
n
41.(c) 42.(c) 43.(a) 44.(a) 45.(c) 46.(d) 47.(b) 48.(b) 49.(a) 50.(a)
51.(c) 52.(c) ja53.(b) 54.(d) 55.(c) 56.(c) 57.(a) 58.(b) 59.(d) 60.(c)
R s
a th
61.(a) 62.(b) 63.(b) 64.(b) 65.(b) 66.(b) 67.(d) 68.(c) 69.(a) 70.(a)
71.(a) 72.(d) 73.(c) 74.(d) 75.(b)
ty a
di M
A
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Unit Digit/ bdkbZ vad
( Practice Sheet With Solution)
Level-01 9. What is the units digit of 1! + 2! + 3! + .....+99!
+ 100! is?
1. Find the unit digit in (6736)32567
1! + 2! + 3! + .....+99! + 100! dk bdkbZ vad D;k
(6736) 32567 esa bdkbZ dk vad Kkr dhft,A gksrk gS\
(a) 3 (b) 6 (a) 3 (b) 1
(c) 4 (d) 2 (c) 5 (d) 6
2. Find the unit digit of 795 – 358 10. What is the unit of 973234! × 234973!?
795 – 358 dk bdkbZ vad Kkr dhft,A 973234! × 234973! dk bdkbZ vad D;k gS\
r
(a) 4 (b) 7 (a) 2 (b) 6
si
(c) 3 (d) 0 (c) 7 (d) 9
3. Find the unit digit of (17)1999 + (11)1999 –71999 11. Find the unit digit of/dk bdkbZ vad D;k gS\
(17) 1999
an by
+ (11) 1999
–7 1999
dk bdkbZ vad Kkr dhft,A
32 37 38
n
(a) 7 (b) 1
(c) 9 (d) 5 (a) 4 (b) 3
4. ja
Find the unit digit in the square root of 15876 (c) 9 (d) 5
R s
12. Find the unit digit of/bdkbZ vad D;k gksxk\
15876 ds oxZewy esa bdkbZ dk vad Kkr dhft,A
2942! + 4421!
a th
(a) 8 (b) 6
(a) 6 (b) 7
(c) 4 (d) 2
(c) 9 (d) 3
5. Find the unit digit of 111!
13. Find the unit digit of/bdkbZ vad D;k gksxk\
ty a
111! dk bdkbZ vad Kkr dhft;sA 55552345 + 66665678
(a) 8 (b) 5 (a) 1 (b) 3
di M
(c) 1 (d) 0 (c) 5 (d) 7
169 337
6. If x = (164) + (333) – (727)726, then what 14. The unit digit in the sum of (124)372 + (124)373 is.
is the unit digit of x?
(124)372 + (124)373 ds ;ksx esa bdkbZ vad gSA
;fn x = (164) + (333)
169 337
– (727) 726
rksx dk bdkbZ (a) 5 (b) 4
vad Kkr djsaA (c) 20 (d) 0
(a) 5 (b) 9 15. What is the unit digit in the product of
(c) 8 (d) 7 (658 × 539 × 476 × 312)?
7. What is the unit digit of the sum of first 150 (658 × 539 × 476 × 312) ds xq.kuiQy esa bdkbZ
whole numbers? vad D;k gS\
çFke 150 iw.kZ la[;kvksa ds ;ksx dk bdkbZ vad D;k gS\ (a) 4 (b) 2
A
(a) 9 (b) 5 (c) 8 (d) None of these
(c) 0 (d) 1 16. Find the last digit of 2136?
8. If in a two digit number, the digit at units 2136 dk vafre vad Kkr dhft,\
place is z and the digit at tens place is 8, then (a) 6 (b) 3
the number is (c) 7 (d) 9
;fn ,d nks vadksa dh la[;k esa] bdkbZ ds LFkku ij
zzvad
17. What is the unit digit of/dk bdkbZ vad D;k gS\
gS vkSj ngkbZ ds LFkku ij vad 8 gS] rks la[;k gS [45231632×22241632×32251632]
(a) 80z + z (b) 80 + z (a) 1 (b) 0
(c) 8z + 8 (d) 80 z + 1 (c) 4 (d) 5
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18. The unit digit of/dk bdkbZ vad D;k gS\ 28. If x is a positive integer, what is the unit digit of
(13713)47 (24)2x+1 × (33)x+1 × (17)x+2 × (9)2x
(a) 1 (b) 3
(c) 5 (d) 7 ;fn x ,d /ukRed iw.kkZad gS](24)rks 2x+1 × (33)x+1 ×
19. What is the unit digit of/dk bdkbZ vad D;k gS\ (17)x+2 × (9)2x dk bdkbZ vad D;k gS\
234! 973! (a) 2 (b) 4
973 × 234
(a) 2 (b) 6 (c) 6 (d) 8
(c) 9 (d) 7 29. What is the ten's digit in 7400?
20. What is the unit digit in the product of ? 7400 esa ngkbZ dk vad D;k gS\
(3547)153 × (251)72 (a) 1 (b) 0
(3547)153 × (251)72 ds xq.kuiQy esa
bdkbZ vadD;k gSA (c) 2 (d) 9
(a) 5 (b) 6
30. Find unit digit of/dk bdkbZ dk vad D;k gS\
(c) 7 (d) 1
100! 99! 98! 1!
1! + 2! + 3! + ...+100!
Level-02
(a) 6 (b) 4
21. Find the unit digit of (23)21× (24)22 × (26)23 ×
r
(27)24 × (25)25 (c) 9 (d) 7
si
(23)21× (24)22 × (26)23 × (27)24 × (25)25 dk bdkbZ 31. What is the unit digit of/bdkbZ vad Kkr dhft,%
vad Kkr dhft, 1.(1!)1! + 2.(2!)2!+ 3.(3!)3!+.......+ 101.(101!)101!
(a) 5
(c) 0
an by (b) 2
(d) 3
(c) 6 (d) 2
n
(c) 0 (d) 1
22. Find the unit digit of (235) 215 + (314) 326 +
(6736)213 + (3167)112. 32. Find the unit digit of/dk bdkbZ vad Kkr dhft,A
ja dk bdkbZ 4 198
+6 12345
+ 348 66
+ 24 11
+ 1.
R s
215 326 213 112
(235) + (314) + (6736) + (3167)
vad Kkr dhft, (a) 3 (b) 2
a th
(a) 6 (b) 5 (c) 1 (d) 0
(c) 0 (d) 8 33. What is the right most integer of the
23. Find the unit digit of/dk bdkbZ vad Kkr dhft,A expression 65776759 + 54697467 + 65776759 + 54697467
ty a
1! + 2! + 3! + 4! + 5! + .......... + 3333! O;atd dk lcls nkfguk iw.kkZad D;k gSA
(a) 1 (b) 9 65776759 + 54697467 + 65776759 + 54697467
di M
(c) 3 (d) 4
(a) 4 (b) 6
24. The unit digit of [(2543 × 5642) + 45625 + 2342 +
(c) 9 (d) 0
7623] is-
[(2543 × 5642) + 45625 + 2342 + 7623] dk bdkbZ vad gS&34. What is the unit digit of/dk bdkbZvad D;k gSA
413 547 624 812
(a) 1 (b) 2 217 ×819 ×414 × 342 .
(c) 3 (d) 4 (a) 2 (b) 4
25. The unit digit in the square root of 66049 is (c) 6 (d) 8
66049 ds oxZewy esa bdkbZ vad gS 35. How many total tens digit in the calculation
(a) 3 (b) 7 from series 1 to 99?
(c) 8 (d) 2 J`a[kyk 1 ls 99 rd dh x.kuk esa dqy fdrus ngkbZ vad gSa\
26. The unit digit in the expression :/O;atd esa bdkbZ (a) 98 (b) 90
A
vad gSA (c) 99 (d) 100
(36234) (33512) (39180) – (5429) (25123) (31512)
36. What is the unit digit of
(a) 8 (b) 0
167 × 2183 × 497 × 839 × 235 × 111 × 1039
(c) 6 (d) 5
× 251 × 563?
27. Find the unit digit of the expression/O;atd esa
bdkbZ vad D;k gksxk\ 167 × 2183 × 497 × 839 × 235 × 111 × 1039
312 + 322 + 332 + 342 + 352 + 362 + 372 + 382 + 392 × 251 × 563 dk bdkbZ vad D;k gS\
(a) 1 (b) 4 (a) 0 (b) 5
(c) 5 (d) 9 (c) 1 (d) 7
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37. The unit digit of/dk bdkbZ vad D;k gS\ 44. The digit in the unit place of 15 + 25 + ... + 995
13 + 23 + 33 + 43 + ....+1013 15 + 25 + ... + 995 ds bdkbZ LFkku esa vad
(a) 1 (b) 3
(a) 0 (b) 5 (c) 0 (d) 2
(c) 6 (d) 1
45. The unit digit of/dk bdkbZ vad D;k gksxk\
38. Find the unit digit of/dk bdkbZ vad D;k gksxk\
184 + 284 + 384 + 484 + 584 +...........7584
33
22566 (a) 0 (b) 5
(a) 0 (b) 3 (c) 2 (d) 1
(c) 4 (d) 5 46. What is the ten's digit in the expression/O;atd
39. Find the unit digit of/dk bdkbZ vad D;k gksxk\ esa ngkbZ dk vad D;k gS\
64 64 64 943268 × 147347 × 9164 × 3285 × 1139
(a) 3 (b) 4 (a) 5 (b) 0
(c) 5 (d) 6 (c) 6 (d) 8
If a and b are positive integers and x = 4a and
40. Find the unit digit of/dk bdkbZ vad D;k gksxk\ 47.
y = 96, which of the following is a possible unit
r
5251
54 53 digit of xy?
;fn a vkSj b èkukRed iw.kk±d gSax vkSj = 4a vkSj
si
(a) 3 (b) 4
(c) 5 (d) 6 y = 96] rks fuEufyf[kr esa ls dkSuxy lk dk laHkkfor
41. Find the unit digit of
an by 45
67
89
10
is?
bdkbZ vad gS\
(a) 1 (b) 4
n
910 (c) 8 (d) 7
56
78
dk bdkbZ vad Kkr dhft;s\
4 48. What is the unit digit of/ dk bdkbZ vad D;k gksxk\
(a) 4 ja
(b) 6
R s
4 5 6 7 8 9
(c) 0 (d) 2 23 34 45 56 67 78
Find the unit digit of/dk bdkbZ vad D;k gksxk\ (a) 3 (b) 0
a th
42.
(c) 6 (d) 6
(217)413 × (519)547 × (414)624 × (342)812
87
(a) 2 (b) 4 49. Find the unit digit of 5773941 is?
(c) 6 (d) 8
ty a
57739 dk bdkbZ vad Kkr dhft;s\
4187
Level-03
(a) 4 (b) 6
di M
43. Find the unit digit of (888)9235! + (222)9235! + (c) 0 (d) 3
(666)2359! + (9999)9999!
50. What is the unit digit of/dk bdkbZ vad D;k gS\
(888)9235! + (222)9235! + (666)2359! + (9999)9999! dk
bdkbZ vad Kkr dhft,A 2 ×334 45
×4 56
×567
×6 78 89
×7 ×8 9 10
?
(a) 6 (b) 0 (a) 2 (b) 1
(c) 9 (d) 4 (c) 0 (d) 3
ANSWER KEY
1.(b) 2.(a) 3.(b) 4.(b) 5.(d) 6.(c) 7.(b) 8.(b) 9.(a) 10.(b)
A
11.(a) 12.(b) 13.(a) 14.(d) 15.(a) 16.(d) 17.(b) 18.(b) 19.(b) 20.(c)
21.(c) 22.(d) 23.(c) 24.(a) 25.(b) 26.(c) 27.(c) 28.(d) 29.(b) 30.(c)
31.(d) 32.(c) 33.(a) 34.(d) 35.(b) 36.(b) 37.(d) 38.(d) 39.(d) 40.(b)
41.(a) 42.(d) 43.(c) 44.(c) 45.(a) 46.(c) 47.(b) 48.(b) 49.(d) 50.(c)
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SOLUTIONS
1 (b) 8. (b)
unit digit of (0,5,6,1)n is same Unit place = z
unit digit of (6736)32567 is 6. then number 10x + z and ten's digit is 8
2. (a) x=8
795 – 358 it is of the form 80 + z
= 73 – 32 9. (a)
=3–9=4 All numbers starting with 5! will end in zero.
3. (b) unit digit of 1! + 2! + 3! + ... + 100! is same
171999 + 111999 – 71999 as unit digit of 1!+2!+3!+4! = 1 + 2 + 6 + 24 = 33.
= 73 + 1 – 73 unit digit = 3
=3+1–3=1 10. (b)
4. (b) 973234! × 234973!
r
Square root of 15876 is 126 234 !
si
unit digit 6 = Remainder 0
4
'OR'
We know,
42 = 16
an by 973!
4
= Remainder 0
n
62 = 36 unit digit = 34 × 44
1×6=6
4 ja
R s
15876 11. (a)
6 We know, (even)even/odd = even.
a th
158 > 122 3738
32 even
124 or 126
We know, 1252 = 15625 < 15876 even unit digit
ty a
from options only option (a) is even.
15876 = 126
12. (b)
unit digit is 6.
di M
2942! + 4421!
5. (d)
94 + 44 = 1 + 6 = 7
111!
13. (a)
Since, any factorial number greater than or
equal to 5! has zero at unit's place. 55552345 + 66665678
'O' is the required answer. 5 + 6 = 11 unit digit = 1
6. (c) 14. (d)
x = (164)169 + (333)337 – (727)726 124372 + 124373
4169 + 3337 –7726 124372 [1 + 124]
41 + 31 –72 124372 [125]
4 + 3 – 9 = 7 – 9 17 – 9 = 8 Even no. × multiple of 5 = unit digit '0'
A
7. (b) '0' is the unit digit.
n n 1 15. (a)
Sum = ATQ,
2
Multiply each unit digit in the series by the
whole numbers = 0, 1, 2, 3, ..., 149
unit digit
150 = 8 × 9 × 6 × 2 = 72 × 12
Sum = 149 × = 11175
2 Now,
unit digit = 5 unit digit = 2 × 2 = 4
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16. (d) 25. (b)
unit digit of 2136 3
660 49
2132 32
7
unit digit = 9 2
660 > 25
17. (b)
ATQ, 66049 is either 253 or 257
Consider
45231632 × 22241632 × 32251632
2552 = 65025 < 66049
34 × 44 × 54 number is 257
unit digit = 0 unit digit = 7
18. (b) 26. (c)
ATQ, (36)234 (33512) (39180) – (54)29 (25123) (31512)
(13713)47 = (7)13 × 47 6 × 30 × 1 – 4 × 5 × 1
Now, 6–0=6
r
Power divided by 4 27. (c)
(7)1×3 = (7)3 = 343
si
312 + 322 + 332 + 342 + 352 + 362 + 372 + 382 +
unit digit = 3 392
19. (b)
We know
an by =1+4+9+6+5+6+9+4+1
= 45
n
234! & 973! both are divisible by 4 unit digit = 5
then, 28. (d)
(973)4n × (234)4nja
R s
(24)2x+1 × (33)x+1 × (17)x+2 × (9)2x
=1×6=6 = 41 × 3x+1 × 7x+2×1 = 4 × (3 × 7)x+1.7×1
a th
Unit digit = 6 = 4×(21)x+1 × 7
20. (c) = 4 × 1 × 7 = 28
(3547)153 + (251)72 Unit digit = 8
ty a
Unit digit = 71 × 14 29. (b)
=7×1=7 Divide 7400 by 100
di M
21. (c) 100 100
(23)21 × (24)22 × (26)23 × (27)24 × (25)25 7400 74 2401 01
= 31 × 42 × 6 × 74 × 5 100 100 100 100
=3×6×6×1×5=0 tens digit = '0'.
22. (d) 30. (c)
(235)215 + (314)326 + (6736)213 + (3167)112 (1!)100! + (2!)99! + (3!)98! + ........... (99!)1!
= 5 + 42 + 6 + 74 1100! + 299! + 698! + 2497! + (120)96! + .......
=5+6+6+1=8
23. (c)
1! + 2! + 3! + 4! + 5! + ........... + 3333!
A
unit digit = 0
5! onwards unit digit will zero. unit digit = 1 + 2 + 6 + 44 + 0
4 4
unit digit = 1 + 2 + 6 + 4 + 0 + 0 ...... =1+6+6+6=9
=3 31. (d)
24. (a) 1.(1!) 1! + 2.(2!) 2! + 3.(3!) 3! + ........... +
(2543 × 5642) + (456)25 + 2342 + 7623 101.(101!)101!
= (5 × 6) + 6 + 9 + 6 5! onwards unit digit will zeoro.
= 30 + 6 + 9 + 6 = 51 1×1 + 2×(2)2 + 3×(6)6 + 4 ×(24)24
unit digit = 1 unit digit = (1 + 8 + 8 + 4) = 1
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32. (c) 39. (d)
ATQ, We know
(4)198 + (6)12345 + (348)66 + (24)11 + 1 (4)odd = unit digit = 4
(4)2 + (6)1 + (8)2 + (4)3 + 1 (4)even = unit digit = 6
64 64
6 + 6 + 4 + 4 + 1 = 21 then, 64 64 = 4 64 = unit digit = 6
unit digit = 1 40. (b)
33. (a)
5251 5251
65776759 + 54697467 + 65776759 + 54697467 54 53 = 4 53 = (4)odd = unit digit = 4
6759 41. (a)
= 3 Remainder 10
4 Unit digit of 67
89
45
7467 When the power of the base of a number is odd,
= 3 Remainder
4 then even if it is raised to an even or odd
Unit digit = 7³ + 9³ + 7³ + 9³ power, the result will remain odd.
then,
r
=3+9+3+9=4
unit digit = (4)odd = 4
34. (d)
si
42. (d)
217413 × 819547 × 414624 × 342812
ATQ,
413
4
= 1,
547
2
an by
= 1,
624
2
= 0,
812
4
=0
All powers divided by 4
n
413 547 624 812
Unit's digits are:- 7 4 9 4 4 4 2 4
= 71 × 93 × 44 × 24
7×9×6×6 ja
R s
unit digit = 8 =7×9×6×6
= unit digit = 8
a th
35. (b)
43. (c)
1 to 10 tenth's digit = 1
(888)9235! + (222)9235! + (666)2359! + (9999)9999!
11 to 99 tenth's digit = 89
= 84 + 24 + 64 + 94 [factorials are multiple of 4]
ty a
Total (1+ 89) = 90
=6+6+6+1=9
36. (b)
44. (c)
di M
We know
15 + 25 + ... + 995
When, we multiply by 5 to any odd number we = (1 + 2 + 3 + ... + 9) + (1 + 2 + 3 + ... + 9)+ ..
get unit digit 5 10 times
37. (d)
91
We know sum of cube of a natural number 9× × 10 = 450
series 2
unit digit = 0
n n +1 2 45. (a)
=
, here n = 101
2 Given that,
184 + 284 + 384 + 584 + .......7584
101102 2 All powers multiple of 4
A
= (101 × 51)2
2 14×n + 24×n + 34×n + 44×n + 54×n + 64×n + 74×n + 84×n
+ 94×n + 104×n....... 754×n
(5151)2 = unit digit = 1 14 + 24 + 34 + 44 + 54 + 64 + 74 + 84 + 94 + 104
38. (d) 1+6+1+6+5+6+1+6+1+0
We know Unit digit for 1 to 10 = 3
(5)odd/even = unit digit = 5 Unit digit for 1 to 70 = 3 × 7 = 1
then, Unit digit for 1 to 75
33
6633
= 1 + 14 + 24 + 34 + 44 + 54
22566 = 5 = unit digit = 5
= (1 + 1 + 6 + 1 + 6 + 5) = 0
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46. (c) 48. (b)
For ten's digit divide given expression by 100, ATQ,
the remainder will be ten's digit
23 3 3 3 × (3)multiple of 4 (5)even × (6)odd × (7)multiple of 4
943268 147347 9164 3285 1139
=2×1×5×6×1
100
= unit digit = 0
943268 147347 2291 657 1139 49. (d)
=
5
Given that,
3 2 1 2 4 3 20 60 4187
= 57739
5 5 20 100
ATQ,
60 is remainder ten's digit = 6
41 devided by 4
47. (b)
39
4187 187
x = 4a , y = 9b 57739 739 74
r
unit digit of x = 4 or 6
= 73 = 343
si
unit digit of y = 9 or 1
unit digit = 3
possibilities are:-
an by 50. (c)
4×1=4
ATQ,
4×9=6
n
4 5 6 7 8 9 10
6×9=4 23 34 4 5 56 67 78 89
6×1=6 ja (2)81 × (3)4n × (4)odd × (5)even × (6)odd × (7)4n × (8)1
R s
The unit digits of xy will be 4 or 6 =2×1×4×5×6×1×8
a th
by options, (B) is correct. = unit digit = 0
ty a
di M
A
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DIVISIBILITY (foHkkT;rk
)
(CLASSROOM SHEET)
1. 210102 can be divided exactly by 7. What is the least value of x so that number
8x5215 becomes divisible by 9?
210102 dks iw.kZr% fdlds }kjk foHkkftr fd;k tk ldrk gS\
SSC CPO 16/03/2019 (Shift- 02)
x dk U;wure eku D;k gksxk rkfd la[;k
x5215]
8 9 ls
(a) 7 (b) 3
foHkkT; gks\
SSC CPO 09/11/2022 (Shift-01)
(c) 4 (d) 8
2. Number 30744 is divisible by which one digit (a) 3 (b) 1
number? (c) 5 (d) 6
la[;k 30744 ,d vad okyh fdl la[;k ls foHkkT; gS\ 8. A 9-digit number 846523X7Y is divisible by 9, and
r
SSC CPO 16/03/2019 (Shift- 03) Y – X = 6. Find the value of 2X 4Y .
si
(a) All the other numbers except 5 and 7 ,d 9-vadh; la[;k gS846523X7Y tks 9 ls foHkkT; gS]
(b) only 2, 3 and 6 vkSjY – X = 6 gSA rc] 2X 4Y dk eku Kkr dhft,A
an by
(c) only 2, 3, 6 and 9
(d) All the other numbers except 5
SSC CGL 26/07/2023 (Shift-02)
n
(a) 4 (b) 2
3. Number 106974 is divisible by which one digit
(c) 6 (d) 8
number?
ja 9. If the 4-digit number x67y is exactly divisible
R s
la[;k 106974 ,d vad okyh fdl la[;k ls foHkkT; gS\ by 9, then the least value of (x + y) is _____.
SSC CPO 15/03/2019 (Shift- 03) ;fn x67y ,d ,slh 4&vadh; la[;k gS tks 9 ls iw.kZr%
a th
(a) only 2, 3, 4 and 7 (b) only 2 and 3 foHkkT; gS] (xrks+y) dk U;wure eku_________gSA
(c) only 2, 3 and 4 (d) only 2, 3 and 7 SSC CGL 02/12/2022 (Shift-01)
4. Which of the following is NOT divisible by 6. (a) 9 (b) 0
ty a
fuEufyf[kr esa ls dkSu lk 6 ls foHkkT; ugha gSA (c) 5 (d) 3
(i) 1,97,232 (ii) 9,72,132 10. The number 1254216 is divisible by which of
di M
the following numbers?
(iii)8,00,552 (iv) 17,90,184
SSC CHSL 02/08/2023 Shift-02
la[;k 1254216 fuEufyf[kr esa ls fdl la[;k ls foHkkT; gS\
SSC CGL MAINS 06/03/2023
(a) (i) (b) (iii)
(c) (iv) (d) (ii) (a) 16 (b) 5
5. Find the value of k such that the number (c) 8 (d) 11
k53206k is divisible by 6. 11. If the 8-digit number 123456xy is divisible by
k dk eku Kkr dhft,] ftlls k53206k, 6 ls foHkkT; 8, then the total possible pairs of (x,y) are:
gksA ;fn 8 vadksa dh la[;k123456xy, 8 ls foHkkT; gS]
SSC CGL 19/04/2022 (Shift- 03) rks(x, y) ds dqy fdrus laHkkfor ;qXe gksaxs\
(a) 4 (b) 1 SSC CGL 03/12/2022 (Shift-04)
A
(c) 2 (d) 7 (a) 8 (b) 13
6. If the number 87m6203m is divisible by 6, then (c) 10 (d) 11
find the sum of all possible values of ‘m’. 12. If the four-digit number 463y is divisible by 7,
then what is the value of y?
;fn la[;k 87m6203m, 6 ls foHkkT; gks] 'm'
rks ds ;fn pkj vadksa dh la[;k463y, 7 ls foHkkT; gS] rks
y
lHkh laHkkfor ekuksa dk ;ksxiQy Kkr djsaA dk eku D;k gksxk\
SSC CHSL 05/08/2021 (Shift- 3) SSC CGL MAINS 07/03/2023
(a) 10 (b) 15 (a) 4 (b) 6
(c) 16 (d) 20 (c) 3 (d) 5
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13. Which number among 24963, 24973, 24983 20. Find the greatest possible value of (a + b) for
and 24993 is divisible by 7? which the 8-digit number 143b203a is divisible
24963, 24973, 24983 vkSj24993 esa ls dkSu&lh by 15.
la[;k 7 ls foHkkT; gS\
(a + b)dk vf/dre laHko eku Kkr djsa] ftlds fy, 8-
SSC CGL 18/07/2023 (Shift-02)
(a) 24973 (b) 24983
vadh; la[;k 143b203a, 15 ls foHkkT; gS
(c) 24963 (d) 24993 SSC CHSL 09/06/2022 (Shift- 02)
14. Which of the following numbers is divisible by (a) 15 (b) 17
11? (c) 16 (d) 14
fuEufyf•r esa ls dkSu lh la[;k 11 ls foHkkT; gS\ 21. What are the values of R and M, respectively,
SSC CGL MAINS 07/03/2023 if the given number is perfectly divisible by
16 and 11?
(a) 5214341 (b) 5648741
(c) 6598321 (d) 2378965
;fn nh xbZ la[;k 16 vkSj 11 ls iw.kZr% foHkkT; gS] r
R vkSjM ds eku Øe'k% D;k gSa\
15. Which of the following numbers is NOT divisible
by 11? 34R05030M6
r
fuEufyf•r esa ls dkSu lh la[;k 11 ls foHkkT; ugha gS\ SSC CPO 10/11/2022 (Shift-01)
(a) 4 and 6 (b) 7 and 5
si
SSC CHSL 02/08/2023 Shift-01
(c) 5 and 5 (d) 5 and 7
(a) 1735624 (b) 752563
16.
(c) 1661308 an by (d) 1904529
Find the smallest value of a so that 42a48b (a
22. An 11-digit number 7823326867X is divisible
by 18. What is the value of X?
,d 11&vadh; la[;k 7823326867X, 18 ls foHkkT; gSA
n
X
> b) is divisible by 11.
a dk U;wure eku Kkr djsa ftlesa
42a48b (a > b) la[;k
dk eku D;k gS\
ja
11 ls foHkkT; gksA SSC CGL 19/07/2023 (Shift-01)
R s
SSC CGL 17/08/2021 (Shift 02) (a) 6 (b) 4
a th
(a) 4 (b) 5 (c) 8 (d) 2
(c) 0 (d) 9 23. A six-digit number 11p9q4 is divisible by 24.
Then the greatest possible value for p and q is:
17. Find the greatest value of b so that 30a68b (a
;fn Ng vadksa dh la[;k11p9q4, 24 ls foHkkT; gS]
ty a
> b) is divisible by 11.
b dk vf/dre eku Kkr djsa] ftlls 30a68b (a > b) la[;k rksp vkSjq vfèkdre laHkkfor eku Kkr dhft,A
di M
11 ls foHkkT; gksA SSC CGL MAINS 26/10/2023
SSC CGL 13/08/2021 (Shift-03) (a) 42 (b) 32
(a) 4 (b) 9 (c) 56 (d) 68
(c) 3 (d) 6 24. The 6-digit number 439xy5 is divisible by 125.
How many such 6-digit numbers are there?
18. If a 4-digit number 273x is divisible by 12 and a
7-digit number y854z 06 is divisible by 11 , then 16&vadh; la[;k439xy5 125 ls foHkkT; gSA ,slh fdruh
what is the value of (x + y + z) ? 6&vadh; la[;k,¡ gSa\
;fn 4& vadksa dh ,d la[;k273x,12 ls foHkkT; gS vkSj SSC PHASE IX 2022
7- vadks dh ,d la[;k y854z06, 11 ls foHkkT; gS] rks (a) 4 (b) 2
(x + y + z) dk eku D;k gksxk\ (c) 5 (d) 3
A
SSC PHASE IX 2022
25. The six-digit number 537xy5 is divisible by
(a) 18 (b) 12 125. How many such six-digit numbers are
(c) 6 (d) 20 there?
19. Which of the following numbers is divisible by N% vad okyh la[;k
537xy5, 125 ls foHkkT; gSA ,sls N%
15? vadksa okyh fdruh la[;k,a gSa\
fuEufyf[kr esa ls dkSu&lh la[;k 15 ls foHkkT; gS\ SSC CHSL 19/04/2021 (Shift- 1)
SSC CHSL 01/06/2022 (Shift- 2)
(a) 4 (b) 2
(a) 1,65,485 (b) 3,06,045
(c) 3 (d) 5
(c) 2,12,695 (d) 2,95,145
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26. Which of the following numbers is divisible (a) 4 (b) 2
by both 7 and 11? (c) 3 (d) 6
fuEufyf[kr esa ls dkSu&lh la[;k 7 vkSj 11 nksuksa ls\ foHkkT;
32. IfgSa
the number 5X8146Y is divisible by 88, then
CHSL 2019 19/03/2020 (Shift- 01) what is the value of X?
(a) 16,324 (b) 10,098 ;fn la[;k 5X8146Y, 88 ls foHkkT; gS] X
rksdk eku
(c) 10,108 (d) 10,087 D;k gS\
27. Which of the following numbers is divisible SSC CHSL 30/05/2022 (Shift- 3)
by both 9 and 11? (a) 3 (b) 1
fuEufyf[kr esa ls dkSu&lh la[;k 9 vkSj 11 nksuksa ls foHkkT;
(c) 2 (d) 4
gSa\ 33. If 4M37094267N is divisible by both 8 and
CHSL 2019 19/03/2020 (Shift- 03) 11, where M and N are single digit integers,
then the valued M and N is :
(a) 10,089 (b) 10,098
(c) 10,108 (d) 10,087
;fn la[;k 4M37094267N, 8 vkSj11 nksuksa ls foHkkT;
gS] tgk¡M vkSjN ,dy vad iw.kk±d gSa]M
rksvkSjN dk
r
28. If a 10-digit number 75462A97B6 is divisible
eku gS &
si
by 72, then the value of 8 A – 4B is:
CHSL 2019 20/10/2020 (Shift- 01)
75462A97B6 ,d ,slh 10&vadh; la[;k gS tks 72 ls
an by (a) M = 5, N = 6 (b) M = 2, N = 5
foHkkT; gS] rks
8 A – 4B dk eku Kkr djsaA (c) M = 5, N = 2 (d) M = 5, N = 4
n
34. What is the least value of x+y, if 10 digit
SSC CGL MAINS 29/01/ 2022
number 780x533y24 is divisible by 88?
(a) ja (b) ;fn 10 vadksa dh la[;k780x533y24, 88 ls foHkkT; gS]
R s
28 21
rksx+y dk U;wure eku D;k gS\
a th
(c) 30 (d) 27 SSC CHSL 03/08/2023 Shift-04
29. If a nine-digit number 789x6378y is divisible (a) 4 (b) 3
by 72, then the value of xy is: (c) 1 (d) 2
ty a
;fn ukS & vadh; ,d la[;k 789x6378y la[;k 72 ls 35. If the nine-digit number 7p5964q28 is
foHkkT; gS] xy
rksdk eku fdruk gksxk\ completely divisible by 88, what is the value
di M
SSC CGL MAINS 03/02/ 2022 of (p² – q), for the largest value of q, where p
and q are natural numbers ?
(a) 10 (b) 12
;fn ukS vad okyh la[;k7p5964q28, 88 ls iw.kZr%
(c) 08 (d) 15
foHkkT; gS]qrks
ds vf/dre eku ds fy, (p² – q) dk eku
30. If the 7-digit number 612x97y is divisible by Kkr djsa] tgk¡
p vkSjq izkÑfrd la[;k,a gSaA
72 , and the 6 - digit number 34z178 is divisible
SSC CGL 16/08/2021 (Shift 02)
by 11 , then the value of (x- 2y + 3z) is:
(a) 72 (b) 9
;fn 7 vadksa dh la[;k
612x97y, 72 ls foHkkT; gS] vkSj
(c) 0 (d) 81
6 vadksa dh la[;k
34z178,11 gS] rks
(x- 2y + 3z) dk
36. If the nine-digit number 9m2365n48 is
eku gS%
completely divisible by 88, what is the value
A
SSC PHASE IX 2022 of (m2 × n2) for the smallest value of n, where
(a) 6 (b) 5 m and n are natural numbers?
(c) 2 (d) 7 9m2365n48 ,d ,slh ukS vadksa dh la[;k gS] tks 88 ls
31. If the number 6336633P is divisible by 132, then iw.kZr% foHkkT; ngS]
ds rks
lcls NksVs eku ds fy,
(m2 ×
the value of P is: n2) dk eku D;k gksxk] tgk¡
m vkSjn izkÑr la[;k,¡ gSa\
;fn la[;k 6336633P, 132 }kjk foHkkT; gS]Prks
dk SSC CGL 13/04/2022 (Shift- 03)
eku D;k gksxk\ (a) 32 (b) 64
SSC CPO 10/11/2022 (Shift-02) (c) 20 (d) 36
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37. If the number 55p1067q9 is exactly divisible 43. If the 6 – digit number 57zxy8 is divisible by
by 99, then pq is equal to: each of 7,11 and 13 , then (x – 2y + z) is:
;f n la[;k 55p1067q9 la[;k 99 ls iwjh rjg foHkkT; ;f n 6 vadksa dh la[;k57zxy8 7,11 vkSj13 eas ls
gS] rkspq dk eku fdruk gksxk\ izR;sd ls foHkkT; gS](xrks
– 2y + z) dk eku gSA
SSC CPO 11/11/2022 (Shift-03) SSC PHASE IX 2022
(a) 35 (b) 28 (a) –1 (b) 2
(c) 36 (d) 42 (c) 1 (d) –2
38. If the 9-digit number 4x92y6588 is exactly 44. Which of the following is the least 6-digit
number that is divisible by 93?
divisible by 99(x + y < 10), then what is the
value of 2(y – x) ? fuEufyf[kr esa ls dkSu&lh 6 vadksa dh og lcls NksVh la[;k
gS] tks 93 ls foHkkT; gS\
;f n 9& vadksa dh la[;k
4x92y6588 99(x + y < 10)
SSC CHSL 10/06/2022 (Shift- 3)
99 ls iw.kZr% foHkkT; gS]
2(y rks
– x) dk eku D;k gS\
(a) 100068 (b) 100070
SSC PHASE IX 2022
(c) 100075 (d) 100065
(a) –2 (b) 1
r
45. Which of the following numbers is NOT divisible
(c) 2 (d) –1 by 150?
si
39. Which of the following numbers is divisible by fuEufyf[kr esa ls dkSu&lh la[;k 150 ls foHkkT; ugha gS
55? SSC CHSL 07/06/2022 (Shift 01)
an by
fuEUkfy[kr esa ls dkSu&lh la[;k 55 ls foHkkT; gS\ (a) 320550 (b) 333300
n
SSC CHSL 06/06/2022 (Shift- 03) (c) 453750 (d) 201300
(a) 178765 (b) 185625 46. The largest six-digit number exactly divisible
(c) 171125 ja (d) 164485
by 243 is:
R s
6 vadksa dh og cM+h ls cM+h la[;k dkSu&lh gS tks 243
40. If the 5-digit number 535ab is divisible by 3,
iw.kZr% foHkkT; gS\
a th
7 and 11, then what is the value of
SSC CHSL 11/08/2021 (Shift- 3)
(a2 – b2 + ab) ?
(a) 999947 (b) 999949
;f n 5-vadh; la[;k 535ab, 3, 7 vkSj11 ls foHkkT; gS] (c) 999943 (d) 999945
ty a
rks(a2 – b2 + ab) dk eku D;k gS\ 47. What is the greatest five-digit number that
SSC CGL MAINS 15/11/ 2020 is completely divisible by 8, 15, 16, 21 and
di M
(a) 95 (b) 83 5?
(c) 89 (d) 77 8] 15] 16] 21 vkSj 5 ls iwjh rjg ls foHkkT; gksus okyh lcls
41. If the 5-digit number 235xy is divisible by 3,
cM+h ikap&vadh; la[;k fuEu eas ls dkSu&lh gS\
SSC CHSL 04/08/2021 (Shift- 1)
7 and 11, then what is the value of
(3x – 4y) ? (a) 98320 (b) 99120
(c) 95760 (d) 92680
;f n 5-vadh; la[;k 235xy, 3, 7 vkSj11 ls foHkkT; gS]
48. 225 + 226 + 227 is divisible by
rks(3x – 4y) dk eku D;k gS\
225 + 226 + 227 fd lls foHkkT; gS\
SSC CGL MAINS 16/11/ 2020
CHSL 2019 19/10/2020 (Shift- 3)
(a) 8 (b) 9 (a) 6 (b) 7
A
(c) 5 (d) 10 (c) 5 (d) 9
42. The number 823p2q is exactly divisible by 7, 49. Which of the following numbers will completely
11 and 13. What is the value of (p–q)? divide 7 81 + 7 82 + 7 83 ?
la[;k 823p2q, 7, 11 vkSj13 ls iw.kZr% foHkkT;
(p–gSA fuEufyf[kr esa ls dkSu lh la[;k
7 81 + 7 82 + 7 83 dks iwjh
q) dk eku Kkr djsaA rjg ls foHkkftr djsxh\
SSC CGL 20/08/2021 (Shift 03) CHSL 2019 17/03/2020 (Shift- 01)
(a) 399 (b) 389
(a) 8 (b) 3
(c) 387 (d) 397
(c) 5 (d) 11
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50. 571 + 572 + 573 + 574 + 575 is divisible by which of 56. The sum of 3-digit numbers abc, bca and cab
the following number? is always divisible by:
571 + 572 + 573 + 574 + 575 fn, x, fodYiksa esa ls fdl 3 vadksa dh la[;k
abc, bca vkSjcab dk ;ksx ges'kk --
la[;k ls foHkkT; gS\ ------ ls foHkkT; gksrk gS
SSC CGL MAINS 03/02/2022 SSC PHASE IX 2022
(a) 71 (b) 69 (a) 35 (b) 41
(c) 89 (d) 73 (c) 37 (d) 31
51. If a positive integer 'n' is divisible by 3, 5 and 57. A 3-digit number ABC, where A is at the
7, then what is the next larger integer divisible hundreds place, B is at the tens place and C is
by all these numbersd? at the unit's place, is so-written as ABCABC and
is divided by the LCM of 7, 11 and 13. What
;fn ,d ldkjkRed iw.kk±dn' 3] 5 vkSj 7 ls foHkkT; gS]
will be the result?
rks lHkh la[;kvksa ds }kjk vxys dkSu&lk cM+k iw.kk±d
foHkkT; gksxk\ 3 vadksa dh ,d la[;kABC gS ftlesaA lSadM+s ds LFkku ij]
CHSL 2019 19/10/2020 (Shift- 01) B ngkbZ ds LFkku ij rFkk
C bdkbZ ds LFkku ij gSA bl la[;k
r
(a) n + 21 (b) n + 35 dksABCABC :i esa iqu% fy•k x;k rFkk bls 7] 11 vkSj 13
si
(c) n + 105 (d) n + 110
ds y?kqÙke lekioR;Z ls foHkkftr fd;k x;kA ifj.kke D;k
52. The greatest whole number by which the
an by
expression n4 + 6n3 + 11n2 + 6n + 24 is divisible gksxk \
for every natural number n, is :
n
UP Constable 28/01/2019 (Shift-02)
lcls cM+h iw.kZ la[;k ftlls O;atd
n4 + 6n3 + 11n2 + (a) ABC (b) CBA
6n + 24, n ds izR;sd eku ds fy, iw.kZr% foHkkT; gS\ (c) BCA
ja (d) AAB
R s
(a) 6 (b) 24 58. Consider a 6-digit number of the form XYXYXY.
a th
(c) 132 (d) 48 The number is divisible by:
53. If n is a whole number greater than 1, then XYXYXY tSlh 6&vadh la[;k ds ckjs esa fopkj djsaA ;g
n2(n2 – 1) is always divisible by : la[;k HkkT; gS%
;fn n ,d 1 ls cM+h iw.kZ la[;k gksn2rks
(n2 – 1) ges'kk
ty a
[CDS - 2023 (I)]
foHkkT; gksxk % (a) 3 and 7 only (b) 7 and 13 only
di M
(a) 16 (b) 2 (c) 3, 13 and 37 only (d) 3, 7, 13 and 37
(c) 10 (d) 8 59. What will be the greatest number 32a78b,
54. A six digit number is formed by repeating a which is divisible by 3 but NOT divisible by 9?
three digit number, for example 256256 or (Where a and b are single tigit number).
678678 etc. Any number of this form is always lcls cM+h la[;k32a78b dkSu lh gksxh] tks 3 ls foHkkT;
exactly divisible by :
gS ysfdu 9 ls foHkkT; ugha gS\ a(tgk¡
vkSjb ,dy 'kh"kZd
rhu vadksa dh la[;k dh iqujko`fÙk ls ,d N%vadh; la[;k la[;k gSa)A
curh gS] tSls & 256256 vFkok 678678 vkfnAbl izdkj SSC CHSL 09/08/2023 Shift-02
dh dksbZ la[;k ges'kk iw.kZr% foHkkftr gksxh % (a) 324781 (b) 329787
(a) 7 only (b) 11 only (c) 326787 (d) 329784
A
(c) 13 only (d) 1001 60. What is the value of x in the number 3426x if
55. The sum of 3-digit numbers abc, cab and bca the number is divisible by 6 but not divisible
is not divisible by: by 5?
3- vad okyh la[;kvksaabc, cab vkSjbca dk ;ksxiQy - la[;k 3426x esax dk eku Kkr dhft,] ;fn la[;k 6 ls
-------- ls foHkkT; ugha gS\ foHkkT; gS ysfdu 5 ls foHkkT; ugha gSA
SSC CGL 24/08/2021 (Shift 01) SSC CHSL, 10/08/2023 (Shift-4)
(a) a + b + c (b) 37 (a) 3 (b) 4
(c) 31 (d) 3 (c) 6 (d) 8
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61. What is the number of possible pairs of 68. Find the number of three digit natural numbers
(P, Q), if the number 357P25Q is divisible by divisible by 8, 12 and 15.
both 3 and 5?
8] 12 vkSj 15 ls foHkkftr gksus okyh rhu vadksa dh
ds laHkkfor ;qXeksa dh la[;k D;k gS] ;fn la[;k izkÑfrd la[;kvksa dh la[;k Kkr djsaA
(P, Q)
357P25Q, 3 vkSj 5 nksuksa ls foHkkT; gS\
(a) 3 (b) 5
(a) 7 (b) 6
(c) 8 (d) 9
(c) 5 (d) None of these
62. How many numbers between 3 and 200 are 69. How many numbers are there between 1 and
divisible by 7 ? 900 which are not divisible by 2, 3
or 5?
3 vkSj 200 ds chp fdruh la[;k 7 ls foHkkT; gSa\
(a) 27 (b) 28
1 ls 900 ds chp ,slh fdruh la[;k,¡ gSa tks 2] 3 vFkok
(c) 29 (d) 36
5 ls foHkkftr ugha gS\
63. How many numbers between 100 and 1000 are (a) 240 (b) 245
divisible by 17?
r
(c) 250 (d) None of these
100 vkSj 1000 ds chp fdruh la[;k,¡ 17 ls foHkkT; gSa\
70. How many numbers are there from 1 to 100
si
SSC CHSL 01/06/2022 (Shift- 1) which are neither divisible by 3 nor by 5?
(a) 51
(c) 52
an by (b) 53
(d) 54
1 ls 100 rd ,slh fdruh la[;k,¡ gSa tks uk rks 3 ls vkSj
uk gh 5 ls foHkkftr gS\
n
64. How many positive two-digit numbers can be (a) 53 (b) 54
divided by 6?
ja (c) 55 (d) None of these
R s
nks vadksa dh fdruh /ukRed la[;kvksa dks 6 ls fOkHkkftr
71. How many numbers are there from 700 to 950
fd;k tk ldrk gS\
a th
which are neither divisible by 3 nor by 7 ?
SSC CHSL 06/06/2022 (Shift 01)
(a) 12 (b) 15 700 ls 950 rd ,slh fdruh la[;k,¡ gSa tks uk rks 3 ls vkSj
(c) 14 (d) 13
uk gh 7 ls foHkkftr gS\
ty a
65. How many numbers between 800 and 2000 are SSC CGL 04/03/ 2020 (Shift- 03)
di M
divisible by 13? (a) 144 (b) 143
800 vkSj 2000 ds chp fdruh la[;k,¡ 13 ls foHkkT; gS\ (c) 146 (d) None of these
CHSL 2019 19/10/2020 (Shift- 01) 72. How many numbers are there from 200 to 800
(a) 90 (b) 92 which are neither divisible by 5 nor by 7 ?
(c) 91 (d) 93
200 ls 800 rd ,slh fdruh la[;k,¡ gSa tks uk rks 5 ls vkSj
66. How many numbers between 1000 and 5000
uk gh 7 ls foHkkftr gS\
are exactly divisible by 225?
SSC CGL 04/03/2020 (Shift- 03)
1000 vkSj 5000 ds chp fdruh la[;k,¡ gS tks 225 ls
iw.kZr% foHkkftr gS\ (a) 407 (b) 410
(a) 16 (b) 18 (c) 413 (d) 411
A
(c) 19 (d) 12 73. How many numbers are there from 500 to 650
67. How many numbers between 300 and 700 are (including both) which are neither divisible by
divisible by 5, 6 and 8? 3 nor by 7?
500 vkSj 700 ds chp fdruh la[;k 5] 6 vkSj 8 ls foHkkT; 500 ls 650 rd (nksuksa dks lfEefyr djrs gq,) ,slh fdruh
gSa\ la[;k,a gSa tks 3 vkSj 7 nkuksa ls foHkkT; ugha gSa\
CPO 2019 25/11/2020 (Shift- 02) SSC CGL 11/04/2022 (Shift- 02)
(a) 20 (b) 2 (a) 21 (b) 121
(c) 5 (d) 3
(c) 87 (d) 99
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74. Let m and n be natural numbers. What is the 75. How many five-digit numbers of the form
minimum value of (m + n) such that 33m + 22n XXYXX is/are divisible by 33?
is divisible by 121? XXYXX ds :i dh ik¡p vadks dh fdruh la[;k,¡ 33 ls
eku yhft, m vkSjn /u iw.kZ la[;k,¡ gSaA
(m + n) U;wure foHkkftr gksrh gSa\
eku D;k gS ftlls fd 33m + 22n, 121 ls HkkT; gks\ [CDS - 2018 (II)]
[CDS - 2023 (I)] (a) 1 (b) 3
(a) 3 (b) 4 (c) 5 (d) Infinite
(c) 5 (d) 10
ANSWER KEY
1.(b) 2.(d) 3.(d) 4.(b) 5.(a) 6.(a) 7.(d) 8.(c) 9.(c) 10.(c)
r
11.(b) 12.(a) 13.(b) 14.(a) 15.(b) 16.(b) 17.(c) 18.(a) 19.(b) 20.(d)
si
21.(c) 22.(d) 23.(c) 24.(a) 25.(a) 26.(a) 27.(b) 28.(a) 29.(c) 30.(c)
31.(d) 32.(a)an by
33.(c) 34.(d) 35.(b) 36.(b) 37.(a) 38.(c) 39.(b) 40.(d)
n
41.(c) 42.(c) 43.(a) 44.(a) 45.(c) 46.(d) 47.(b) 48.(b) 49.(a) 50.(a)
51.(c) 52.(c) ja53.(b) 54.(d) 55.(c) 56.(c) 57.(a) 58.(b) 59.(d) 60.(c)
R s
a th
61.(a) 62.(b) 63.(b) 64.(b) 65.(b) 66.(b) 67.(d) 68.(c) 69.(a) 70.(a)
71.(a) 72.(d) 73.(c) 74.(d) 75.(b)
ty a
di M
A
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Divisibility/foHkkT;rk
( Practice Sheet With Solution)
1. If the 4-digit number x67y is exactly divisible 7. A six-digit number is formed by repeating a
by 9, then the least value of (x + y) is _____. three-digit number for example 347347. The
resultant number will be divisible by?
; fn x67y , d ,slh 4&vadh; la[;k gS tks 9 ls iw.kZr%
rhu vadksa dh la[;k dks nksgjkdj ,d Ng vadksa dh la[;k
foHkkT; gS] (xrks+y) d k U;wure eku
_________gSA
cukbZ tkrh gSA tSls 347347 ifj.kkeh la[;k fdlds }kjk
(a) 9 (b) 0
foHkkT; gksxh\
(c) 5 (d) 3
(a) Only 7 (b) Only 11
2. Find the greatest value of (x + y) such that a
(c) Only 13 (d) 1001
7-digit number 33x920y is divisible by 15.
r
8. What is the smallest number that should be
(x + y) dk lcls cM+k eku Kkr dhft, rkfd 7 vadksa dh
added to 4567 so that the sum is divisible
si
la[;k 33x920y 15 ls foHkkT; gksA by 7?
(a) 7 (b) 11
an by 4567 esa og NksVh ls NksVh dkSu&lh la[;k tksM+h
(c) 12 (d) 13
fd ;ksxiQy 7 ls foHkkT; gks\
n
3. If N5921 is divisible by 11, find the value of
smallest natural number N. (a) 7 (b) 5
(c) 6 (d) 4
;fn N5921, 11 ls foHkkT; gS] rks lcls NksVh izkd`frd
ja
R s
9. What is the smallest perfect square which is
la[;k N dk eku Kkr djsaA
completely divisible by each of 16, 18 and 36?
a th
(a) 9 (b) 7
og lcls NksVk iw.kZ oxZ dkSu lk gS] tks 16] 18 vkS
(c) 6 (d) 8
4. What is the product of the largest and the
36 esa ls izR;sd ls iw.kZr% foHkkT; gS\
smallest possible values of m for which a (a) 144 (b) 81
ty a
number 5m83m4m1 is divisible by 9? (c) 196 (d) 169
m ds lcls cM+s vkSj lcls NksVs laHkkfor ekuksa dk xq.kuiQy D;k should be subtracted from 246837 to
10. What
di M
gS ftlds fy, ,d la[;k 5m83m4m1, 9 ls foHkkT; gS\ make it divisible by 13?
(a) 16 (b) 40 246837 esa ls fdruk ?kVk;k tk, fd og 13 ls foHkkT;
(c) 80 (d) 10 gks tk,xk\
5. A number 4 16 + 1 is divisible by x. Which (a) 4 (b) 5
among the following is also divisible by x? (c) 3 (d) 6
, d la[;k 416 + 1, x ls foHkkT; gSA fuEufyf•r esa11.
ls In a 7-digit number 89476*2, what is the
dkSux l s Hkh foHkkT; gS\ smallest possible value of * such that the
(a) 496 + 1 (b) 432 + 1 number is divisible by 8?
8
(c) 4 + 1 (d) 448 + 1 , d 7& vadh; la[;k 89476*2 esa* d k U;wure laHko
6. If the six-digit number 479xyz is exactly eku D;k gS ftlls la[;k 8 ls foHkkT; gks\
A
divisible by 7, 11 and 13, then {(y + z) ÷ x} is
(a) 2 (b) 1
equal to:
(c) 4 (d) 3
N vadksa okyh la[;k
479xyz ;fn 7] 11 vkSj 13 ls
12. If the four-digit number 463y is divisible by
iw.kZr% foHkkT;{(y
gS+rks
z) ÷ x} dk eku cjkcj gSA 7, then what is the value of y?
7 ;fn pkj vadksa dh la[;k463y, 7 ls foHkkT; gS] yrks
(a) (b) 4
13 dk eku D;k gksxk\
13 11 (a) 4 (b) 6
(c) (d)
7 9 (c) 3 (d) 5
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13. A 6-digit number has digits as consecutive 20. Which of the following pairs of non-zero values
natural numbers. The number is always of p and q make 6-digit number 674pq0
divisible by____. divisible by both 3 and 11?
,d 6 vadh; la[;k esa vad Øekxr izkÑr la[;kvksa ds :i p vkSjq ds 'kwU;srj ekuksa dk fuEufyf[kr esa ls dkSu&
esa gSA ;g la[;k lnSo
___ ls foHkkT; gksxhA ;qXe 6 vadksa dh la[;k
674pq0 dks 3 vkSj 11 nksuksa
(a) 4 (b) 5 ls foHkkT; cukrk gS\
(c) 2 (d) 3 (a) p = 2 and q = 2 (b) p = 5 and q = 4
14. Which number among 11368, 11638, 11863 (c) p = 4 and q = 2 (d) p = 5 and q = 2
and 12638 is divisible by 11? 21. If the 9–digit number 83p93678Q is divisible
11368, 11638, 11863 vkSj 12638 esa ls dkSu&lh by 72, then what is the value of
la[;k 11 ls foHkkT; gSA P 2 + Q 2 + 12 ?
(a) 11368 (b) 12638
;fn 9 vadksa dh la[;k83p93678Q, 72 ls foHkkT;
(c) 11638 (d) 11863
15. What is the value of x so that the seven-digit gS] rks P 2 + Q 2 + 12 dk eku D;k gS\
r
number 8439x53 is divisible by 99? (a) 6 (b) 7
dk eku D;k gksxk fd lkr vadksa dh la[;k
si
x 8439x53, (c) 8 (d) 9
99 ls foHkkT; gS\ 22. Find the smallest number that can be
(a) 9
an by (b) 4
subtracted from 148109326 so that it becomes
divisible by 8.
n
(c) 3 (d) 6
og NksVh ls NksVh la[;k Kkr dhft, ftls 148109326 esa ls
16. If the 9-digit number 97x4562y8 is divisible
?kVkus ij izkIr la[;k 8 ls foHkkT; gksxhA
ja
by 88, what is the value of (x² + y²) for the
R s
smallest value of y, given that x and y are (a) 4 (b) 8
natural numbers? (c) 6 (d) 10
a th
;fn 9&vadksa dh la[;k97x4562y8] 88 ls foHkkT; gS]23. A 11-digit number 7823326867X is divisible
by 18. What is the value of X?
rksy ds lcls NksVs eku ds fy,(x² + y²) dk eku D;k
,d 11&vadh; la[;k 7823326867X, 18 ls foHkkT;
gS] fn;k x;k gS fdx vkSjy çkÑr la[;k,¡ gSa\
gSAX dk eku D;k gS\
ty a
(a) 64 (b) 68 (a) 6 (b) 4
di M
(c) 76 (d) 80 (c) 8 (d) 2
17. 8A5146B is divisible by 88, then what is the 24. Which of the following numbers is divisible by
value of A × B? 44?
8A5146B, 88 ls foHkkT; gS] A
rks× B dk eku D;k gS\ fuEufyf[kr esa ls dkSu&lh la[;k 44 ls foHkkT; gS\
(a) 4 (b) 16 (a) 32802 (b) 54736
(c) 8 (d) 12 (c) 93472 (d) 27048
25. Find the largest number which exactly divides
18. What will be the least number which when every number of the form (n³ – n) (n – 2) Where
doubled will be exactly divisible by 15, 18, n is a natural number greater than 2
25 and 32?
og vf/dre la[;k Kkr djsa tks (n³ – 2) (n – 2) fd
og NksVh ls NksVh la[;k D;k gksxh ftls nksxquk djus rjg
ij dh izR;sd la[;k dks foHkkftr djsa tgka
n ,d izkÑr
A
og 15] 18] 25 vkSj 32 ls iw.kZr% foHkkT; gks tk,xh\ la[;k gS] tks 2 ls vf/d gSA
(a) 3600 (b) 7200 (a) 6 (b) 12
(c) 6400 (d) 3200 (c) 24 (d) 48
19. Find the largest number of 3 digits divisible 26. If 111...1 (n digit) is divisible by 9, then find
the minimum value of n.
by 4 and 7.
;fn 111------1 (n vad) 9 ls foHkkT; gS] rks
n dk
4 vkSj 7 ls foHkkT;vadksa
3 okyh lcls cM+h la[;k Kkr
djsaA U;wure eku Kkr dhft,A
(a) 960 (b) 980 (a) 18 (b) 9
(c) 990 (d) 970 (c) 12 (d) 3
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27. If the 5-digit number 750PQ is divisible by 3, 35. If the 5-digit number 505xy is divisible by 11
7 and 11, then what is the value of P + 2Q? and 21, then what is the value of (7x – 5y)?
;fn 750PQ ,d ,slh 5&vadh; la[;k gS] tks 3] 7 ;fn 5 vadksa dh la[;k505xy, 11 vkSj 21 ls foHkkT;
vkSj 11 ls foHkkT; gS]Prks
+ 2Q dk eku Kkr dhft,A gS] rks
(7x – 5y) dk eku D;k gS\
(a) 17 (b) 15 (a) 16 (b) 30
(c) 18 (d) 16 (c) 31 (d) 11
36. If the 5-digit number 593ab is divisible by 3, 7
28. Which of the following numbers are divisible
and 11, then what is the value of (a2 – b2 + ab)?
by 2, 3 and 5?
;fn 5 vad okyh la[;k 593ab 3, 7 vkSj 11 ls foHkkT;
fuEufyf[kr esa ls dkSu&lh la[;k 2] 3 vkSj 5 ls foHkkT; gS\
gS] rks
(a2 – b2 + ab) dk eku Kkr djsaA
(a) 5467760 (b) 1345678
(a) 35 (b) 31
(c) 2345760 (d) 2456732
(c) 25 (d) 29
29. The nearest number which is greater to 87501,
37. A 9-digit number 846523X7Y is divisible by 9,
and is completely divisible by 765 is :
and Y – X = 6. Find the value of 2X + 4Y .
og fudVre la[;k Kkr dhft, tks 87501 ls cM+h gS
846523X7Y ,d ,slh 9 vadh; la[;k gS] tks9 ls foHkkT;
r
vkSj 765 ls iw.kZr% foHkkT; gSA
gS vkSj
Y – X = 6 gSA rc
] 2X + 4Y dk eku Kkr dhft,A
si
(a) 88975 (b) 87975
(c) 87966 (d) 87775 (a) 4 (b) 2
30. an by
The largest 5-digit number that is exactly (c) 6 (d) 8
divisible by 88 is: Level-02
n
5 vadksa dh lcls cM+h la[;k
______ gS] tks 88 ls 38. How many three digit numbers are divisible by
5 or 9?
iw.kZr% foHkkT; gSA
ja rhu vadksa dh fdruh la[;k,¡ 5 ;k 9 ls foHkkT; gSa\
R s
(a) 99968 (b) 99689
(a) 260 (b) 280
(c) 68999 (d) 66698
a th
(c) 200 (d) 180
31. The least number that should be added to
39. The number A39K2 is completely divisible by
35460 so that the sum is exactly divisible by
both 8 and 11. Here both A and K are single
3, 4, 5 and 7 is:
digit natural numbers.
ty a
35460 esa dkSu&lh lcls NksVh la[;k tksM+h tkuh pkfg,Which of the following is a possible value of
rkfd ;ksxiQy 3] 4] 5 vkSj 7 ls iw.kZr% foHkkT; gks\ A + K ?
di M
(a) 84 (b) 420 la[;k A39K2, 8 vkSj 11 nksuksa ls iwjh rjg ls foHkkT; gS
(c) 240 (d) 180 ;gk¡A vkSjK nksuksa ,d vad okyh çkÑfrd la[;k,¡ gSaA
32. A six-digit number is divisible by 33. If 54 is fuEufyf•r esa lsA + K dk dkSu lk laHkkfor eku? gS
added to the number, then the new number (a) 8 (b) 10
formed will also be divisible by: (c) 12 (d) 14
Ng vadksa dh ,d la[;k 33 ls foHkkT; gSA ;fn la[;k esa 54 tksM+
40. How many number are there from 700 to 950
fn;k tk,] rks fufeZr ubZ la[;k Hkh &&&&&& ls foHkkT; gksxhA (including both) which are neither divisible by
(a) 3 (b) 2 3 nor by 7?
(c) 5 (d) 7 700 ls 950 (nksuksa dks feykdj) ,slh fdruh la[;k,¡ gS]
33. Which of the following numbers is divisible by 30? tks u rks 3 ls u gh 7 ls foHkkftr gksrh gS\
fuEu esa ls dkSu lh la[;k 30 ls foHkkT; gS\ (a) 107 (b) 141
A
(c) 144 (d) 145
(a) 34560 (b) 23450
41. When 335 is added to 5A7 the result is 8B2.
(c) 12340 (d) 45670 8B2 is divisible by 3 what is the largest
34. If 11-digit number 88p554085k6, k p, is divisible possible value of A.
by 72, then what is the value of (3k + 2p)? tc 335 dks 5A7 eas tksM+k tkrk gS] rks ifj.kke
8B2
;fn 11 vadksa dh la[;k88p554085k6] tgk¡ k p izkIr gksrk 8B2,
gS 3 ls foHkkT; gS] rks
A dk egÙke
gS] 72 ls foHkkT; gS] 3k
rks+( 2p) dk eku D;k gksxk\ laHkor eku D;k gSA
(a) 13 (b) 12 (a) 8 (b) 5
(c) 23 (d) 7 (c) 1 (d) 4
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42. If the 8-digit number 123456xy is divisible 50. How many three-digit numbers are divisible by 5?
by 8, then the total possible pairs of (x,y) are: rhu vadksa dh fdruh la[;k,¡ 5 ls foHkkT; gSa\
;fn 8 vadksa dh la[;k123456xy, 8 ls foHkkT; gS] (a) 150 (b) 160
rks(x, y) ds dqy fdrus laHkkfor ;qXe gksaxs\ (c) 170 (d) 180
(a) 8 (b) 13 Level-03
(c) 10 (d) 11 51. What is the number of even factors of 36000
43. Find the greatest 5-digit number which is which are divisible by 9 but not by 36?
divisible by 11, 33, 99 and 121. 36000 ds le xq.ku•aMksa dh la[;k D;k gS tks 9 ls foHkkT;
5 vadksa dh lcls cM+h la[;k Kkr dhft, tks 11] 33] gS ysfdu 36 ls ugha\
99 vkSj 121 ls foHkkT; gksA (a) 20 (b) 4
(a) 90099 (b) 99990 (c) 10 (d) 12
(c) 99099 (d) 90909 52. How many 3-digit positive integers, with digits
44. 350 + 926 + 2718 + 928 + 929 is divisible by which a, b and c exist such that a < b and c < b?
of the following integers? Assume that a is in hundred’s place, b is in
fuEufyf[kr esa ls fdl ten’s place, c is in unit’s place and a is a non-
r
350 + 926 + 2718 + 928 + 929
zero digit.
iw.kk±d ls foHkkT; gS\
si
(a) 11 (b) 5 vad a, b vkSj c ds lkFk fdrus 3&vadh; ldkjkRed
(c) 7 (d) 2 iw.kkZad ekStwn gSaa tSls
< b vkSj
fd c < b eku ysa fda
45.
an by
Numbers divisible by 9 between 43 and 481 are:
43 vkSj 481 ds chp esa fdruh la[;k,¡ 9 }kjk foHkkT; gSa\
lSdM+k ds LFkku ijbgS]
ds LFkku ij gS vkSj
ngkbZ ds LFkku ij cgS]
'kwU;srj vad gSA
bdkbZ
n
a
(a) 51 (b) 49 (a) 450 (b) 240
(c) 48 ja (d) 50 (c) 364 (d) 648
R s
46. How many numbers from 1 to 430 are divisible 53. If a number K = 42 × 25 × 54 × 135 is divisible
by 7 and 11 both? by 3a, then find the maximum value of a.
a th
1 ls 430 rd fdruh la[;k,¡ 7 vkSj 11 nksuksa ls foHkkT; gSa\ ;fn ,d la[;k K = 42 × 25 × 54 × 135, 3a }kjk
(a) 5 (b) 11 foHkkT; gS] arks
dk vf/dre eku Kkr dhft,A
(c) 9 (d) 7 (a) 6 (b) 7
ty a
47. 9435 is added to 7593, then 2607 is
(c) 4 (d) 5
subtracted from the sum. The result is
di M
54. If 7-digit number 678p37q is divisible by 75
divisible by:
and p is not a composite, number then the
9435 dks 7593 esa tksM+k tkrk gS] vkSj fiQj 2607 dks buds values of p and q are:
;ksxiQy ls ?kVk;k tkrk gSA ifj.kke fdlls foHkkT; gksxk\ ;fn 7 vadksa dh la[;k678p37q, 75 ls foHkkT; gS
(a) 4 (b) 10
vkSjp ,d HkkT; la[;k ugha gS] p rksvkSjq ds eku
(c) 3 (d) 5
Kkr dhft,A
48. If the seven-digit number 52A6B7C is divisible
(a) p = 5, q = 5 (b) p = 3, q = 0
by 33, and A, B, C are primes, then the
(c) p = 3, q = 5 (d) p = 2, q = 5
maximum value of 2A+3B+C is:
55. A four-digit pin, say abcd, of a lock has different
;fn lkr vadksa dh la[;k 52A6B7C, 33 ls foHkkT; non-zero digits. The digits satisfy b = 2a, c
gS] vkSjA, B, C vHkkT; gSa] 2A rks + 3B + C dk = 2b, d = 2c. The pin is divisible by.
vfèkdre eku gS% ,d ykWd ds pkj vadksa ds fiu dks] abcd dgrs gSa]
A
(a) 32 (b) 23 esa vyx&vyx xSj&'kwU; vad gksrs gSaA vadc
b = 2a,
(c) 27 (d) 34 = 2b, d = 2c dks larq"V djrs gSaA fiulsfdl
foHkkT; gS\
49. The largest three-digit number that gives the (a) 2, 3, 5 (b) 2, 3, 7
same remainder 2 when divided by 3, 5 and 9 (c) 2, 3, 13 (d) 2, 3, 11
is ________. 56. Which of the following numbers is divisible by
rhu vadksa dh lcls cM+h la[;k tks 3] 5 vkSj 9 ls foHkkftr 7, 11 and 13?
gksus ij leku 'ks"kiQy 2 nsrh gS] ----------------- gSA fuEufyf•r esa ls dkSu lh la[;k 7] 11 vkSj 13 ls foHkkT; gS\
(a) 999 (b) 984 (a) 1002001 (b) 1003001
(c) 998 (d) 992 (c) 1005001 (d) 1004001
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57. If x and y are two digits of the number 115xy 59. A four-digits number abba is divisible by 4 and
such that the number is divisible by 90, then a < b. How many such numbers are there ?
find the value of x + y. abba ,d ,slh pkj vadh; la[;k gS] tks 4 ls foHkkT; gS
;fn x vkSjy la[;k 115xy ds nks vad gSa tSls fd la[;k vkSja < b gSA ,slh fdruh la[;k,¡ gS\
90 ls foHkkT; gS] xrks
+ y dk eku Kkr djsaA (a) 10 (b) 8
(a) 3 (b) 2 (c) 12 (d) 6
(c) 6 (d) 5 60. How many five-digit numbers of the form
xxyxx is/are divisible by 33?
58. 202020 + 162020 – 32020 – 1 is divisible by:
xxyxx ds :i dh ik¡p vadksa dh fdruh la[;k,¡ 33 ls
202020 + 162020 – 32020 – 1 fdlls foHkkT; gS\ foHkkT; gS@gSa\
(a) 317 (b) 91 (a) 1 (b) 3
(c) 253 (d) 323 (c) 5 (d) Infinite
ANSWER KEY
r
1.(c) 2.(d) 3.(d) 4.(a) 5.(d) 6.(b) 7.(d) 8.(d) 9.(a) 10.(d)
si
11.(d) 12.(a) 13.(d) 14.(c) 15.(b) 16.(d) 17.(d) 18.(a) 19.(b) 20.(d)
21.(c)
an by
22.(c) 23.(d) 24.(b) 25.(c) 26.(b) 27.(a) 28.(c) 29.(b) 30.(a)
n
31.(c) 32.(a) 33.(a) 34.(a) 35.(d) 36.(d) 37.(c) 38.(a) 39.(b) 40.(c)
ja
R s
41.(b) 42.(b) 43.(c) 44.(a) 45.(b) 46.(a) 47.(c) 48.(b) 49.(d) 50.(d)
a th
51.(b) 52.(b) 53.(b) 54.(c) 55.(c) 56.(a) 57.(b) 58.(d) 59.(b) 60.(b)
ty a
di M
A
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SOLUTIONS
1. (c) 7. (d)
Given, x67y is exactly divisible by 9. A six digit no. formed by repeating a three
digit no. as 347347
x 67y the no. is divisible by 3 × 11 × 7 = 1001
=0
9 8. (d)
Least value of (x + y) = 5 When, we divided 4567 by 7
2. (d) we get the remainder 3
Given that, 33x920y is divisible by 15 So, the number which is divisible by 7 after
it is also divisible by 3, 5 4567 = (4567 – 3 + 7) = 4571
Required number = (4571 – 4567) = 4
33x 920 y 9. (a)
=0
3 Among the options (a) the smallest perfect
square which is divisible by 16, 18, 36 is 144.
r
for largest value of (x + y)
10. (d)
x = 8, y = 5
si
(x + y) = 13 246837
= 6 (Remainder)
3. (d) an by
Divisibility rule of 11
13
To make 246837 divisible by 13 we should
n
Odd place even place = 0 or multiple of 11 subtract 6 from this no.
ATQ, 11. (d)
ja Given 89476*2 is divisible by 8
R s
For smallest possible value of *
(N + 9 + 1) (5 + 2) = 11
a th
take N = 8 6*2
8
=3
11 = 11
LHS = RHS 12. (a)
ty a
So, option (d) is correct 463y is divisible by 7
4. (a) y=4
di M
13. (d)
Given, 5m83m4m1 is divisible by 9.
Any 6-digit number formed by a consecutive
5 m 8 3 m 4 m 1 natural number is always divisible by 3 be-
So, =0 cause the sum of digits of any such number
9
is divisible by 3.
Largest value of m = 8 1+ 2 + 3 + 4 + 5 + 6
smallest possible value of m = 2 Let 123456 =
3
So product = 8 × 2 = 16 14. (c)
5. (d) 11638 1+ 6 + 8 – 1+ 3
(c) =
Given 416 + 1 is divisible by x. 11 11
Here (416)³ + 1 is the multiple of 416 + 1 15 – 4 11
A
= =
So (416)³ + 1 is also divisible by x 11 11
448 + 1 15. (b)
6. (b) Given, 8439x53 is divisible by 99
Given, 479xyz is exactly divisible by 7, 11, 13 it is also divisible by 9, 11
it is also divisible by 1001 (7, 11, 13) 8 4 3 9 x 53
So, 0
So, x = 4, y = 7, z = 9 9
{(y + z) ÷ x} and, (8+3+x+3) – (4+9+5) = 0, multiple of 11
(16 ÷ 4) = 4 x = 4 satisfies for both
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16. (d) 22. (c)
The no. 97x4562y8 is divisible by 88 For divisible by 8 we consider only last three
it is also divisible by 8, 11 digits of the no 148109326
2y8 326
for minimum value of y, =0, y = 4
8 8
for value of x, (9 + x + 5 + 2 + 8) – (7 + 4 + 6 + 4) Remainder = 6
= 0, multiple of 11 So, 6 should be subtracted from this no.
x = 8, so (x² + y²) = (64 + 16) = 80 23. (d)
17. (d) Divisibility of 18 = divisibility of 2 and 9
the number 8A5146B is divisible by 88, 7823326867X/18
It is also divisible by 8, 11 From divisibility of 2, X should be even number.
7+8+2+3+3+2+6+8+6+7+X
464
For value of B =0 X=2
8
24. (b)
B = 4 54736 is divisible by 44 because it is also
r
for value of A (8 + 5 + 4 + 4) – (A + 1 + 6) divisible by 4 and 11.
si
= 0, multiple of 11 25. (c)
A = 3 Given form = (n³ – n) (n – 2)
18. (a)
an by
So, A × B = 3 × 4 = 12 Let n = 4
So, (4³ – 4) (4 – 2) = 60 × 2 = 120
n
L.C.M of (15, 18, 25 and 32) = 7200 the largest number which exactly divides this
from is 24.
=
7200
= 3600 ja
R s
2 26. (b)
Let minimum value of n = 999999999
a th
19. (b)
ATQ,
Largest no. of 3 digit = 999
This number also divisible by 9
on dividing 999 by 28
27. (a)
remainder = 19
ty a
Given, the no. 750PQ is divisible by 3, 7, 11
So the no. which is divisible by 4 and 7
it is also divisible by 231 (3 × 7 × 11)
= (999 – 19) = 980
di M
20. (d) Let us consider largest no. 75099
By option (d) So, on dividing 75099 by 231
Let, P = 5, q = 2 Remainder = 24
Number ® 674520 So, required no. = 75099 – 24 = 75075
For divisibility rule of 3, sum of number should P = 7, Q = 5
be divisible by 3 P + 2Q = (7 + 10) = 17
(6 + 7 + 4 + 5 + 2 + 0) = 24, it is divisible by 28. (c)
3 for divisiblity rule of 11 Divisibility of 2 = last digit of no. divisible by 2.
(6 + 4 + 2) – (7 + 5 + 0) = 11 or 0 Divisibility of 3 = sum of digits is divisible by 3.
It is divisible by 11 Divisibility of 5 = last digit is 0 or 5
both conditions are satisfied
A
Among the given options only 2345760 follows
21. (c) these conditions.
Given no. 83p93678Q is divisible by 72 So, the required no. is 2345760
It is also divisible by 9, 8 29. (b)
for value of Q = last 3 digit divisible by 8 We have to find the no. which is divisible by 765
So, Q = 4
765 = 3 × 3 × 5 × 17
for value of P = 83p936784
So, among the options the no. which is
P=6
divisible by 9, 5, 17 is also divisible by 765.
P² Q² 12 = (6)² (4)² (12) = 64 = 8 Only 87975 follows all conditions.
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30. (a) 36. (d)
Largest 5-digit no. = 99999 Given that 593ab is divisible by 3, 7, 11
divide 99999 by 88 the no. is also divisible by 231
remainder = 31 Let us consider the no. 59300
So, required no. = 99999-31 = 99968 So no. dividing 59300 by 231
31. (c) Remainder = 164
L.C.M of (3, 4, 5, 7) = 420 So required no. (59300 – 164 + 231) = 59367
When we divided 35460 by 420 we get the so, a = 6, b = 7
remainder 180 (a2 – b2 + ab) = (62 – 72 + 42) = 29
So required no. is (420 – 180) = 240 37. (c)
32. (a) Given no. 846523X7Y is divisible by 9.
Let, A be six digit no that is divisible by 33. x + y = 10
A.T.Q y–x=6
A y=8
+ 54
r
33 x=2
si
= 33 = 11 × 3 2x + 4y
54 is also divisible by 3.
an by 36 = 6
So the no obtained on adding 54 is also divisible
38. (a)
by 3.
n
Three digit number divisible by 5 100, 105
33. (a)
...... 995
Divisibility rule of 3 = Sum of the digits must
ja 995 – 100
R s
be divisible by 3
N= +1
Divisibility rule of 10 = Last two digits must 5
a th
be divisible by 10
895
Option1:- = 1
5
3 + 4 + 5 + 6 + 0 = 18 which is divisible by 3
= 179 + 1 = 180
ty a
34. (a)
divisible by 9 108, 117 ......... 999
Given no. 88p554085k6 is divisible by 72
di M
It is also divisible by 8, 9 999 – 108
= +1
For value of k = last 3 digit divisible by 8 9
So, k = 3
891
Value of p = 88P55408536 = 1
9
p=2
= 99 + 1 = 100
Now,
divisible by 45 135, 180 ......... 990
(3k + 2p) = (3 × 3 + 2 × 2) = 13
35. (d) 990 – 135
N= 1 = 20
LCM (11, 21) = 231 45
By taking the largest 5-digit number 50599 and No. 180 + 100 – 20
A
divide it by 231 = 280 – 20 = 260
50599 = 231 × 219 + 10 39. (b)
50599 = 50589 + 10 A39K2 is divisible by both 8 and 11.
50599 – 10 = 50589 (Completely divisible by 9K2
231) =0
8
50589 = 505xy(where x = 8, y = 9) K=9
Now the value of (7x – 5y) (A + 9 + 2) – (3 + K) = 0, multiple of 11
(7x – 5y) = 7 × 8 – 5 × 9 A=1
(56 – 45) = 11 So, A + K = (9 + 1) = 10
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40. (c) 46. (a)
Total numbers between 700 and 950 No. are
= 950 – 700 + 1 = 251 77, 154, .........., 385
251 We know that
Numbers divisible by 3 = = 83 (Quotient)
3 Last term – first term
n= +1
251 Difference
Numbers divisible by 7 = = 35 (Quotient)
7 385 – 77
n= +1 = 5
251 77
Numbers divisible by 21 = = 11 (Quotient) 47. (c)
21
ATQ,
Number which are not divisible by 3 or 7
= 251 – (83 + 35 – 11) = 144 (9435 + 7593 – 2607) = 14421
41. (b) Given no. is multiple of 3
ATQ, 335 + 5A7 = 8B2 So no. is divisible by 3.
48. (b)
882
r
since, =0 Given, 52A6B7C is divisible by 33
3 it is also divisible by 11 and 3.
si
So, largest possible value of A = 5 Maximum 2A + 3B + C
42. (b)
an by
Given, 123456xy is divisible by 8,
5 + A + B + C = 15
A = 3, B = 5, C = 2
n
6 xy 2×3+3×5+2
then 8 =0
(6 + 15 + 2) = 23
So, ja 49. (d)
R s
largest 3 digit no. = 999
x 0 0 1 2 3 4 4 5 6 7 8 8 9
L.C.M of 3, 5 and 9 = 45
a th
y 0 8 6 4 2 8 0 6 4 2 0 8 6
on dividing 999 by 45
total pairs (x, y) = 13 remainder = 9
43. (c) So, required no. = (999 – 9 + 2) = 992
ty a
Greatest 5-digit no. = 99999 50. (d)
LCM of 11, 33, 99 and 121 = 11 × 9 × 11 Numbers are,
di M
= 1089 100, 105, 110 .......... 995
Divide 99999 by 1089 We know that
Remainder = 900
So required no. = (99999 – 900) = 99099 Last term – first term
n= +1
44. (a) Difference
Given 350 + 926 + 2718 + 928 + 929 (995 – 100)
350 + 352 + 354 + 356 + 358 n= +1
5
350 (1 + 9 + 81 + 729 + 6561)
350 (7381) 895
n= +1
So this no. is divisible by 11. 5
45. (b) n = 180
A
No. are 51. (b)
45, 54, 63, .........., 477 36000 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5
We know that since we are talking about even factors,
Last term – first term there must be at least one 2 in the required
n= +1 factors:
Difference
since the number is divisible by 9, we have
477 – 45 both the three.
n= +1 = 49
9 So, we have 1 way of choosing 2, 1 way of
So numbers between 43 and 481 which are choosing 3, 4 ways of choosing 5
divisible by 9 are 49. required factors = 1 × 1 × 4 = 4
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52. (b) 56. (a)
We note that b is largest among these. LCM of 7, 11, 13 = 1001
So, for b = 1 no numbers possible because When 3 digits of a number are repeated twice
then that number will be divisible by 7, 11
a0
and 13.
for b = 2, a = 1, c = 0, 1 total possible no. 1002001 is divisible by 7, 11 and 13
=1×2=2
57. (b)
for b = 3, a = 1, 2, c = 0, 1, 2 total possible Given no 115xy is divisible by 90
no. = 2 × 3 = 6
It is also divisible by 10, 9
for b = 4, a = 1, 2, 3, c = 0, 1, 2, 3 total possible
For value of y = 0
no. = 3 × 4 = 12
(1 + 1 + 5 + x + y) must be divisible by 9
in same way for b = 9, total possible no.
= 8 × 9 = 72 (7 + x) is divisible by 9
The smallest value of y by which (7 + x) will
So, total no. = 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +
be divisible by 9 is x = 2
5×6+6×7+7×8+8×9
r
x+y=2+0=2
= 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 = 240
si
Hance, the value of x + y is 2
53. (b) 58. (d)
an by
Given, K = 42 × 25 × 54 × 135 is divisible 3a.
42 × 25 × 54 × 135
Ist Case
(162020 – 12020) + (202020 – 32020)
n
= 3 × 14 × 25 × 3³ × 2 × 3³ × 5 an – bn n is even [(a – b) (a + b)]
37 × 14 × 25 × 5 = 17 common factor
ja
R s
So maximum value of a = 7. IInd Case
54. (c) (162020 – 32020) + (202020 – 12020)
a th
Given, 678p37q is divisible by 75 an – bn n is even [(a – b) (a + b)]
it is also divisible by 5, 3 = 19 common factor
So q = 0, 5 Series should be divisible by = 19 × 17 = 323
ty a
In first case, put q = 0, 59. (b)
di M
A B
678p370
So, 2 (3, 9) = 32,52, 72, 92 = 4
3
4 (5, 9) = 64,84 = 2
p=2 6 (7, 9) = 76,96 = 2
In second case, put q = 5, 8 (9) = 0
678p375 Total possible = (4 + 2 + 2) = 8
so, 60. (b)
3
Any number that is divisible by 33 must also be
p = 0, 3, 6, 9 divisible by 3 and 11. A number which is divisible
by 3 has the sum of its digits divisible by 3.
So, p = 3, q = 5
x + x + y + x + x = (4x + y) must be divisible by 3
55. (c)
A
And, for divisibility 11, (x + y + x) – (x + x) = 0
pin = abcd y=0
b = 2a, c = 2b, d = 2c Now, 4x + y = 4x = 0 = 4x will be divisible by 3.
let a = 1 Only when
so, pin = 1248 x = 3, 6 or 9
So the pin is divisible by 2, 3, 13 So the three numbers are 33033, 66066, 99099.
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REMAINDER ('ks"kiQy
)
(CLASSROOM SHEET)
1. On dividing a number by 38, the quotient is 24 6. If 7 divides the integer n, then the remainder is 2.
and the remainder is 13, then the number is : What will be the remainder if 9n is divided by 7?
fdlh la[;k dks 38 ls Hkkx nsus ij HkkxiQy 24 rFkk ;fn 7 iw.kkZad
n dks foHkkftr djrk gS] rks 'ks"kiQy 2 gksrk gS
'ks"kiQy 13 vkrk gSA og la[;k dkSu&lh gS\ ;fn 9n dks 7 ls foHkkftr fd;k tkrk gS] rks 'ks"kiQy D;k gksxk\
SSC CPO 16/03/2019 (Shift- 01) SSC CHSL 09/08/2023 Shift-01
(a) 925 (b) 975 (a) 3 (b) 5
(c) 904 (d) 956 (c) 1 (d) 4
2. On dividing 8675123 by a certain number, the 7. A number 'n' when divided by 6 leaves
r
quotient is 33611 and the remainder is 3485. remainder 2. What will be the remainder when
si
The divisor is (n2 + n + 2) is divided by 6?
8675123 dks ,d fuf'r la[;k ls Hkkx nsus ij HkkxiQy ,d la[;k 'n' dks 6 ls foHkkftr djus ij 2 'ks"kiQy
an by
33611 vkSj 'ks"kiQy 3485 vkrk gSA Hkktd Kkr dhft,A cprk gSA(n2 + n + 2) dks 6 ls foHkkftr djus ij
'ks"kiQy D;k izkIr gksxk\
n
SSC CHSL 25/05/2022 (Shift- 02)
(a) 538 (b) 258 SSC CGL 18/04/2022 (Shift- 03)
(c) 248 ja (d) 356 (a) 6 (b) 2
R s
3. The divisor is 10 times the quotient and 5 times (c) 4 (d) 0
a th
the remainder in a division sum. What is the 8. If a positive integer n is divided by 7, the
dividend if the remainder is 46? remainder is 2. Which of the numbers in the
fdlh Hkkx ds ;ksx esa Hkktd HkkxiQy dk 10 xquk vkSj 'ks"kiQy
options yields a remainder of 0 when it is
dk 5 xquk gksrk gSA ;fn 'ks"kiQy 46
'ks"kiQy
gS rks
D;k gS\ divided by 7?
ty a
SSC CHSL 03/08/2023 (Shift-02) ;fn ,d /ukRed iw.kk±d n dks 7 ls foHkkftr fd;k
tkrk gS] rks 'ks"k 2 gSA fodYiksa esa ls dkSu&lh la[;k
di M
(a) 5972 (b) 4286
(c) 4874 (d) 5336 foHkkftr gksus ij 'ks"k 0 nsrk gS\
4. A number when divided by 44, gives 432 as CHSL 2019 18/03/2020 (Shift- 02)
quotient and 10 as remainder. What will be (a) n + 3 (b) n + 1
the remainder when dividing the same (c) n + 2 (d) n + 5
number by 31?
9. For any integeral value of n, 32n + 9n + 5, when
fdlh la[;k dks 44 ls Hkkx nsus ij HkkxiQy 432 vkSj divided by 3 will leave the remainder :
'ks"k 10 izkIr gksrk gSA mlh la[;k dks 31 ls foHkkftrn ds fdlh iw.kk±d eku ds fy,32n + 9n + 5 dks tc
djus ij 'ks"kiQy D;k izkIr gksxk\ 3 ls foHkkftr fd;k tkrk gS rks izkIr 'ks"kiQy gksxk %
(a) 3 (b) 4
(a) 1 (b) 2
A
(c) 5 (d) 6
(c) 0 (d) 5
5. When a number is divided by 7, the remainder
10. Given n is an integer, what is the remainder
is 1. What will be the remainder when the cube
of this number is divided by 7? when (6n + 3)2 is divided by 9?
fdlh la[;k dks 7 ls Hkkx nsus ij 'ks"kiQy 1 cprk gSA bl fn;k x;k n ,d iw.kk±d gSA tc
(6n + 3)2 dks9 ls Hkkx
la[;k ds ?ku dks 7 ls Hkkx nsus ij] 'ks"kiQy D;k gksxk\ fn;k tkrk gS] rc 'ks"kiQy D;k vk,xk\
SSC CHSL 04/08/2023 Shift-03 SSC CGL 08/07/2019 (Shift- 02)
(a) 2 (b) 4 (a) 3 (b) 2
(c) 1 (d) 3 (c) 1 (d) 0
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11. When positive number x, y and z are divided (a) 9 (b) 0
by 31, the reminders are 17, 24 and 27
respectively. When (4x – 2y + 3z) is divided (c) 3 (d) 8
by 31, the remainder will be : 17. When a positive integer is divided by d, the
tc /ukRed la[;k x, y vkSjz dks 31 ls Hkkx fn;k remainder is 15. When ten times of the same
tkrk gS rks 'ks"kiQy Øe'k%
17, 24 vkSj27 gksrs gSaA tc number is divided by d, the remainder is 6.
(4x – 2y + 3z) dks31 ls foHkkftr fd;k tkrk gS] rks The least possible value of d is :
'ks"kiQy D;k gksxk\ tc ,d /ukRed iw.kk±d dksd ls Hkkx fn;k tkrk gS]
SSC CGL 2019 Tier-II (15/11/2020) rks 'ks"kiQy
15 vkrk gSA tc blh la[;k ds 10 xquk dks
d
(a) 9 (b) 8 ls Hkkx fn;kt krk gS] rks 'ks"kiQy
6 vkrk gSA
d dk U;wure
(c) 16 (d) 19 laHko eku gks ldrk gS %
12. When positive number a, b and c are divided
SSC CGL 05/03/2020 (Shift- 03)
by 13, the reminders are 9, 7 and 10
respectively. What will be the remainder when (a) 9 (b) 12
(a + 2b + 5c) is divided by 13? (c) 16 (d) 18
tc /ukRed la[;k a, b vkSjc dks 13 ls Hkkx fn;k 18. If a number is divided by 899, the remainder
r
tkrk gS rks 'ks"kiQy Øe'k%
9, 7 vkSj10 gksrs gSaA (atc is 63. If the same number is divided by 29,
si
+ 2b + 5c) dks13 ls foHkkftr fd;k tkrk gS] rks 'ks"kiQy the remainder will be :
D;k gksxk\ ;fn ,d la[;k dks 899 ls foHkkftr fd;k tkrk gS]
(a) 10
an by
SSC CGL 2019 Tier-II (16/11/2020)
(b) 5
rks 'ks"k
63 cprk gSA ;fn mlh la[;k dks29 ls foHkkftr
fd;k tkrk gS] rks 'ks"k la[;k D;k cpsxk\
n
(c) 9 (d) 8
CHSL 2019 26/10/2020 (Shfit- 01)
13. What is the remainder when the product of
ja (a) 10 (b) 2
R s
335, 608 and 853 is divided by 13?
(c) 4 (d) 5
335] 608 vkSj 853 ds Xkq.kuiQy dks 13 ls foHkkftr
a th
19. When an integer K is divided by 3, the
djus ij izkIr 'ks"kiQy D;k gksXkk\
remainder is 1 and when K + 1 is divided by
SSC CGL 12/04/2022 (Shift- 01)
5, the remainder is 0. Of the following a
(a) 11 (b) 12
ty a
possible value of a K is :
(c) 6 (d) 7
tc K dks3 ls foHkkftr fd;k tkrk gS rks 'ks"k 1 izkIr
14. The remainder when 75 × 73 × 78 × 76 is
di M
gksrk gS vkSj K tc + 1 dks5 ls foHkkftr fd;k tkrk gS
divided by 34 is :
rks 'ks"k
0 izkIr gksrk gSA
K dk laHko eku Kkr djsaA
75 × 73 × 78 × 76 dks34 ls foHkkftr fd;k tkrk
(a) 62 (b) 60
gS] rks 'ks"k Kkr djsaA
(c) 64 (d) 65
SSC CPO 2019 23/11/2020 (Shift- 03)
20. When two number are seperately divided by
(a) 18 (b) 12
33, the remainder are 21 and 28 respectively.
(c) 22 (d) 15
If the sum of the two numbrs is divided by 33,
15. If 1433 × 1433 × 1422 × 1425 is divided by
the remainder will be :
12, then what is the remainder?
;fn 1433 × 1433 × 1422 × 1425 dks12 ls foHkkftr tc nks la[;kvksa dks vyx&vyx 33 ls foHkkftr fd;k
fd;k tkrk gS] rks izkIr 'ks"kiQy Kkr djsaA tkrk gS rks 'ks"k Øe'k% 21 vkSj 28 izkIr gksrk gSA ;
A
SSC CPO 23/11/2020 (Shift-1) nksuksa la[;kvksa ds ;ksxiQy dks 33 ls foHkkftr fd;
(a) 3 (b) 9 tk, rks 'ks"k izkIr gksxk %
(c) 8 (d) 6 (a) 10 (b) 12
16. If 1433 ×1433 × 1422 × 1425 is divided by 10, (c) 14 (d) 16
then what is the remainder? 21. When two numbers are separately divided by
;fn 1433 × 1433 × 1422 × 1425 dks10 ls foHkkftr 44, the remainders are 11 and 38, respectively.
fd;k tkrk gS] rks izkIr 'ks"kiQy Kkr djsaA If the sum of the two numbers is divided by
SSC CPO 25/11/2020 (Shift-1) 44, then the remainder will be:
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nks la[;kvksa dks 44 ls foHkkftr djus ij izkIr 'ks"kiQy (a) 1539 (b) 539
Øe'k% 11 vkSj 38 gSaA ;fn mu nksuksa la[;kvksa ds ;ksx(c)dks 359 (d) 1359
44 ls foHkkftr fd;k tk,] rks izkIr 'ks"kiQy -------- gksxkA
26. When a positive integer N divided by 16, 17
SSC CHSL 30/05/2022 (Shift- 01) and 18 leaves remainders of 6, 7 and 8
(a) 16 (b) 44 respectively. Find the remainder when
N2 + 5N + 6 is divided by 11?
(c) 33 (d) 5
tc ,d /ukRed iw.kk±d la[;k dks16, 17 vkSj18 ls
22. Two positive numbers differ by 2001. When the
larger number is divided by the smaller number, foHkkftr fd;k tkrk gS rks Øe'k%
6, 7 vkSj8 'ks"k cprk
the quotient is 9 and remainder is 41. The sum gSA tcN + 5N + 6 dks11 ls foHkkftr fd;k tkrk gS
2
of the digits of the larger number is : rks 'ks"kiQy Kkr djsaA
nks /ukRed la[;kvksa esa 2001 dk varj gSA tc cM+h 1 (a) (b) 2
la[;k dks NksVh la[;k ls Hkkx fn;k tkrk gS] rks HkkxiQy
(c) 4 (d) 7
27. The
9 vkrk gS rFkk 'ks"kiQy 41 vkrk gSA cM+h vadksa ds vadksa sum of the digits of the least number
which when divided by 36, 72, 80 and 88
dk ;ksx gS %
r
leaves the remainders 16, 52, 60 and 68,
SSC CGL Tier-II, 13/09/2019
respectively, is:
si
(a) 15 (b) 11
lcls NksVh la[;k ds vadksa dk ;ksx ftls 36] 72] 80
(c) 10 (d) 14
an by vkSj 88 ls foHkkftr djus ij Øe'k% 16] 52] 60 vkSj
23. Two positive number differ by 3951. When the
larger number is divided by the smaller 68 izkIr gksrk gS\
n
number, the quotient is 12 and the remainder SSC CGL MAINS 03/02/2022
is 13. The sum of the digits of the larger (a) 17 (b) 11
number is: ja (c) 14 (d) 16
R s
nks /ukRed la[;kvksa esa 3951 dk varj gSA tc 28.cM+hFind the greatest number which divides 108,
la[;k dks NksVh la[;k ls foHkkftr fd;k tkrk gS] rks 124 and 156, leaving the same remainder:
a th
HkkxiQy 12 vkrk gS vkSj 'ks"k 13 cprk gSA cM+h la[;k
og cM+h ls cM+h la[;k Kkr dhft,] ftlls 108] 124 vkSj
ds vadksa dk ;ksx D;k gksxk\ 156 dks foHkkftr djus ij leku 'ks"kiQy izkIr gksrk gSA
ty a
SSC CPO 11/11/2022 (Shift-03)
SSC CGL 11/04/2022 (Shift- 02)
(a) 12 (b) 16 (a) 18 (b) 10
di M
(c) 18 (d) 14 (c) 12 (d) 16
24. If two number x and y is separately divided by 29. What is the greatest number by which when
a number the ramainder is 4376 and 2986 156, 181 and 331 are divided, the remainder
respectively. If the sum of that numbers i.e. is 6 in each case?
(x + y) is divided by that number 'd' then the og lcls cM+h la[;k dkSu lh gS] ftlls 156] 181 vkSj 331
remainder is 2361. The value of number d is :
dks foHkkftr djus ij izR;sd fLFkfr esa 'ks"kiQy 6 vkrk gS\
;fn nks la[;kvksa
x rFkky dks fdlh la[;k 'd' ls vyx&vyx
SSC CGL 11/04/2022 (Shift- 03)
Hkkx fd;k tk, rks 'ks"k Øe'k% 4376 rFkk2986 izkIr (a) 26 (b) 17
gksrk gSA ;fn ml la[;kvksa ds ;ksx] vFkkZr~ (x + y) dks (c) 25 (d) 13
mlh la[;k 'd' ls Hkkx fn;k tk, rks 'ks"k2361 izkIr gksrk30. When a number is successively divided by 3,
gSA la[;kd dk eku gS & 4 and 7, the remainder obtained are 2, 3 and
A
5 respectively. What will be the remainder
(a) 7362 (b) 5000
when 84 divides the same number?
(c) 4000 (d) 2542
25. The number which is divided by 10 leaves 9
tc dksbZ la[;k Øfed :i ls 3] 4 vkSj 7 ls foHkkftr
as remainder when it is divided by 9 leaves gksrh gS] rks izkIr 'ks"k Øe'k% 2] 3 vkSj 5 gksrs gSaA t
8 as remainder and when it is divided by 8 la[;k dks 84 ls foHkkftr djrs gSa rks 'ks"k D;k gksxk\
leaves 7 as remainder is : SSC CPO 2019 24/11/2020 (Shift- 03)
og la[;k] ftls 10 ls Hkkx nsus ij 9] 9 ls Hkkx nsus ij (a) 71 (b) 53
8 rFkk 8 ls Hkkx nsus ij 7 'ks"k cprk gS] og gS % (c) 30 (d) 48
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31. When a number is successively divided by 3, 36. What is the remainder when (12797 + 9797) is
4 and 7, the remainder obtained is 2, 3 and divided by 32?
5 respectively. What will be the remainder tc (12797 + 9797) dks32 ls Hkkx fn;k tkrk gS] rks
when 42 divides the same number?
'ks"kiQy D;k vk,xk \
tc dksbZ la[;k Øfed :i ls 3] 4 vkSj 7 ls foHkkftr SSC CGL Tier-II 13/09/2019
gksrh gS] rks izkIr 'ks"k Øe'k% 2] 3 vkSj 5 gksrh gSA(a)tc 4 (b) 2
42 mlh la[;k dks foHkkftr djrs gSa rks 'ks"k D;k gksxk\ (c) 7 (d) 0
SSC CPO 2019 25/11/2020 (Shift- 03) 37. What is the remainder when 2727 – 1527 is
(a) 31 (b) 41 divided by 6?
(c) 30 (d) 29 2727 – 1527 dks 6 ls foHkkftr djus ij 'ks"kiQy D;k
32. A number when successively divided by 3, 5 gksxk\
and 8 leaves remainder 1, 4 and 7 [CDS - 2021 (I)]
respectively. Find the respective remainder (a) 0 (b) 1
when the order of the divisors is reversed.
(c) 3 (d) 4
,d la[;k dks tc Øfed :i ls 3] 5 vkSj 8 ls foHkkftr
r
38. What is the remainder when we divide 570 +
fd;k tkrk gS rks 'ks"kiQy Øe'k% 1] 4 vkSj 7 izkIr gksrk770 by 74?
si
gSA tc Hkktd dk Øe myV fn;k tkrk gS] rks lacaf/r tc ge 570 + 770 dks74 ls Hkkx nsrs gSa] rks 'ks"kiQ
'ks"kiQy Kkr dhft,A
an by D;k vk,xk\
(a) 8 , 5, 3 (b) 4, 2, 1
SSC CGL 07/03/2020 (Shift- 03)
(c) 3, 2, 1 (d) 6, 4, 2
n
33. A number when successively divided by 4 and (a) 7 (b) 1
5 leaves remainder 1 and 4 respectively. When (c) 0 (d) 5
ja
R s
it is successively divided by 5 and 4, the 39. When (77 77 + 77) is divided by 78, the
respective remainder will be : remainder is :
a th
,d la[;k dks tc Øfed :i ls 3 vkSj 5 ls foHkkftr tc (7777 + 77) dks78 ls foHkkftr fd;k tkrk gS] rks
fd;k tkrk gS rks 'ks"kiQy Øe'k% 1 vkSj 4 izkIr gksrk gSA
'ks"kiQy fdruk vk,xk\
tc bls Øfed :i ls 5 vkSj 4 ls foHkkftr fd;k tkrk CHSL 2019 12/10/2020 (Shift- 02)
gS rks lacaf/r 'ks"kiQy gksxk %
ty a
(a) 75 (b) 77
(a) 4, 1 (b) 3, 2
(c) 76 (d) 74
di M
(c) 2, 3 (d) 1, 2
40. What will be the remainder when 2727 + 27
34. A number on being divided by 3, 4 and 5
is divided by 28?
successively the remainder 2, 1 and 2
respectively. Find the remainders when the tc 27 27$27 ds eku dks 28 ls foHkkftr fd;k tk,] rks
number successively divided by 5, 4 and 3. 'ks"kiQy fdruk gksxk\
,d la[;k dks Øfed :i ls 3] 4 vkSj 5 ls foHkkftr SSC CGL 01/12/2022 (Shift-03)
djus ij Øe'k% 2] 1 vkSj 2 'ks"k izkIr gksrk gSA tc mlh (a) 28 (b) 27
la[;k dks Øfed :i ls 5] 4 vkSj 3 ls foHkkftr fd;k (c) 25 (d) 26
83 83
tkrk gS rks 'ks"k Kkr djsaA 41. If 71 + 73 is divided by 36, the remainder is:
(a) 4, 0, 1 (b) 4, 2, 1 ;fn 7183 + 7383 dks 36 ls foHkkftr fd;k tk,] rks
(c) 4, 1, 1 (d) 4, 1, 2, 'ks"kiQy D;k gksxk\
If x is the remainder when 361284 is divided by
A
35. SSC CHSL 09/08/2021 (Shift- 01)
5 and y is the remainder when 4 96 (a) 0 (b) 13
is divided by 6, then what is the value of
(c) 9 (d) 8
(2x – y) ? 20
;fn x, 361284 dks 5 ls foHkkftr djus ij vkus okyk 42. Given that N = 5 , what will be the remainder
when N is divisible by 7 ?
'ks"kiQy gS rFkk
y, 496 dks6 ls foHkkftr djus ij vkus
okyk 'ks"kiQy gS](2xrks– y) dk eku D;k gS\ fn;k x;k gS N = 520, N dks 7 ls foHkkftr djus ij
SSC CGL Tier-II 13/09/2019
D;k 'ks"k izkIr gksxk\
(a) – 4 (b) 4 (a) 4 (b) – 4
(c) – 2 (d) 2 (c) – 1 (d) None of these
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43. Given that N = 200 1000 , what will be the 51. 96 – 11 is divided by 8 would leave a remainder
remainder when N is divisible by 17 ? of :
fn;k x;k gSN = 2001000, N dks17 ls foHkkftr djus 96 – 11 dks 8 ls foHkkftr djus ij 'ks"kiQy izkIr gksxk %
ij D;k 'ks"k izkIr gksxk\ (a) 0 (b) 1
(a) 1 (b) 2
(c) 2 (d) 6
(c) 3 (d) 4 n
52. If n is even (6 – 11) is divisible by
44. Given that N = (2222)5555 + (5555)2222. What will
be the remainder when N is divided by 7? ;fn n le la[;k gks rks(6n – 11) foHkkftr gksxh %
fn;k x;k gSN = (2222)5555 + (5555)2222. tc N dks (a) 37 (b) 35
7 ls foHkkftr fd;k tkrk gS rks D;k 'ks"k cpsxk\ (c) 30 (d) 6
(a) 0 (b) 1 53. When 200 is divided by a positive integer x,
(c) 2 (d) 3 the remainder is 8. How many values of x are
45. What is the remainder when 6599 is divided by there?
11?
tc 200 dks fdlh /ukRed iw.kk±d x ls Hkkx fn;k tkrk
6599 dks 11 ls foHkkftr djus ij 'ks"kiQy D;k gksxk\ gS] rks 'ks"kiQy
8 vkrk gSA
x ds fdrus eku gSa\
r
[CDS – 2023 (I)]
SSC CGL 03/03/2020 (Shift- 02)
si
(a) 0 (b) 5
(a) 7 (b) 5
(c) 9 (d) 10
46.
an by
What is the remainder after dividing the
number 371000 by 9? 54.
(c) 8 (d) 6
When 732 is divided by a positive integer x,
n
the remainder is 12. How many value of x are
la[;k 37 1000
dks 9 ls Hkkx nsus ij 'ks"kiQy D;k gksxk\
there?
ja [CDS – 2021 (I)]
tc 732 dks ,d /ukRed iw.kk±d x ls Hkkx fn;k tkrk gS]
R s
(a) 1 (b) 3
rks 'ks"kiQy
12 vkrk gSA
x ds fdrus eku gSa\
a th
(c) 7 (d) 9
47. What is the remainder when 21000000 is divided SSC CGL 03/03/2020 (Shift- 01)
by 7? (a) 19 (b) 20
21000000
dks 7 ls foHkkftr djus ij 'ks"kiQy D;k gksxk\ (c) 18 (d) 16
ty a
[CDS – 2021 (I)] 55. 'a' divides 228 leaving a remainder 18. The
biggest two-digits value of 'a' is :
di M
(a) 1 (b) 2
(c) 4 (d) 6 228 dks'a' ls foHkkftr djus ij 'ks"k 18 izkIr gksrk gSA
48. What is the remainder when 38 is divided by 7? 'a' dk nks vadh; vf/dre eku gS %
tc 3 8 dks 7 ls foHkkftr fd;k tkrk gS rks 'ks"kiQy D;k gksrk gS\(a) 70 (b) 21
SSC CHSL 08/08/2023 Shift-02 (c) 35 (d) 30
(a) 5 (b) 4 56. 64329 is divided by a certain number. While
(c) 6 (d) 2 dividing the numbers, 175, 114 and 213
49. What will be the remainder when (265)4081 + 9 is appear as the three successive remainders. The
divided by 266? divisor is :
(265)4081 + 9 dks266 ls foHkkftr djus ij izkIr 'ks"kiQy 64329 dks ,d fuf'pr la[;k ls foHkkftr fd;k tkrk gSA
Kkr dhft,A
A
la[;kvksa dks foHkkftr djrs le;] 175] 114 vkSj 213
SSC CGL 14/07/2023 (Shift-01) yxHkx rhu Øekxr 'ks"kiQy gSaA Hkktd gS %
(a) 8 (b) 6 (a) 184 (b) 224
(c) 1 (d) 9
(c) 234 (d) 256
50. Find the remainder when 88 + 6 is divided by 7.
57. If 243 gives a remainder of 4 when divided
88 $ 6 dks 7 ls foHkkftr djus ij izkIr 'ks"kiQy Kkr dhft,A by a certain natural number n, what will be
SSC CGL 25/07/2023 (Shift-01) the remainder when 482 is divided by 2n
(a) 0 (b) 2 assuming that the quotient in both cases is
(c) 3 (d) 1 the same?
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243 dks fdlh fuf'pr izkÑfrd la[;k n l s Hkkx nsus X
61. If X = (163 + 173 + 183 + 193), then leaves
ij 4 'ks"k cprk gSA 482 dks
2n ls Hkkx nsus ij 'ks"kiQy 70
a remainder of
D;k gksxk] ;g ekurs gq, fd HkkxiQy nksuksa n'kkvksa esa
leku gSaA ;fn X = (163 + 173 + 183 + 193)] rks ls Hkkxnsus
(a) 1 (b) 4 ij 'ks"kiQyD;k cpsxk\
(c) 5 (d) 8 (a) 0 (b) 1
58. N = 1234567......55. Find the remainder when (c) 69 (d) 35
N is divided by 16.
62. What is the remainder when (11 + 22 + 33 +
N = 1234567......55, N dks16 ls Hkkx nsus ij 'ks"kiQy 44+ ...... + 100100) is divided by 4?
Kkr djsaaA
(11 + 22 + 33 + 44+ ...... + 100100) dks 4 ls Hkkx nsus
(a) 13 (b) 15
(c) 18 (d) 22 ij 'ks"kiQy D;k cpsxk\
59. R = Remainder when (1! + 2! + 3! + ......120!) (a) 0 (b) 1
is divided by 15. find R. (c) 2 (d) 3
(1! + 2! + 3! + ......120!) dks tc 15 l s Hkkx fn;k63. Two numbers 11284 and 7655 is divided by
r
tkrk gS rks
R 'ks"k cprk gSA
R dk eku gS % three digits numbers leaves the same
si
(a) 0 (b) 1 remainder. That three digit number is :
(c) 2 (d) 3 nks la[;k,¡ 11284 vkSj 7655 dks fdlh 3 vadh; la[;k ls
60.
an by
If X = (12 + 13 + 14 + 153), then what is the
3 3 3
remainder when X is divided by 9?
Hkkx nsus ij leku 'ks"k jgrs gSaA ;g rhu vadh; la[;k gS &
n
(a) 199
;f n X = (123 + 133 + 143 + 153) g S] rks
X dks 9 ls (b) 197
foHkkftr djus ij 'ks"kiQy D;k gS\
ja (c) 193
R s
(a) 0 (b) 1 (d) 191
a th
(c) 27 (d) None of these
ANSWER KEY
ty a
1.(a) 2.(b) 3.(d) 4.(c) 5.(c) 6.(a) 7.(b) 8.(d) 9.(b) 10.(d)
di M
11.(b) 12.(d) 13.(d) 14.(b) 15.(d) 16.(b) 17.(c) 18.(d) 19.(c) 20.(d)
21.(d) 22.(d) 23.(b) 24.(b) 25.(c) 26.(b) 27.(d) 28.(d) 29.(c) 30.(a)
31.(d) 32.(d) 33.(c) 34.(c) 35.(c) 36.(d) 37.(a) 38.(c) 39.(c) 40.(d)
41.(a) 42.(a) 43.(a) 44.(a) 45.(d) 46.(a) 47.(b) 48.(b) 49.(a) 50.(a)
51.(d) 52.(b) 53.(c) 54.(b) 55.(a) 56.(c) 57.(b) 58.(b) 59.(d) 60.(a)
A
61.(a) 62.(a) 63.(d)
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Remainder @'ks"kiQy
(Practice Sheet With Solution)
8. What is the remainder when (x17 + 1) is divided
Level-01 by (x + 1)?
1. Find the remainder:/'ks"kiQy Kkr dhft,% tc (x17 + 1) dks (x + 1)ls foHkkftr fd;k tkrk gS] rks
59637 'ks"kiQy D;k gksxk\
58 (a) x (b) x –1
(a) 1 (b) 2 (c) 0 (d) 1
(c) 7 (d) 10 9. A number when divided by 78 gives the
2. Find the remainder:/'ks"kiQy Kkr dhft,% quotient 280 and the remainder 0. If the same
r
number is divided by 65, what will be the value
96132 of the reminder?
si
97 ,d la[;k dks 78 ls foHkkftr djus ij HkkxiQy 280 vkSj
(a) 1
(c) 5
an by (b) 3
(d) 2
'ks"kiQy 0 çkIr gksrk gSA ;fn mlh la[;k dks 65 ls foHkkftr
fd;k tk,] rks vuqLekjd dk eku D;k gksxk\
n
3. Find the remainder:/'ks"kiQy Kkr dhft,% (a) 1 (b) 3
181 753 (c) 0 (d) 2
ja
R s
182 10. If the number 123456789 is divided by 9, then
(a) 0 (b) 181 the remainder is:
a th
(c) 50 (d) 180 ;fn la[;k 123456789 dks 9 }kjk foHkkftr fd;k tkrk
4. Find the remainder:/'ks"kiQy Kkr dhft,% gS] rks 'ks"kiQy D;k gksxk\
(a) 0 (b) 1
6867 – 1
ty a
(c) 2 (d) 3
68
11. The remainder when 1919+20 is divided by 18 is:
di M
(a) 0 (b) 67
(c) 62 (d) 65
tc 1919 + 20 dks 18 ls foHkkftr fd;k tk,] rks 'ks"k
Kkr dhft,A
5. Find the remainder:/'ks"kiQy Kkr dhft,%
(a) 3 (b) 2
153153 153 (c) 1 (d) 0
154 12
12. When m – 1 is divided by m + 1, the
(a) 100 (b) 191 remainder is:
(c) 0 (d) 152 tc m12–1 dks m + 1 }kjk foHkkftr fd;k tkrk gS]
6. Find the remainder:/'ks"kiQy Kkr dhft,% rks 'ks"kiQy D;k gksxk\
8889 26 (a) 1 (b) 2
89 (c) 0 (d) –1
A
(a) 10 (b) 20 13. Find the Remainder when (12 13 + 23 13) is
(c) 25 (d) 35 divided by 11.
7. When m is divided by 7, the remainder is 5. (1213 + 2313) dks 11 ls foHkkftr djus ij 'ks"kiQy Kkr
When 3m is divided by 7, the remainder is: dhft,A
tc m dks 7 ls foHkkftr fd;k tkrk gS] rks 'ks"kiQy
5 (a) 2 (b) 1
vkrk gSA tc3m dks 7 ls foHkkftr fd;k tk,xk] rks (c) 0 (d) 3
'ks"kiQy D;k gksxk\ 14. A number when divided by 7 leaves remainder
(a) 3 (b) 2 of 4. If the square of the same number is
(c) 1 (d) 0 divided by 7, then what is the remainder?
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fdlh la[;k dks 7 ls foHkftr djus ij 4 'ks"kiQy cprk 21. Find the remainder when 1 2 3 4 5 _____ 41
gSA ;fn mlh la[;k ds oxZ dks 7 ls foHkkftr fd;k digits is divided by 4.
tk,] rks 'ks"kiQy D;k gksxk\ 1 2 3 4 5 _________ 41 vadksa dks 4 ls foHkkftr
(a) 3 (b) 1 djus ij 'ks"kiQy Kkr djsaA
(c) 4 (d) 2 (a) 1 (b) 3
15. On dividing a certain number by 363, we get (c) 2 (d) 0
17 as the remainder. What will be the remainder 22. If each of the two numbers 516 and 525 are
when the same number is divided by 11? divided by 6, the remainders are R1 and R2,
,d la[;k dks 363 }kjk foHkkftr fd;k tkrk gS rks respectively. What is the value of 1
R R2
?
'ks"kiQy 17 izkIr gksrk gSA tc mlh la[;k dks 11 }kjk R2
foHkkftr fd;k tk,xk rks 'ks"kiQy D;k gksxk\ ;fn nks la[;kvksa16 5vkSj 525 esa ls izR;sd dks 6 ls foHkkftr
(a) 7 (b) 8
R R2
(c) 6 (d) 9 fd;k tkrk gS] rks 'ks"kiQy Øe'k%
R1 vkSjR2 gSaA1
R2
16. When an integer n is divided by 6, the
remainder is 5. What is the remainder if 9n is dk eku D;k gS\
r
divided by 6?
1 5
tc iw.kk±dn dks 6 ls foHkkftr fd;k tkrk gS] rks 'ks"kiQy (a) (b)
si
6 6
5 izkIr gksrk gSA 9n;fn dks 6 ls foHkkftr fd;k tkrk
(a) 4
an by
gS] rks 'ks"kiQy Kkr djsaA
(b) 3
(c)
1
5
(d)
6
5
n
(c) 5 (d) 2 23. When a number is divided by 3, the remainder
is 2. Again, when the quotient is divided by 7,
17. If a number is divisible by 624, the remainder
ja the remainder is 5. What will be the remainder
will be 53. If the same number is divisible by
R s
when the original number is divided by 21?
16, then the remainder will be:
tc fdlh la[;k dks 3 ls foHkkftr fd;k tkrk gS] rks
;fn fdlh la[;k dks 624 ls foHkkftr fd;k tkrk gS] rks
a th
'ks"kiQy 2 vkrk gSA fiQj] tc HkkxiQy dks 7 ls foHkkftr
'ks"kiQy 53 izkIr gksrk gSA ;fn mlh la[;k dks 16 ls foHkkftr
fd;k tkrk gS] rks 'ks"kiQy 5 vkrk gSA ewy la[;k dks 21 ls
fd;k tk,] rks 'ks"kiQy D;k gksxk\
foHkkftr djus ij 'ks"kiQy fdruk gksxk\
ty a
(a) 5 (b) 4
(a) 13 (b) 16
(c) 7 (d) 6
(c) 14 (d) 17
di M
18. If 3 divided the integer n, the remainder is 2.
24. If 3147 + 4347 is divided by 37, the remainder is:
Then what will be the remainder when 7n
divided by 3. ;fn 3147 + 4347 dks 37 ls foHkkftr fd;k tk,] rks
;fn iw.kkZad
n dks 3 ls foHkkftr fd;k tk,] rks 'ks"kiQy 'ks"kiQy gS%
2 vkrk gSA rksn7dks 3 ls foHkkftr djus ij 'ks"kiQy (a) 1 (b) 3
D;k gksxk\ (c) 0 (d) 2
(a) 3 (b) 2 25. When (224 – 1) is divided by 7, the remainder is:
(c) 6 (d) 4 (224 – 1) dks 7 ls foHkkftr djus ij fdruk 'ks"kiQy cpsxk\
19. Find the remainder when 2787 × 2345 × 1992 is (a) 4 (b) 2
divided by 23. (c) 3 (d) 0
2787 × 23 45 × 19 92dks 23 ls foHkkftr djus ij 'ks"kiQy 26. What is the remainder when (341218 – 156218)
A
Kkr djsaA is divisible by 259?
(a) 1 (b) 2 tc (341 218 & 156218) dks 259 ls foHkkftr fd;k tkrk
(c) 0 (d) 5 gS] rks fdruk 'ks"kiQy cprk gS\
20. Find the remainder when 163 + 173 +183 +193 (a) 0 (b) 2
is divided by 70. (c) 7 (d) 3
163 + 173 +183 +193 dks 70 ls foHkkftr djus ij 'ks"k 2n 2n
27. (167) – 103 is exactly divisible by:
Kkr djsaA (167)2n – 1032n foHkkT; gS%
(a) 69 (b) 0 (a) 144 (b) 170
(c) 68 (d) 51 (c) 106 (d) 225
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28. (68)n + 1 is exactly divisible by 23 when n is? 37. Find the remainder:/'ks"kiQy Kkr dhft,%
(68)n + 1, 23 ls foHkkT; gS] ;fn
n:
8317 – 5217 8417 – 5317
(a) Any natural number
62
(b) Odd number
(a) 0 (b) 3
(c) Even number
(c) 2 (d) 4
(d) Only prime number
38. Find the remainder:/'ks"kiQy Kkr dhft,%
29. When (218 – 1) is divided by 9, the remainder is:
tc (2 18 &1) dks 9 ls foHkkftr fd;k tkrk gS] rks 'ks"kiQy 137 71 77 93
fdruk gksxk% 15
(a) 2 (b) 0 (a) 4 (b) 10
(c) 8 (d) 1 (c) 2 (d) 5
30. Find the remainder when 341 +782 is divided by 52. 39. Find the remainder:/'ks"kiQy Kkr dhft,%
3 + 7 dks 52 ls foHkkftr djus ij 'ks"kiQy Kkr djsaA
41 82
132 135 136 138 142
r
(a) 0 (b) 1 137
(c) 2 (d) 3
si
(a) 0 (b) 87
31. Find the remainder when 23 10 – 1024 is (c) 10 (d) 82
divided by 7. an by 40. Find the Remainder when 273 + 375 + 478 +
2310 – 1024 dks 7 ls foHkkftr djus ij 'ks"kiQy Kkr djsaA 657 + 597 is divided by 25.
n
(a) 6 (b) 5 273 + 375 + 478 + 657 + 597 dks 25 ls foHkkftr
(c) 0 (d) 3 djus ij 'ks"kiQy Kkr dhft,A
ja
R s
32. What is the remainder when 3040 is divided by 7? (a) 5 (b) 10
3040 dks 7 ls foHkkftr djus ij 'ks"kiQy D;k gksxk\ (c) 9 (d) 8
a th
(a) 2 (b) 3 41. What is the remainder when the product of
(c) 4 (d) 5 335, 608 and 853 is divided by 13?
33. Find the remainder when 21875 divided by 17? 335] 608 vkSj 853 ds xq.kuiQy dks 13 ls foHkkftr djus
ty a
21 dks 17 ls foHkkftr djus ij 'ks"kiQy Kkr djsa\
875 ij izkIr 'ks"kiQy D;k gksxk\
(a) 11 (b) 12
di M
(a) 12 (b) 14
(c) 6 (d) 7
(c) 16 (d) 13
42. Find the remainder when 73 + 75 + 78 + 57 +
34. What will be the remainder when 2727 + 27
197 is divided by 34
is divided by 28?
73 + 75 + 78 + 57 + 197 dks 34 ls foHkkftr djus
tc 27 27$27 ds eku dks 28 ls foHkkftr fd;k tk,]
ij 'ks"kiQy Kkr dhft,
rks 'ks"kiQy fdruk gksxk\
(a) 3 (b) 4
(a) 28 (b) 27
(c) 5 (d) 6
(c) 25 (d) 26
35. What will be the remainder when 742 is divided 43. Find the remainder:/'ks"kiQy Kkr dhft,%
by 48? 6397 9398 7007 3111
tc 7 42
dks 48 ls foHkkftr fd;k tk,] rks 'ks"k Kkr dhft,A
A
100
(a) 2 (b) 3 (a) 62 (b) 0
(c) 1 (d) 0 (c) 65 (d) 63
36. Find the Remainder when 77 × 85 × 73 is 44. Find the remainder:/'ks"kiQy Kkr dhft,%
divided by 9.
77 × 85 × 73 dks 9 ls foHkkftr djus ij 'ks"kiQy Kkr 816 825 823 827
dhft,A 819
(a) 1 (b) 2 (a) 0 (b) 243
(c) 4 (d) 7 (c) 34 (d) 321
Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs3
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45. 9 62 – 11 when divided by 8 would leave a
remainder of: Level-02
9 & 11 tc 8 ls foHkkftr fd;k tkrk gS] rks 'ks"kiQy51. Find the remainder when 10 1 + 10 2 +10 3 +
62
D;k gksxk% 104 + __________ + 10100 is divided by 6.
(a) 5 (b) 6 101 + 102 +103 + 104 + __________ + 10100, 6 ls
(c) 2 (d) 3 foHkkftr djus ij 'ks"kiQy Kkr djsaA
46. A number when divided successively by 4 and (a) 3 (b) 1
5 leaves remainder 1 and 4 respectively.
When it is successively divided by 5 and 4, (c) 5 (d) 4
the respective remainders will be: 52. Find the remainder when 7 7 + 7 77 + 7 777 +
,d la[;k tc Øfed :i ls 4 vkSj 5 ls foHkkftr fd;k 77777 +............+ 7777777777 is divided by 6.
tkrk gS] rks Øe'k% 1 vkSj 4 'ks"k jgrk gSA tc bls Øfed 77 + 777 + 7777 + 77777 + ......... + 7777777777 dks 6
:i ls 5 vkSj 4 ls foHkkftr fd;k tkrk gS] rks Øfed ls foHkkftr djus ij 'ks"kiQy Kkr djsaA
'ks"kiQy fuEu gksaxs% (a) 2 (b) 3
(a) 4, 1 (b) 3, 2 (c) 5 (d) 6
(c) 2, 3 (d) 1, 2 53. What is the remainder when 1! + 2! + 3! + ........
+ 100! is divided by 5.
r
47. When a number N is successively divided by
6, 5 and 7 the remainder obtained are 3, 1 and tc 1! + 2! + 3! + ........ + 100! dks 5 ls foHkkftr
si
4 respectively. What is the remainder when N
fd;k tkrk gS rks 'ks"k D;k gSA
is divided by 15?
(a) 2 (b) 4
tc ,d la[;k N Øfed :i ls 6] 5 vkSj 7 ls foHkkftr
an by dks gSaA(c) 1
gksrh gS] rks izkIr 'ks"k Øe'k% 3] 1 vkSj 4Ngksrs (d) 3
15 ls foHkkftr djus ij D;k 'ks"k gS\ 54. Find the remainder when 1! + 2! + 3! +4! + 5!
n
+ _________ 1000! is divided by 8.
(a) 9 (b) 6
1! + 2! + 3! + 4! + 5! + _________ 1000! dks8 ls
(c) 12 ja (d) 7
foHkkftr djus ij 'ks"k Kkr dhft,A
R s
48. If the number is successively divided by 7, 4
and 3 leaves remainder 5, 3 and 2 respectively. (a) 1 (b) 2
a th
If the order of divisor are reversed then what (c) 3 (d) 4
will be the remainder?
55. What is the remainder when 1! + 2! + 3! + …
;fn la[;k dks Øfed :i ls 7] 4 vkSj 3 ls foHkkftr ..+100! is divided by 18?
fd;k tkrk gS] rks Øe'k% 5] 3 vkSj 2 'ks"k cprk gSA ;fn 'ks"kiQy D;k gksxk 1!
tc + 2! + 3! +…...+100!, 18
ty a
Hkktd ds Øe dks myV fn;k tk, rks 'ks"kiQy D;k gksxk\ ls foHkkftr fd;k tkrk gS\
(a) 1, 3, 6 (b) 2, 4, 3
di M
(a) 10 (b) 11
(c) 0, 2, 6 (d) 1, 3, 5
(c) 9 (d) 8
49. Find the least number which when divided by 4,
9, 12 and 15, leaves the remainder 3 in each case. 56. Find the remainder when 51203 is divided by 7.
og U;wure la[;k Kkr dhft, ftls 4] 9] 12 vkSj 15 ls 51203dks 7 ls foHkkftr djus ij 'ks"kiQy Kkr djsaA
foHkkftr djus ij çR;sd fLFkfr esa 3 'ks"k cprk gSA (a) 2 (b) 3
(a) 193 (b) 183 (c) 4 (d) 5
(c) 360 (d) 180 57. A number when divided by 18 leaves a
50. When a number is divided by a certain divisor, remainder 7. The same number when divided
the remainder is 46. If another number is by 12 leaves a remainder n. How many values
divided by same divisor, it leaves remainder can n take?
31. If the sum of both the numbers is divided ,d la[;k dks tc 18 ls Hkkx fn;k tkrk gS rks 'ks"kiQy 7
by the same divisor, it gives remainder 19. cprk gSA mlh la[;k dks tc 12 ls foHkkftr fd;k tkrk
A
Find divisor.
gS rks 'ks"kiQy
n jgrk gSAn fdrus eku ys ldrk gS\
tc fdlh la[;k dks ,d fuf'pr Hkktd }kjk foHkkftr
(a) 2 (b) 0
fd;k tkrk gS] rks 'ks"k 46 gksrk gSA ;fn fdlh vU;
(c) 1 (d) 3
la[;k dks mlh Hkktd ls foHkkftr fd;k tkrk gS] rks
58. N leaves a remainder of 4 when divided by 33,
og 'ks"k 31 NksM+ nsrk gSA ;fn nksuksa la[;kvksa dk ;ksxare the possible remainders when N is
what
mlh Hkktd ls foHkkftr fd;k tkrk gS] rks ;g 'ks"k 19 divided by 55?
nsrk gSA Hkktd dk irk yxk,aA N dks 33 ls foHkkftr djus ij 4 'ks"k cprk gS] tc
(a) 56 (b) 50 N dks 55 ls foHkkftr fd;k tkrk gS] rks 'ks"kiQy D;k
(c) 58 (d) 64 gks ldrk gS\
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(a) 3 (b) 5 66. Find the remainder:/'ks"kiQy Kkr dhft,%
(c) 4 (d) 2
2763
59. Given a number, N = 553 + 173 – 723, then
192
which of the following is true?
(a) 2 (b) 0
,d la[;k nh xbZ gS]N = 553 + 173 – 723 rks fuEu esa
(c) 3 (d) 7
ls dkSu lk lR; gS\
67. Find the remainder:/'ks"kiQy Kkr dhft,%
(a) N is divisible by both 7 and 13
(b) N is divisible by both 3 and 17 101 102 103 104 ........ 10100
(c) N is divisible by 17 but not 3 6
(d) N is divisible by 11 but not 17 (a) 0 (b) 4
60. (31 20
– 1024) is not divisible by : (c) 10 (d) 5
O;atd (3120 – 1024) ls foHkkftr ugh gS\ 68. What is the remainder when 7 + 77 + 777 + 7777
+ ......... + (till 100 terms) is divided by 8?
(a) 137 (b) 107
7 + 77 + 777 + 7777 + ......... + (100 inks rd)
r
(c) 9 (d) 32
dks 8 ls foHkkftr djus ij D;k 'ks"kiQy gS\
61. Find the remainder:/'ks"kiQy Kkr dhft,%
si
(a) 6 (b) 0
23 23 23 23
1 2 3
an by ........ 70 (c) 1 (d) 7
71 69. What is the remainder when 727272727272
........... till 999 terms is divided by 101?
n
(a) 0 (b) 1
(c) 4 (d) 2 727272727272 ........... 999 inkas rd dks 101 ls
ja foHkkftr djus ij D;k 'ks"kiQy gS\
62. Find the remainder:/'ks"kiQy Kkr dhft,%
R s
(a) 0 (b) 20
a th
7 7 7 7
1 2 3 ........ 100 (c) 27 (d) 7
202 70. What is the smallest number which leaves
remainder 3 when divided by any of the
(a) 0 (b) 1 numbers 5, 8 and 9, but leaves no remainder
ty a
(c) 2 (d) 3 when it is divided by 11?
63. What is the remainder when 91 × 92 × 93 og lcls NksVh la[;k dkSu lh gS ftls 5] 8 vkSj 9 esa ls
di M
......... 99, is divided by 1261? fdlh Hkh la[;k ls foHkkftr djus ij 'ks"kiQy 3 cprk gS
tc 91 × 92 × 93 ......... 99 dks 1261 ls foHkkftr ysfdu 11 ls foHkkftr djus ij dksbZ 'ks"k ugha cprk gS\
fd;k tkrk gS] rks 'ks"kiQy D;k gksxk\ (a) 363 (b) 563
(a) 0 (b) 1 (c) 463 (d) 663
(c) 2 (d) 3 71. What is the remainder obtained when a prime
number greater than 7 is divided by 6?
64. Find the remainder:/'ks"kiQy Kkr dhft,%
7 ls cM+h vHkkT; la[;k dks 6 ls foHkkftr djus ij
1! 2! 3! 4 ! ....... 10000 ! D;k 'ks"kiQy çkIr gksrk gS\
18 (a) 1 or 5 (b) 2 or 3
(a) 0 (b) 3 (c) 3 or 5 (d) 2 or 5
72. Find the remainder/'ks"kiQy Kkr dhft,%
A
(c) 9 (d) 8
65. If x is the remainder when 361284 is divided by
11937 – 8037 +10337 – 6837
5 and y is the remainder when 496 is divided =?
37
by 6, then what is the value of (2xy)?
(a) 0 (b) 4
;fn 3 61284 dks 5 ls foHkkftr fd;k tkrk gS rks 'ks"kiQy
x
(c) 6 (d) 8
cprk gS vkSj ;fn 4 dks 6 ls foHkkftr fd;k tkrk gS rks
96
73. Two positive numbers differ by 1280. When
'ks"kiQyy cprk gSA(2x – y) dk eku D;k gS\
the greater number is divided by the smaller
(a) 8 (b) – 2 number, the quotient is 7 and the remainder
(c) 2 (d) – 4 is 50. The greater number is:
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nks /ukRed la[;kvksa dk varj 1280 gSA tc cM+h la[;k Level-03
dks NksVh la[;k ls foHkkftr fd;k tkrk gS] rks HkkxiQy 7
80. The sum of the digits of two-digit number is
vkSj 'ks"kiQy 50 izkIr gksrk gSA cM+h la[;k Kkr dhft,A
(a) 1458 (b) 1485 1
of the number. The unit digit is 4 less than
(c) 1585 (d) 1558 7
the tens digit. If the number obtained on
74. A number when divided by 442 gives a reversing its digit is divided by 7, The
remainder of 49. Find the remainder when it remainder will be.
is divided by 39.
1
,d la[;k dks 442 ls Hkkx nsus ij 49 'ks"kiQy vkrk nks vadksa dh la[;k ds vadksa dk ;ksx la[;k dk
gSA
7
gSA Kkr dhft, fd 39 ls foHkkftr djus ij fdruk bdkbZ dk vad ngkbZ ds vad ls 4 de gSA ;fn mlds vad
'ks"kiQy cpsxk\ dks myVus ij çkIr la[;k dks 7 ls foHkkftr fd;k tk,]
(a) 10 (b) 9 rks 'ks"kiQy gksxkA
(c) 11 (d) CND (a) 4 (b) 5
75. What will be the remainder when 201 + 202 + 203 (c) 1 (d) 6
r
+ 204 + 205 + 206 +...... 260 is divided by 9? 81. The integers 573921 and 575713 when divided
by a 3 digit number leave the same remainder.
tc 201 + 202 + 203 + 204 + 205 + 206 +...... 260
si
What is that 3 digit number?
dks 9 ls foHkkftr fd;k tkrk gS] rc 'ks"kiQy D;k gksxk\ iw.kkZad 573921 vkSj 575713 dks tc 3 vadksa dh la[;k
(a) 8
(c) 6
an by (b) 4
(d) 2
ls foHkkftr fd;k tkrk gS rks leku 'ks"k cprk gSA og 3
vadksa dh la[;k D;k gS\
n
76. Find the smallest 4-digit number which if divided (a) 206 (b) 256
by 10, 11, 20, 22 leaves remainder as 8. (c) 274 (d) 189
ja
4-vadksa dh og NksVh ls NksVh la[;k Kkr dhft,] 82. ftls If N = (243 + 253 + 263 + 273), then N divided
R s
10, 11, 20, 22 ls foHkkftr djus ij 'ks"kiQy
8 jgrk gSA by 102 leaves a remainder of?
a th
(a) 1110 (b) 1108 vxj N = (243 + 253 + 263 + 273)] rksN dks 102 ls
(c) 1092 (d) 1100 Hkkx nsus ij D;k 'ks"k cprk gS\
77. 0! + 1! + 2! + 3! .......... + 98! What will be the (a) 18 (b) 12
ty a
remainder when divided by 5? (c) 1 (d) 0
0! + 1! + 2! + 3! .......... + 98! dks 5 ls Hkkx nsus ij83. The remainder when 29 29 divided by 9 is.
29
di M
'ks"kiQy D;k gksxk\
(a) 2 (b) 5 2929
29
dks 9 ls foHkkftr djus ij 'ks"kiQy gksrk gSA
(c) 3 (d) 4 (a) 1 (b) 2
78. When x is divided by 8 , remainder obtained is (c) 3 (d) 4
3. Find the remainder when (x4 + x3 + x2 + x + 1) 84. 25102 divided by 17 gives the remainder.
is divided by 8 ? dks 17 ls foHkkftr djus ij 'ks"kiQy çkIr gksrk gSA
25102
tc x dks8 ls foHkkftr fd;k tkrk gS] rks 'ks"kiQy
3 izkIr (a) 2 (b) 6
gksrk gSA 'ks"kiQy Kkr dhft,(xtc
4
+ x3 + x2 + x + 1) (c) 1 (d) 4
dks8 ls foHkkftr fd;k tkrk gS\ 85. What is the remainder when 21040 is divided by 131?
(a) 3 (b) 2 21040 dks 131 ls foHkkftr djus ij 'ks"kiQy D;k gksxk\
A
(c) 5 (d) 1 (a) 130 (b) 1
79. If a number 'x ' is divided by 21, it leaves a (c) 125 (d) 0
remainder of 5, if (5x2 +2x – 1) divided by 13, 86. Find the remainder when we divide 3x4 – 2x²
then what is the remainder? +4x – 1 by 2x – 1.
;fn ,d la[;k 'x ' dks21 ls foHkkftr fd;k tkrk gS rks 3x4– 2x² +4x – 1 dks 2x – 1 }kjk foHkkftr djus
5 'ks"k cprk gS] ;fn(5x2 +2x – 1) dks 13 ls foHkkftr ij izkIr 'ks"kiQy Kkr dhft,A
fd;k tkrk gS] rks 'ks"kiQy D;k gS\ (a) 2 (b) 3
(a) 2 (b) 3 11 15
(c) (d)
(c) 4 (d) 1 16 16
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87. What is the remainder of 15 + 25 + 35 + 45 + 55 92. What is the largest value of the positive integer
+ 65 + 75 + ........ + 505 when divided by 5 k such that k divides n²(n² – 1) (n² – n – 2) for
15 + 25 + 35 + 45 + 55 + 65 + 75 + ........ + 505 dks every natural number n?
5 ls foHkkftr djus ij 'ks"kiQy D;k gksxk\ ldkjkRed iw.kkZad
k dk lcls cM+k eku D;k gS tSls fd k
(a) 3 (b) 4 çR;sd çkÑfrd la[;k n ds fy, n²(n² – 1) (n² – n – 2)
(c) 2 (d) 0 dks foHkkftr djrk gS\
88. If a and b are two odd positive integers, by (a) 6 (b) 12
which of the following integers is (a4 – b4) (c) 24 (d) 48
always divisible?
93. If 'a' is a natural number, then (7a² + 7a) is
;fn a vkSjb nks fo"ke /ukRed iw.kk±d gSa] rks fuEu esa
always divisible by.
fdl iw.kk±d ls(a4 – b4) ges'kk foHkkT; gS\
;fn 'a' ,d izkÑr la[;k gS] rks(7a² + 7a) buesa ls fdl
(a) 3 (b) 6 la[;k ls lnSo foHkkT; gS\
(c) 8 (d) 12
(a) 7 and 14 both (b) 7 only
89. Find the remainder:/'ks"kiQy Kkr dhft,%
r
(c) 14 only (d) 21 only
si
10 1 10 2 10 3 10 30
10 10 10 ........ 10 94. Which of the following numbers leaves the
7 remainder equal to the highest common factor
(a) 1
(c) 0
an by (b) 2
(d) 4
of 6, 8 and 9, when divided by 6, 8 and 9?
fuEufyf•r esa ls dkSu lh la[;k dks 6] 8 vkSj 9 ls foHkkftr
n
90. Find the remainder:/'ks"kiQy Kkr dhft,% djus ij 'ks"kiQy 6] 8 vkSj 9 ds mPpre lkekU; xq.ku•aM
ja ds cjkcj cprk gS\
R s
99132
171 (a) 506 (b) 575
a th
(a) 182 (b) 0 (c) 291 (d) 433
(c) 127 (d) 144 95. Find the remainder when 13456789101112.....
91. Find the remainder:/'ks"kiQy Kkr dhft,% is 97 digit number divided by 16?
ty a
255504
13456789101112..... 97 vadksa dh la[;k 16 ls
foHkkftr gksus ij 'ks"kiQy Kkr dhft,\
di M
84
(a) 58 (b) 79 (a) 5 (b) 6
(c) 57 (d) 50 (c) 7 (d) 8
A
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ANSWER KEY
1.(a) 2.(a) 3.(b) 4.(b) 5.(d) 6.(c) 7.(c) 8.(c) 9.(c) 10.(a)
11.(a) 12.(c) 13.(a) 14.(d) 15.(c) 16.(b) 17.(a) 18.(b) 19.(c) 20.(b)
21.(a) 22.(d) 23.(d) 24.(c) 25.(d) 26.(a) 27.(a) 28.(b) 29.(b) 30.(a)
31.(c) 32.(a) 33.(d) 34.(d) 35.(c) 36.(b) 37.(a) 38.(a) 39.(b) 40.(a)
41.(d) 42.(b) 43. (a) 44. (b) 45.(b) 46.(c) 47.(a) 48.(a) 49.(b) 50.(c)
51.(d) 52.(b) 53.(d) 54.(a) 55.(c) 56.(c) 57.(a) 58.(c) 59.(b) 60.(d)
r
61.(a) 62.(a) 63.(a) 64.(c) 65.(a) 66.(a) 67.(b) 68.(a) 69.(b) 70.(a)
si
71.(a) 72.(a)an by
73.(b) 74.(d) 75.(c) 76.(b) 77.(d) 78.(d) 79.(c) 80.(d)
n
81.(b) 82.(d) 83.(b) 84.(d) 85.(b) 86.(c) 87.(d) 88.(c) 89.(a) 90.(d)
ja
R s
91.(c) 92.(d) 93.(a) 94.(d) 95.(d)
a th
ty a
di M
A
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SOLUTIONS
1. (a) 10. (a)
If the no. 1 2 3 4 5 6 7 8 9 is divided by 9
59637 (1)637
1 Remainder 1 2 3 4 5 6 7 8 9
58 58 So,
9
2. (a)
= 0 Remainder
132 132
(96) (–1) 11. (a)
= 1 Remainder
97 97 ATQ,
3. (b) 1919 20
= 1 + 2 = 3 Remainder
18
181753 (–1)753 –1
= 181 Remainder 12. (c)
182 182 182
(m12 – 112) divided by (m + 1)
r
4. (b) we know that, (an – bn) is always divisible by
si
(a – b) and (a + b) when n = even.
6867 – 1 0 – 1
67 Remainder So, = 0 Remainder
5.
68
(d)
68 an by 13. (a)
Given, (1213 + 2313)
n
153153 153 (–1)153 – 1 (1213 + 2313 )
154 154 ja 11
R s
–2 1+1 2
= =
a th
152 11 11
154
So, = 2 Remainder
6. (c)
14. (d)
ty a
8889 26 (–1)89 26 Let no. is k.
89 89 ATQ,
di M
= 25 Remainder k
4 Remainder
7. (c) 7
M k² (4)²
R5
= 2 Remainder
7 7 7
15. (c)
3M 15 Let no. be k
= 1 Remainder
7 7 ATQ,
8. (c) k
17 Remainder
17 363
x 1
A
x 1
k 17
= 6 Remainder
11 11
Remainder = 0
Because (xn + 1n) is always divisible by x +1 (n = odd) 16. (b)
ATQ,
9. (c)
Dividend = Divisor × Quotient + remainder n
5 Remainder
N = 78 × 280 + 0 6
78 280 9n 9×5
Now, R=0 = 3 Remainder
65 6 6
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17. (a) 24. (c)
n = 624d + 53
3147 4347
n 624d 53 37
=
16 16
We know that an + bn is always divisible by
624d 48 5 (a + b) [when n = odd]
= = 5 Remainder So, (a + b) = (31 + 43) = 74
16
18. (b) 74 is also a multiple of 37
So, = 0 Remainder
n
= 2 Remainder 25. (d)
3
ATQ,
7n
= 2 Remainder 8
224 – 1 2 – 1 1 – 1
3
3
= = = 0 Remainder
19. (c) 7 7 7
2787 2345 1992 26. (a)
r
23 341218 – 156218
si
(4)87 0 (– 4)92 259
So, = 0 Remainder (an – bn) is divisible by (a – b) & (a + b) if n is
20. (b)
23an by even
(341 – 156) = 185 = 5 × 37
n
16³ 17³ 18³ 19³ (341 + 156) = 497 = 7 × 71
70
ja also, 259 = 7 × 37
R s
(16 17 18 19)k 70k 341218 – 156218 is completely divisibe by 259
= =0
a th
70 70 0 Remainder
21. (a) 27. (a)
12345678910 ....... 2425 is divided by 4 (167)2n – (103)2n
ty a
25 (167 – 103) (167 + 103)
on dividing by 4 we take last 2 digit =
4 (64) (270) It has factors 144
di M
So, = 1 Remainder It is completely divisible by 144
22. (d) 28. (b)
68n + 1
516
= R1 = 1 (68 + 1) = 69 is exactly divisible 23
6
n should be odd
516 R an + bn is divisible by (a + b) if n is odd
= 2
6 29. (b)
–1
=5 218 – 1 (23 )6 – 1 –16 – 1
6
9 9 9
R1 R 2 1 5 6 = 0 Remainder
A
= =
R2 5 5 Remainder 30. (a)
23. (d)
341 782
Divisor Dividend Remainder
52
3 38 2
7 12 5 341 (7²)41
1 52
38 341 4941
= 17 Remainder So, = 0 Remainder
21 52
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31. (c) 38. (a)
2310 – 1024 137 71 77 93
=
7 15
210 – 2 (2³)³ 2 – 2 137 71 77 31
= = = 0 Remainder =
7 7 5
32. (a)
2 1 2 1
= 4 Remainder
3040 5
7 39. (b)
40 13
2 (2³) 2 1 2 132 135 136 138 142
= = = 2 Remainder
7 7 7 137
33. (d)
(–5) (–2) (–1) (1) (5)
21875 =
137
r
17
– 50
si
4 875 (4²)437 4 = = 87 Remainder
= = 137
17 17
=
(–1)437 4 – 4
=
an by
= 13 Remainder
40. (a)
n
273 375 478 657 597
17 17
25
34. (d)
ja =–2+0+3+7–3
R s
2727 27 = 5 Remainder
a th
28 41. (d)
=–1–1
335 608 853
= – 2 + 28 = 26 Remainder
13
ty a
35. (c)
ATQ, 10 10 8
=
di M
13
742 (7²)21
=
48 48 20 40
= = 7 × 1 = 7 Remainder
13
(49)21
= 1 Remainder 42. (b)
48
36. (b) 73 75 78 57 197
34
77 85 73
9 5 7 10 23 – 7 38
= = = 4 Remainder
34 34
5 4 1 20
= = = 2 Remainder
43. (a)
A
9 9
37. (a) 6397 9398 7007 3111
100
8317 – 5217 8417 – 5317
62 6397 9398 7007 3111
=
8317 – 5217 is divisible by 83 – 52 = 31 100 100 100 100
Also, 8417 – 5317 is divisible by 84 – 53 = 31 Remainder = (– 3) (– 2) (7) (11)
31 31 6 77
0 Remainder = = 62 Remainder
62 100
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44. (b) 49. (b)
816 825 823 827 Consider,
819 L.C.M (4, 9, 12, 15) = 180
–3 6 4 8 Least such number is 183.
=
819 50. (c)
–576 We know,
= = 243 Remainder Divisor = First Remainder + Second Remainder
819
45. (b) – Remainder of sum of both number
= (46 + 31 – 19) = 58
962 – 11 (92 )31 – 11 131 – 3
51. (d)
8 8 8
–2 101 10² 10³ ........ 1099 10100
= 6 Remainder 6
8
46. (c) We know,
Ist Case 10n ÷ 6
r
Divisor Dividend Re mainder then, the remainder = 4
ATQ,
si
4 37 1
5 9 4 4 100
an by
1
52. (b)
6
= 4 Remainder
n
2nd Case
Divisor Dividend Remainder 77 777 .....7777777777
5× 37 ja +2 6
R s
4× 7 +3 We know,
a th
1 7n ÷ 6
47. (a) then, the remainder = 1
ATQ,
Divisor Dividend Remainder
ty a
6 129 3 1 9
=
5 21 1 6
di M
7 4 4 So, = 3 Remainder
0 53. (d)
Observed that in the series 5! onwards every
129 number is divisible by 5 i.e. the remainder in
= = 9 Remainder each case is 0.
15
48. (a) So, the required remainder is obtained by
dividing only the first 4 number i.e.
Ist Case
1! 2! 3! 4 ! 1 2 6 24 33
Divisor Dividend Remainder = =
5 5 5
7 166 5
= 3 Remainder
4 23 3
A
54. (a)
3 5 2
1 1! 2! 3! .........1000 !
8
Ind Case
Divisor Dividend Remainder 1 2 6 24 ..........
3 166 1 8
4 55 3 After 3! all No. divisible by 8
7 13 6 1 2 6
So, = = 1 Remainder
1 8
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55. (c) 59. (b)
1! 2! 3! ......... 100! N = 55³ + 17³ – 72³
18 Let a = 55, b = 17, c = – 72
then a + b + c = 0
1 2 6 24 120 720 ........
a³ + b³ + c³ = 3abc
18
N = – 3 × 55 × 17 × 72
All term after 120 divisible by 18
it is divisible by 3 and 17 both
153 60. (d)
So, = = 9 Remainder
18
3120 – 1024
56. (c) odd – even = odd number
51203 which is not divisible by even number
7 It is not divisible by 32
61. (a)
(49 2)203 (0 2)203
=
7 7 123 223 323 ... 7023
r
71
(2³)67 .2² 867.2²
si
= = 4 Remainder
7 7 (123 70 23 ) (2 23 69 23 ) ... (3523 36 23 )
=
57. (a)
N = 18d + 7
an by 71
= 0 + 0 + ... + 0 = 0 Remainder
n
N (18d 7) 62. (a)
= = n Remainder
12 12 17 27 37 ... 1007
ja
R s
(n) 202
25
Put d = 1 = 1
a th
12 17 1007 27 997 ... 507 517
202
43
d=2 =7
12 101x 101y ... 101
ty a
=
202
61
d=3 = 1 (now remainder are repeating)
di M
12 50 101 a 202 b
=
So, n Take only 2 values. 202 202
58. (c) 0 Remainder
N = 33d + 4 63. (a)
N 33d 4 91 92 93 ... 99
=
55 55 1261
Put d = 1
91 92 93 ... 99
37 =
= 37 (but not in option) 13 97
55 = 0 × 92 × 93 × ... × 96 × 98 × 99
70 0 Remainder
A
d = 2, = 15 (but not in option)
55 64. (c)
103 1! 2! 3! 4 ! ... 10000 !
d = 3, = 48 (but not in option)
55 18
136 1 2 6 24 120 0 0 0... 0
d = 4, = (26) not in option
55 18
169 153
d = 5, = 4 Remainder = 9 Remainder
55 18
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65. (a) 71. (a)
ATQ,
361284
Required prime numbers
5
11, 13, 17, 19.......
The unit digit of 361284 = 1
When we devide these prime numbers by 6 we
Remainder = 1 = x get the remainder 1 or 5
We know, 72. (a)
4n ÷ 6 4 Remainder ATQ,
496 119 37
+10337 – 8037 + 6837
Now, = Remainder = 4 = y
6 37
2×x×y=2×1×4=8 We know a + bn always divisible by a + b
n
66. (a) here n is odd
2763 2763 2763 then,
192 3 64 3 26 222 & 148 are multiple of 37
r
So, Remainder = 0
2757 (–1)757 73. (b)
si
= 2 Remainder
3 3 Let the numbers x, y
67. (b) an by ATQ,
101 102 103 104 ... 10100 x – y = 1280 ......(i)
n
x = 7y + 50 ......(ii)
6
From equ. (i) & (ii)
ja
4 4 4 ... 4 y = 205
R s
6 x = 1485
a th
400 74. (d)
= 4 Remainder Let, the number is N
6
ATQ,
68. (a)
N = 442 × Q + 49
ty a
7 77 777 ... 100 terms
N 442 Q 49
8
di M
39 39
7 5 1 1 ... 1 7 5 98 110 39 is not the multiple of 442
=
8 8 8 So, answer can not be determined
= 6 Remainder 75. (c)
69. (b) 201 + 202 + 203 + 204 +....... + 260 = x
72727272... a = 201, n = 60
999 terms d = 202 – 201 = 1
101
7272 ÷ 101 = 0 Remainder n
Sum of n terms Sn [2a (n –1)d]
and 2
727 ÷ 101 = 20 Remainder a = 1st term, n = no. of terms
A
Upto 996 digits it is completely divided by 101, d = difference
for last 3 digits, remainder = 20
70. (a) 60
Sum of x term = [2 × 201 + 59 × 1]
ATQ, 2
L.C.M of 5, 8, 9 = 360 = 30[402 + 59]
360K + 3 = 30 × 461 = 13830
11 Sum of digit = 1 + 3 + 8 + 3 + 0 = 15
Let, K = 1 15
now] = 6 Remainder
Required number = 363 9
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76. (b) 79. (c)
2 10, 11, 20, 22 ATQ,
11 5, 11, 10, 11 21k + 5
2
5 5, 1, 10, 1 5x + 2x – 1
1, 1, 2, 1 5× 52 + 2 × 5 – 1
13
L.C.M = 2 × 11 × 5 × 2 = 220
Smallest no. of 4 digits = 1000 125 +10 – 1
13
220) 1000 (4
880 8 + 10 – 1 17
=
120 13 13
220 – 120 = 100 = 4 Remainder
Now] 1000 + 100 = 1100 80. (d)
r
Since 8 remains Let no. = 10a + b
ATQ,
si
(1100 + 8) = 1108
Required no. = 1108 1
77. (d)
an by (a b) =
7
(10a b)
n
3a = 6b
0! +1! + 2! + ...... + 98!
5 a 2
ja =
R s
4! = 4 × 3 × 2 × 1 = 24 b 1
2x – x = 4
a th
5! = 5 × 4 × 3 × 2 × 1
x=4
5 × 4 × 3 × 2 ×1
= a = 8, b = 4
5
ty a
Required no. = 10 × 8 + 4 = 84
Since we divide this series by 5, after 4! we
get the remainder 0 After reversing digit number is 48
di M
1+1+ 2 + 6 + 24 + 0 + 0.... 48
So, = 6 Remainder
5 7
81. (b)
34
= = 4 Remainder Let N be 3 digit number and r be remainder
5
So, (573921 – r) and (575713 – r) will be
78. (d) divisible by it means their difference is also
ATQ, divisible by N.
8k + 3 Difference = (575713 – r – 573921 + r) = 1792
1792 = 256 × 7
x 4 + x 3 + x 2 + x +1
A
Remainder So, N = 256
8
82. (d)
(3)4 + (3)3 + (3)2 + 3 +1 N = (24³ + 27³) + (25³ + 26³)
Remainder
8
= (51) (24² + 27² – 24 × 27) + (51) (25² + 26² – 25 × 26)
81+ 27 + 9 + 3 +1 = 51 [24² + 27² – 24 × 27 + 25² + 26² – 25 × 26]
Remainder
8 Divisor = 102 = 51 × 2
1 + 3 + 1 + 3 + 1 N is multiple of divisor
1 Remainder So, = 0 Remainder
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83. (b) 87. (d)
2929
29
229
29
15 25 35 4 5 55 ......505
= 5
9 9
we can write this series as
229 (1 + 2 + 3 + 4 + .......... + 50)
We solve in first step
9 5
26×4 × 25 32 Sum of series
= = =5
9 9 50 51
=
2
529
In 2nd step we solve 50 51
9
So, = 0 Remainder
25
53×8 × 55 1 × 55 53 × 52 –25 –7 88. (c)
= = = = =
9 9 9 9 9 (a4 – b4) = (a2 + b2) (a2 – b2)
r
= 2 Remainder = (a2 + b2) (a – b) (a + b)
si
84. (d) If a and b are odd
a – b and a + b are even
25102
17
an by also, a2 and b2 are odd
a2 + b2 is even
n
a4 – b4 is divisible by atleast 23 = 8
25102 (25³)34 (–2)34
= = 89. (a)
17 17 17ja
R s
1 2 3 30
1010 1010 1010 ... 1010
a th
4 8
(–2 ) (–2)² 1 4 7
= =
17 17
1010 310 (32 )5 25 23 22 4
= 4 Remainder
7 7 7 7 7 7
ty a
85. (b)
= 4R
ATQ,
di M
10100 3100 (32 )50 250 248.22
21040 7 7 7 7 7
131
(23 )16 22
4R
By eular theorem 7
30
[2(131–1) ]8 1010
= So, = 1 Remainder 4R
131 7
86. (c) 4 30 120
= 1 Remainder
7 7
1
2x – 1 = 0 x = 90. (d)
2
A
99132 99 99131
4 2 =
1 1 1 171 171
= 3
– 2
4 –1
2 2 2
11 (95 4)131
=
3 1 19
= – 2 –1
16 2
11 4131
=
3 – 8 32 – 16 11 19
= = Remainder
16 16 By eular theorem
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11 418 7 45 93. (a)
19 7a2 + 7a
11 1024 187 = 7a (a + 1)
=
19 19 if a is odd
Remainder = 16 × 9 = 144 a + 1 is even
91. (c) divisible by 7 and 14 both
504 504
255 (12 21 3) If a is even
=
84 84 7 × 2 (2 + 1)
504
(84 3 3) 3504 3503 14(3)
= = =
84 84 28 divisible by both 7 and 14
(28 – 1)167 3² –1 9 94. (d)
= = = 19
28 28 HCF (6, 8, 9) = 1
r
Remainder = 19 × 3 = 57 LCM (6, 8, 9) = 72
92. (d)
si
Required number is of the form:-
n2 (n2 – 1) (n2 – n – 2), n > 2
72k + 1
for n = 3
an by
n2 (n – 1) (n + 1) (n + 1) (n – 2)
from given options:-
n
for k = 6
9 (2) (4) (4) (1) 16 × 9 × 2
for n = 4 72 × 6 + 1 = 433
ja
R s
16 (3) (5) (5) (2) 95. (d)
6 × 16 × 25 Required series
a th
for n = 5 1234.......5354
25 (4) (6) (6) (3) 16
= 25 × 16 × 27
ty a
5354
k = 48 is such number Required remainder = =8
16
di M
A
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FACTOR (xq.ku[kaM
)
(CLASSROOM SHEET)
1. Find the total number of factor of 1470. 9. The sum of odd divisors of 158760 is :
1470 ds dqy xq.ku[k.Mksa dh la[;k Kkr djsaA 158760 ds fo"ke xq.ku[kaMksa dk ;ksxiQy Kkr djsaA
(a) 20 (b) 18 (a) 41211 (b) 618165
(c) 24 (d) 22 (c) 576945 (d) None of these
2. Find the total number of factor of 3600. 10. Consider the following statements in respect
of all factors of 360:
3600 ds dqy xq.ku[k.Mksa dh la[;k Kkr djsaA
360 ds lHkh xq.ku[kaMksa ds lanHkZ esa fuEufyf[kr dFkuksa i
SSC CGL TIER-II 12 /09/2019 fopkj dhft,%
r
(a) 45 (b) 44 1. The number of factors is 24.
si
(c) 43 (d) 42 xq.ku[kaMksa dh la[;k 24 gSA
3. Find the proper factors of 2940. 2. The sum of all factors is 1170.
a n by
2940 dk leqfpr xq.ku[kaMksa dh la[;k Kkr djsaA lHkh xq.ku[kaMksa dk ;ksxiQy 1170 gSA
n
(a) 34 (b) 36 [CDS - 2023 (I)]
(c) 24 (d) 32 Which of the above statements is/are correct?
mijksÙkQ esa ls dkSu lk@ls dFku lgh gS@gSa\
4.
ja
If N = 411 + 412 + 413 + 414, then how many
R s
positive factors of N are there? (a) 1 only (b) 2 only
;fn N = 411 + 412 + 413 + 414
a th
(c) Both 1 and 2 (d) Neither 1 nor 2
11. The sum of even divisors of 4096 is :
rksN ds dqy fdrus /ukRed xq.ku[kaM gksaxs\
4096 ds Lke xq.ku[kaMksa dk ;ksxiQy Kkr djsaA
(a) 92 (b) 48
ty a
(a) 6144 (b) 8190
(c) 50 (d) 51
4 6
(c) 8192 (d) 6142
5. If N = 4 + 6 , then find the number of factors
di M
12. The sum of even factors of 1800 is :
of N.
1800 ds Lke xq.ku[kaMksa dk ;ksxiQy Kkr djsaA
;fn N = 44 + 66 gS] rks
N ds xq.ku[kaMksa dh la[;k Kkr
(a) 403 (b) 6045
djsaA (c) 6448 (d) 5642
(a) 28 (b) 56 13. Consider the number N = 126 × 38 × 53.
(c) 14 (d) 7 la[;k N = 126 × 38 × 53 ij fopkj djsaA
6. Find the number of odd factors of 7200. Which of the following statements is/are
7200 ds fo"ke xq.ku[kaMksa dh la[;k Kkr djsaA correct?
(a) 4 (b) 9 fuEufyf•r esa ls dkSu lk@ls dFku lgh gS@gSa\
(c) 54 (d) 45 1. The number of odd factors of N is 60.
N ds fo"ke xq.ku•.Mksa dh la[;k 60 gSA
A
7. Find the number of even factors of 10500.
2. The number of even factors of N is 720.
10500 ds le xq.ku[kaMksa dh la[;k Kkr djsaA
N ds le xq.ku•aMksa dh la[;k 720 gSA
(a) 48 (b) 16
Select the correct answer using the code given
(c) 32 (d) 46 below :
8. The sum of all factors of 19600 is : uhps fn, x, dksM dk mi;ksx djds lgh mÙkj pqusa%
19600 ds lHkh xq.ku[kaMksa dk ;ksxiQy Kkr djsaA [CDS - 2022 (II)]
(a) 5428 (b) 54777 (a) Only 1 (b) Only 2
(c) 33667 (d) None of these (c) Both 1 and 2 (d) Neither 1 nor 2
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14. What is the sum of reciprocal of all factors of (a) 29, 17, 37 (b) 31, 17, 47
number 360. (c) 19, 37, 13 (d) 23, 31, 37
la[;k 360 ds LkHkh xq.ku[kaM ds O;qRØeksa dk ;ksx
20.D;kThe
gS\ sum of divisiors of 10800 which are
(a) 2.65 (b) 3.25 perfect square.
(c) 3.48 (d) 4.20 10800 ds mu xq.ku[kaMksa dk ;ksx Kkr dhft, tks iw.kZ oxZ gSA
15. Find the product of all factors of 544? (a) 393120 (b) 6240
544 ds lHkh xq.ku[kaMksa dk xq.kuiQy Kkr djsaA (c) 5200 (d) 5460
3
21. The number of factor of 7200 divisible by 40.
(a) 12 (b) 544 2
7200 ds 40 ls foHkkT; xq.ku[kaMksa dh la[;k gS %
6
(c) 558 (d) 544 (a) 18 (b) 54
16. Find the product of all factors of 1800? (c) 9 (d) 20
1800 ds lHkh xq.ku[kaMksa dk xq.kuiQy Kkr djsaA 22. Find the ways to express 240 as product of two
(a) 180018 (b) 18008 factors.
(c) 90018 (d) 9008 240 dks nks xq.k[k.Mksa ds xq.kuiQy ds :i esa O;Dr djus
r
17. If 847 × 385 × 675 × 3025 = 3a × 5b × 7c × 11d, ds rjhdksa dh la[;k Kkr djsaA
si
then the value of ab – cd is: (a) 10 (b) 11
;fn 847 × 385 × 675 × 3025 = 3a × 5b × 7c × 11d rksab (c) 64 (d) 20
a n by
– cd dk eku D;k gksxk%
SSC CGL MAINS 29/01/2022
23. Find the ways to express 11025 as product of
two factors.
n
(a) 4 (b) 5 11025 dks nks xq.k[k.Mksa ds xq.kuiQy ds :i esa O;Dr djus
(c) 1 (d) 7 ds rjhdksa dh la[;k Kkr djsaA
ja
R s
18. Find the number of prime factors in the (a) 13 (b) 14
product (30)5 × (24)5.
a th
(c) 27 (d) 30
(30)5 × (24)5 ds vHkkT; xq.ku[kaMksa dh la[;k Kkr djsaA
24. The number of divisors of the number 38808,
SSC CGL TIER-II (18/11/2020) exclusive of the divisors 1 and itself, is
(a) 45 (b) 35 la[;k 38808 ds Hkktdksa dh la[;k] foHkktdksa 1 vkSj [kqn
ty a
(c) 10 (d) 30 bl la[;k dks NksM+dj] fdruh gS\
19. What are distinct prime factors of the number
di M
[CDS - 2018 (II)]
26381?
(a) 74 (b) 72
la[;k 26381 ds fof'k"V vHkkT; xq.ku•aM D;k gSa\
(c) 70 (d) 68
[CDS - 2021 (II)]
ANSWER KEY
1.(c) 2.(a) 3.(a) 4.(a) 5.(c) 6.(b) 7.(c) 8.(b) 9.(a) 10.(c)
11.(b) 12.(d) 13.(c) 14.(b) 15.(d) 16.(a) 17.(b) 18.(b) 19.(d) 20.(d)
A
21.(a) 22.(a) 23.(b) 24.(c)
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Factor/xq.ku[kaM
( Practice Sheet With Solution)
Level-01 9. Find the number of prime factors of
237× 353 × 510
1. Find the number of factors in 23 × 34 × 56
237 × 353 × 510 esa vHkkT; xq.ku[k.Mksa dh la[;k Kkr dhft,A
23 × 3 4 × 56 e sa xq.ku•.Mksa dh la[;k Kkr dhft,
(a) 95 (b) 96
(a) 120 (b) 125
(c) 99 (d) 100
(c) 130 (d) 140
10. Find the number of even factors of 842 .
2. Find the number of factors in 720
720 esa xq.ku[k.Mksa dh la[;k Kkr dhft,A 842 ds le xq.ku[k.Mksa dh la[;k Kkr djsa\
r
(a) 2 (b) 4
(a) 25 (b) 30
(c) 6 (d) 8
si
(c) 35 (d) 40
3. a and b are two positive integers such that the 11. Find the highest power of 3 that completely
an by
least prime factor of a is 3 and least prime factor
of b is 5. The least prime factor of a + b is.
divides 63!
3 dh mPpre ?kkr dks Kkr djsa ftlls63! dks iwjh rjg
n
a vkSjb nks ldkjkRed iw.kkZad bl izdkj gSA a dkfd ls foHkkftr fd;k tk ldsA
dk lcls NksVk xq.ku[k.M
lcls NksVk xq.ku[k.M 3 gSbvkSj UPSI 14/11/2021 (Shift-02)
5 gS] rks ja
a + b d k lcls NksVk xq.ku[k.M D;k gSA
R s
(a) 26 (b) 24
(a) 2 (b) 3 (c) 28 (d) 30
a th
(c) 5 (d) 8 12. Find the highest power of 7 that completely
4. If product of two prime numbers A and B (A < B) divides 84!
is 221, then what is the value of (4A – 3B)?
7 d h mPpre ?kkr Kkr djsa 84!
tks d ks iwjh rjg ls foHkkftr
; fn nks vHkkT; la[;kvksa
A vkSjB ( A < B) dk xq.kuiQy
ty a
djrh gSA
221 gS] rks4A
( – 3B) dk eku D;k gS\
UPSI 21/11/2021 (Shift-02)
di M
(a) – 1 (b) 2
(a) 13 (b) 11
(c) 1 (d) – 2
(c) 7 (d) 9
5. Find the number of even factors in 34 × 78
13. Find the number of multiples of 11025.
34 × 78 e sa le xq.ku•.Mksa dh la[;k Kkr dhft,
11025 ds xq.kdksa dh la[;k Kkr djsaA
(a) 45 (b) 32
(c) 10 (d) 0 UPSI 13/12/2017 (Shift-01)
6. Find the sum of the factors of 720. (a) 31 (b) 20
720 ds xq.ku•.Mksa dk ;ksx Kkr dhft,A (c) 23 (d) 27
(a) 2018 (b) 2418 Level-02
(c) 2618 (d) 2518 14. Find the number of even factors 2³ × 34 × 56
A
7. Find the sum of factors of 2³ × 3² × 51 2³ × 34 × 56 l e xq.ku•.Mksa dh la[;k Kkr dhft,
2³ × 3² × 5 1
d s xq.ku•.Mksa dk ;ksx Kkr dhft,A (a) 105 (b) 104
(a) 1170 (b) 1070 (c) 110 (d) 115
(c) 950 (d) 1150 15. How many factors of N = 25 × 38 × 56 are perfect
8. Find the number of odd factors of 23 × 34 × 57. squares?
23 × 34 × 57 esa fo"ke&xq.ku[k.Mksa dh la[;k Kkr dhft,A N = 25 × 38 × 56 d s fdrus xq.ku[k.M iw.kZ oxZ gSa\
(a) 40 (b) 42 (a) 50 (b) 60
(c) 45 (d) 50 (c) 70 (d) 80
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16. How many factors of N = 25 × 38 × 56 are perfect (a) 4 (b) 6
cubes? (c) 5 (d) 7
N = 25 × 38 × 56 ds fdrus xq.ku[k.M iw.kZ ?ku gSa\ 26. How many factors of 25 × 36 × 52 are perfect
squares?
(a) 16 (b) 17
25 × 36 × 52 ds fdrus xq.ku[akM iw.kZ oxZ gSa\
(c) 18 (d) 19
(a) 18 (b) 24
17. Find the sum of even factors of 2³ × 32 × 51
(c) 36 (d) 8
2³ × 32 × 51 ds le xq.ku•.Mksa dk ;ksx Kkr dhft,A
(a) 1080 (b) 1092
Level-03
27. Find the number of factors in 1540.
(c) 1093 (d) 1090
18. Find the sum of even factors in 35 × 56 1540 esa xq.ku•aMksa dh la[;k Kkr dhft,A
35 × 56 esa le xq.ku•.Mksa dk ;ksx Kkr dhft,A UPSI 17/11/2021 (Shift-01)
(a) 512 (b) 509 (a) 20 (b) 24
(c) 0 (d) 505 (c) 22 (d) 23
19. Find the sum of odd factors 24 × 33 × 51 28. Find the number of factors of 60.
r
24 × 33 × 51 esa fo"ke xq.ku•.Mksa dk ;ksx Kkr dhft,A 60 ds xq.ku[kaMksa dh la[;k Kkr djsaA
si
(a) 120 (b) 240 UPSI 13/12/2017 (Shift-03)
(c) 360 (d) 480 (a) 14 (b) 10
20.
an by
Find the product of all factors of 200.
200 ds lHkh xq.ku•aMksa dk xq.kuiQy Kkr dhft,A 29.
(c) 15 (d) 12
Given two prime no's P and Q. What could be
n
the minimum no. of factors of P*Q?
(a) 2006 (b) 20000
nks vHkkT; la[;k,
P,Q gSA
P*Q ds U;wUre fdrus xq.ku[akM
(c) 10 4
ja (d) 105
gks ldrs gS\
R s
21. Find the sum of all even factors of 25 × 3² × 51.
(a) 2 (b) 4
esa le&xq.ku[k.Mksa dk ;ksxiQy Kkr dhft,A (c) 3
a th
25 × 3² × 51 (d) 5
(a) 4536 (b) 4836 30. The total no. of 2 digit no's. Which have only
(c) 4260 (d) 4160 3 factores will be:
2 vadksa dh fdruh la[;k, gS ftuds dsoy 3 xq.ku[kaM gS\
ty a
22. Find the sum of all odd factors of 28 × 31 × 53
2 × 3 × 5 esafo"ke&xq.ku[k.Mksa dk ;ksxiQy Kkr dhft,A(a) 3
8 1 3 (b) 2
di M
(a) 624 (b) 660 (c) 5 (d) 4
9
31. If N = 9 , then N is the divisible by how many
(c) 690 (d) 700
positive perfect cubes?
23. Find how many factors of x are there which are
multiple of 20.
;fn N = 99 gS] rks
N fdrus /UkkRed iw.kZ ?kuksa ls foHkkT;
(a) 6 (b) 7
x = 25 × 32 × 53
(c) 4 (d) 5
x = 25 × 32 × 53 esax ds fdrus xq.ku[k.M gSa tks 20 ds
32. If a three digit number 'abc' has 3 factors, how
xq.kt gSa\ many factors does the 6-digit number 'abcabc'
(a) 26 (b) 30 have?
(c) 36 (d) 40 ;fn rhu vadksa dh la[;k'abc' esa 3 xq.ku[akM gSa] rks 6
24. What is the ratio of the total number of factors vadksa dh la[;k ds fdrus xq.ku[kaM gksaxs\
of 216 and 180?
A
(a) 16 (b) 24
216 vkSj 180 ds xq.kuiQyksa dh dqy la[;k dk vuqikr (c) 16 or 24 (d) 20
D;k gS\ 33. If a certain number has 388 factors excluding
1 and itself then maximum number of prime
(a) 2 : 3 (b) 8 : 9
factors it can have?
(c) 5 : 6 (d) 4 : 5
;fn ,d fuf'Pkr la[;k esa 1 vkSj Lo;a dks NksM+dj
25. If the total number of positive factors of 1440
388 xq.ku[kaM gSa rks mlds vf/dre vHkkT; xq.ku[ak
is (x + 14), then, what is the value of x 3 ?
gks ldrs gS
a\
;fn 1440 ds /ukRed xq.ku[akMksa dh dqy la[;k
(x + gS
14) (a) 4 (b) 5
rks x 3 dk eku D;k gS\ (c) 3 (d) None
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34. If a number N has 8 factors then find the 35. N is a two digit number, what is the maximum
minimum value of N? value of number of factors of N?
;fn fdlh la[;k N ds 8 xq.ku[akM gSa
N rks
dk U;wure nks vadksa dh la[;k N
gS]ds xq.ku[akMksa dh la[;k dk
eku Kkr dhft,\ vf/dre Ekku D;k gS\
(a) 15 (b) 45 (a) 12 (b) 14
(c) 30 (d) 55 (c) 15 (d) 8
ANSWER KEY
1.(d) 2.(b) 3.(a) 4.(c) 5.(d) 6.(b) 7.(a) 8.(a) 9.(d) 10.(a)
11.(d) 12.(a) 13.(d) 14.(a) 15.(b) 16.(c) 17.(b) 18.(c) 19.(b) 20.(a)
21.(b) 22.(a) 23.(c) 24.(b) 25.(c) 26.(b) 27.(b) 28.(d) 29.(b) 30.(b)
r
si
31.(b) 32.(c) 33.(a) 34.(c) 35.(a)
an by
n
ja
R s
a th
ty a
di M
A
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SOLUTIONS
1. (d) 2³ × 34 × 56 12. (a) 84! The number of 7's in a number will be
No. of factors = (3 + 1) (4 + 1) (6 + 1) the highest power of 7.
= 4 × 5 × 7 = 140 7 84
2. (b) 720 = 24 × 32 × 51 7 12
No. of factors = (4 + 1) × (2 + 1) × (1 + 1) 1
= 5 × 3 × 2 = 30
The highest power of 7 = (12 + 1) = 13
3. (a) Given, Least factor of (a) = 3
and least factor of (b) = 5 13. (d) 11025 = 32 × 52 × 72
So, (a + b) = 8 Least factor of 8 = 2 No. of factors
4. (c) Given, = (2 + 1) × (2 + 1) × (2 + 1) = 27
A × B = 221 = 3 × 3 × 3 = 27
14. (a) 2³ × 34 × 56
r
Total No. of factors = (3 + 1) × (4 + 1) × (6 + 1)
si
= 4 × 5 × 7 = 140
17 13 No. of odd factors = (4 + 1) × (6 + 1) = 35
(B)
an by
(A)
4A – 3B = 4 × 13 – 3 × 17
No. of even factors = 140 – 35 = 105
'OR'
n
= 52 – 51 = 1 For No. of even factors = 2 × [2² × 34 × 56]
5. (d) 34 × 78 (2 + 1) × (4 + 1) × (6 + 1) = 105
ja
R s
Here No. of 2 = 0 15. (b) N = 25 × 38 × 56
So, No. of even factor = 0 2 × 24 × (3²)4 × (5²)³
a th
6. (b) 720 = 24 × 3² × 51 2 × [4² × 94 × 25³]
Sum of factors No. of perfect square factors
= (20 + 21 + 2² + 2³ + 24) × (30 + 31 + 3²) × (50 + 51) = (2 + 1) × (4 + 1) × (3 + 1)
ty a
= 31 × 13 × 6 = 3 × 5 × 4 = 60
= 186 × 13 = 2418 16. (c) N = 25 × 38 × 56
di M
7. (a) 2³ × 3² × 51 = 2² × 2³ × 3² × (3³)² × (5³)²
Sum of factors
= 4 × 9 [81 × 27² × 125²]
= (20 + 21 + 2² + 2³) × (30 + 31 + 3²) × (50 + 51)
No. of perfect cube factors
= 15 × 13 × 6
= (1 + 1) × (2 + 1) (2 + 1)
= 15 × 78 = 1170
= 2 × 3 × 3 = 18
8. (a) 2³ × 34 × 57
17. (b) 2³ × 3² × 51
No. of odd factors = ( 4 + 1) × (7 + 1)
= 5 × 8 = 40 Sum of factors
9. (d) 237 × 353 × 510 (20 + 21 + 22 + 23) × (30 + 31 + 32) × (50 + 51)
No. of prime factors = (37 + 53 + 10) = 100 = 15 × 13 × 6 = 1170
10. (a) 842 = 4211 × 21 Sum of odd factors = 78
A
No. of even factor 1 × 2 = 2 Sum of even factors = (1170 – 78) = 1092
18. (c) 35 × 56
11. (d) 3 63
Here No. of 2 = 0
3 21
So, No. of even factors = 0
3 7
So, sum of even factors = 0
2
19. (b) 24 × 3³ × 51
The more times 3 occurs, the higher its power Sum of odd factors
will be in the division. = (30 + 31 + 3² + 3³) × (50 + 51)
The highest power of 3 = (21 + 7 + 2) = 30 = 40 × 6 = 240
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20. (a) N = 200 = 2³ × 5² 29. (b) Let, 21 × 31
No. of factors = (3 + 1) × (2 + 1) = 12 No. of factors (1 + 1) × (1 + 1) = 4
12/2 6
Product of factors = (N) = 200 30. (b) 3 factor's means, that no. is perfect square
21. (b) 25 × 3² × 51 (Prime no.)
Sum of factors 52 & 72
(20 + 21 + 22 + 23 +24 + 25) × (30 + 31 + 32) × (50 + 51) 25 & 49
= 63 × 13 × 6 = 4914 31. (b)
Sum of odd factors = 78 N = 99 = (32)9
Sum of even factors = (4914 – 78) = 4836 = 318
22. (a) 28 × 31 × 5³ = (33)6
Sum of odd factors Factor's = (6 + 1) = 7
0 1 0 1
= (3 + 3 ) × (5 + 5 + 5² + 5³) 32. (c) abc in factors 3(means prime no. square)
= 4 × 156 = 624 abcabc = abc(1001)
23. (c) 25 × 3² × 5³
r
= abc (7, 11, 13)
2² × 5 [2³ × 3² × 5²] Let, prime no. x
si
No. of factors which are multiple of 20 Factors x2 × 71 × 111 × 131
= (3 + 1) (2 + 1) (2 + 1) = 36
an by (2 + 1) × (1 + 1) × (1 + 1) × (1 + 1)
24. (b) 216 63 = 23 × 33 = factor = (3 + 1) (3 + 1) 3 × 2 × 2 × 2 = 24
n
Factor = 4× 4 = 16 If x = 11, 13 then
180 9 × 20 = 22 × 32 × 51 112 × 71 × 111 × 131
ja
No. of factor's (2 + 1) × (2 + 1) × (1 + 1)
R s
113 × 71 × 131
= 3 × 3 × 2 = 18 Factors (3 + 1) × (1 + 1) × (1 + 1)
a th
Ratio of factor's = 16 : 18 4 × 2 × 2 = 16
=8 : 9
33. (a) ap × bq × cr
25. (c) 1440 25 × 32 × 51
(p + 1) (q + 1) (r + 1)......= 388 + 2
ty a
Factors (5 + 1) × (2 + 1) × (1 + 1)
(p + 1) (q + 1) (r + 1).....= 390
6×3×2
di M
= 13 × 3 × 2 × 5
x + 14 = 36
ap × bq × cr = a12 × b2 × c1 × d4
x = 22
Hence,
x +3 22 3 4 factors prime
34. (c) N = ap × bq × cr....
25 5
(p + 1) (q + 1) (r + 1).......= 8
26. (b) 25 × 36 × 52
=2×2×2
Factor's (22)2 × (32)3 × (52)1 × 2
N = a1 × b1 × c1 = 21 × 31 × 51
Factors of perfect sq. (2 + 1) × (3 + 1) × (1 + 1)
3×4×2 N = 30
Factors of perfect sq. = 24 35. (a) Let, Prime No. 2, 3, 5, 7
A
Ist = 21 × 31 × 51 = factors = (1 + 1)(1 + 1)(1 + 1) = 8
27. (b) 1540 = 22 × 51 × 71 × 111
= 30 × 2 = 60 = 31 × 22 × 51 factor
No. of factors
= (1 + 1)(2+1)(1 + 1) = 12
= (2 + 1) × (1 + 1) × (1 + 1) × (1 + 1)
IInd = 21 × 31 × 71 = factors = (1 + 1)(1 + 1)(1 + 1) = 8
= 3 × 2 × 2 × 2 = 24 = 42 × 2 = 84 = 22 × 31 × 71
28. (d) 60 = 22 × 31 × 51 Maximum factors = (2 + 1) (1 + 1) (1 + 1)
No. of factors =3×2×2
(2 + 1) × (1 + 1) × (1 + 1) = 3 × 2 × 2 = 12 Maximum factors = 12
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NO. OF ZEROS AT THE END OF AN EXPRESSION
(CLASSROOM SHEET)
1. The number of zeroes at the end of 8. D = 1 × 3 × 5 × 7 ×...........999
21 × 35 × 625 × 8 × 165 is : (a) 246 (b) 199
21 × 35 × 625 × 8 × 165 ds var esa 'kwU;ksa dh la[;k gS%(c) 99 (d) 0
9. E = 1 × 3 × 5 × 7 × 9 ×......× 99 × 128
(a) 1 (b) 3 (a) 10 (b) 12
(c) 5 (d) 7 (c) 7 (d) 19
2. The number of zeroes at the end of 10. H = 168! × 143!
25 × 35 × 40 × 50 × 60 × 65 is : (a) 40 (b) 34
25 × 35 × 40 × 50 × 60 × 65 ds var esa 'kwU;ksa dh (c) 74 (d) 1360
r
la[;k gS% 11. I = 2 × 4 × 6 × 8 ×..........1000
si
(a) 249 (b) 125
(a) 6 (b) 8
(c) 124 (d) 997
(c) 5 (d) 7
a n by
12. J = 5 × 10 × 15 × 20 × .......1000
3. The number of zeroes at the end of (a) 49 (b) 111
1003 × 1001 × 999 × ....... × 123 is : (c) 197 (d) 247
n
1003 × 1001 × 999 × ....... × 123 ds var esa 'kwU;ksa 13. K = 10100 + 1090 + 1080 + ......1020 + 1010
dh la[;k gS% (a) 10 (b) 100
ja
(c) 550 (d) 55
(a) 224
(c) 0
R s (b) 217
(d) 212
14. L = (5!)6!
(a) 120 (b) 720
a th
4. The number 1, 2, 3, 4, ........ 1000 are multipled
(c) 600 (d) 840
together. The number of zeros at the end (on
15. P = 11 × 22 × 33 × 44 × 55 ×......100100
the right) of the product must be: (a) 1050 (b) 1200
la[;kvksa 1] 2] 3] 4 -----------1000 dks ,d lkFk xquk fd;k tkrk (c) 1300
ty a
(d) 1450
gSA xq.kuiQy ds var (nkfguh vksj) esa 'kwU;ksa dh la[;k 16.gS%How many zeros are there in the product :
(a) 30 (b) 200 xq.kuiQy esa fdrus 'kwU; gSa %
di M
(c) 211 (d) 249 150 × 249 × 348 ×------ × 501
Direction: Find the number of zero's at the end [CDS – 2021 (II)]
of followings numbers or expression. (a) 262 (b) 261
funZs'k% fuEufyf•r la[;kvksa ;k O;atdksa ds var esa 'kwU;ksa dh
(c) 246 (d) 235
la[;k Kkr dhft,A 17. Q = 1001 × 992 × 983 ×........299 × 1100
5. A = 300! (a) 1124 (b) 1120
(c) 970 (d) 1121
(a) 74 (b) 72
18. What is the highest power of 4 in 100! ?
(c) 75 (d) 76
100! esa 4 dh mPpre ?kkr D;k gS\
6. B = 625!
(a) 32 (b) 48
(a) 125 (b) 150 (c) 49 (d) None of these
(c) 155 (d) 156 19. What is the highest power of 72 in 100! ?
A
7. C = 2502 × 5205 100! esa 72 dh mPpre ?kkr D;k gS\
(a) 502 (b) 205 (a) 20 (b) 21
(c) 707 (d) None of these (c) 19 (d) 24
ANSWER KEY
1.(b) 2.(a) 3.(c) 4.(d) 5.(a) 6.(d) 7.(b) 8.(d) 9.(c) 10.(c)
11.(c) 12.(c) 13.(a) 14.(b) 15.(c) 16.(a) 17.(a) 18.(b) 19.(d)
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Number of zeroes/'kwU;ksa dh la[;k
( Practice Sheet With Solution)
Level-01 8. Find the Number of zeros at the end of the
expression
1. Find the number of trailing zero's in 475!.
fuEufyf[kr O;atd ds var esa 'kwU;ksa dh la[;k Kkr dhft,A
475! esa vuqxkeh 'kwU;ksa dh la[;k Kkr dhft,A
10 + 100 + 1000 + .........10000000.
UPSI 22/11/2021 (Shift-02) (a) 1 (b) 10
(a) 137 (b) 127 (c) 55 (d) 10000000
(c) 117 (d) 147
Level-02
2. Find no. of zero (1 × 2 × 3 × .......108)
9. Find the number of zeros at the end of the
r
1 × 2 × 3 × .......×108 esa 'kwU;ksa dh la[;k Kkr dhft,A product of following expression.
si
(a) 25 (b) 30 fuEufyf
[kr O;atd ds xq.kuiQy ds var esa 'kwU;ksa dh la[;k
(c) 20 (d) 15 Kkr dhft,A
3.
an by
The numbers 1, 3, 5,....., 25 are multiplied
together. The number of zeros at the right end
241! × 25 × 24 × 35 × 12
(a) 61 (b) 58
n
of the product is: (c) 63 (d) 66
l a[;k 1] 3] 5]-------------] 25 dks ,d lkFk xq.kk fd;k tkrk
ja 10. Find the number of zeros at the end of the
product of expression
R s
gSA mRikn ds nkfgus Nksj ij 'kwU;ksa dh la[;k gSA
[kr O;atd ds xq.kuiQy ds var esa 'kwU;ksa dh la[;k
fuEufyf
a th
(a) 1 (b) 0
Kkr dhft,A
(c) 2 (d) 3 1532! × 481! × 34!
4. Find the number of trailing zeros in 15870! (a) 499 (b) 431
esa vuqxkeh 'kwU;ksa dh la[;k Kkr dhft,A
ty a
15870! (c) 506 (d) 388
(a) 3865 (b) 3665 11. Find the number of zeros at the end of the
di M
product of expression
(c) 3965 (d) 3765
fuEufyf
[kr O;atd ds xq.kuiQy ds var esa 'kwU;ksa dh la[;k
5. Find the number of trailing zeros in 15370!
Kkr dhft,A
15370! esa vuqxkeh 'kwU;ksa dh la[;k Kkr dhft,A 11 × 22× 33 × 44 × ..........× 4949
UPSI 14/11/2021 (Shift-03) (a) 100 (b) 150
(a) 3738 (b) 3538 (c) 200 (d) 250
(c) 3838 (d) 3638 12. Find the number of zeros at the end of the
product of expression
6. Find the number of trailing zeros in 15245!.
fuEufyf
[kr O;atd ds xq.kuiQy ds var esa 'kwU;ksa dh la[;k
15245! e sa vuqxkeh 'kwU;ksa dh la[;k Kkr dhft,A Kkr dhft,A
UPSI 17/11/2021 (Shift-01) 11! × 22! × 33! × 44! × ..........× 1010!
A
(a) 3807 (b) 3507 (a) 2! + 4! + 8! (b) 2! + 5!
(c) 3607 (d) 3707 (c) 5! + 10! (d) 2! + 4! + 8! + 10!
7. Find the number of consecutive zeroes at the 13. Find the number of zeros at the end of the
product of expression is :
end of 649!.
fuEufyf
[kr O;atd ds xq.kuiQy ds var esa 'kwU;ksa dh la[;k
649! d s var esa Øekxr 'kwU;ksa dh la[;k Kkr djsaA Kkr dhft,A
UPSI 24/11/2021 (Shift-01) 10 × 100 × 1000 × ......... × 10000000000
(a) 130 (b) 140 (a) 10 (b) 50
(c) 150 (d) 160 (c) 55 (d) 100
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14. Find the number of zeros at the end of the 21. Find no. of zero in (5 × 10 × 15....200)
following expression
5 × 10 × 15......200 esa 'kwU;ksa dh la[;k Kkr dhft,A
fuEufyf[kr O;atd ds var esa 'kwU;ksa dh la[;k Kkr dhft,A (a) 19 (b) 57
(5!)5! + (20!)20! + (50!)50! + (100!)100! (c) 38 (d) 76
(a) 120 (b) 165 22. Find no. of zero in (10 × 20 × 30 × .....500)
(c) 500 (d) 1025
10 × 20 × 30 ×......500 esa 'kwU;ksa dh la[;k KkrAdhft,
15. Find the maximum value of n such that 3500!
(a) 54 (b) 62
is perfectly divisible by 40n.
(c) 65 (d) 61
n dk og vf/dre eku Kkr dhft, ftlls 3500!, 40n
ls iw.kZr% foHkDr gks tk,A Level-03
(a) 874 (b) 1164 23. The number 1, 3, 5, 7.....99 and 50 128 are
multiplied together.The number of zeros at the
(c) 3493 (d) None of these
end of the product must be:
16. Find the highest power of 63 which can exactly
la[;k;as1, 3, 5, 7.....99 rFkk] 128 dks ijLij xq.kk
r
divide 5335!
dh og vf/dre ?kkr Kkr dhft, tks dks fd;k tkrk gS] rks xq.kuiQy ds var esa 'kwU;ksa dh la[
si
63 5335!
iw.kZr% foHkkftr djsA gksxhA
(a) 887
(c) 2662
an by (b) 1331
(d) None of these
(a) 19
(c) 7
(b) 22
(d) 0
n
17. Find the maximum value of n such that 50! is 24. Find no. of zero in (48! + 49!)
perfectly divisible by 12600n.
ja (48! + 49!) esa 'kwU; dh la[;k Kkr dhft,A
R s
n dk og vf/dre eku Kkr dhft, ftlls 50!, 12600 n
(a) 11 (b) 10
a th
ls iw.kZr% foHkDr gks tk,A (c) 9 (d) 12
(a) 5 (b) 6 25. Find no. of zero in (76! – 75!)
(c) 7 (d) 8 (76! – 75!) esa 'kwU; dh la[;k Kkr dht,A
ty a
18. Find the number of zeroes in n! where n is (a) 30 (b) 25
number between 66 to 69.
di M
(c) 20 (d) 15
n! esa 'kwU;ksa dh la[;k Kkr dhft,
¡n,tgk
66 vkSj69 d s
26. Find the number of zeros (3 29 – 3 28 – 3 27)
chp dh la[;k gSA 31 29 28
(7 –7 –7 )
(a) 12 (b) 15
(329 – 328 – 327)(731 – 729 – 728) esa 'kwU; dh la[;k
(c) 13 (d) 1
Kkr dhft,A
19. Find the number of zeroes in (a × b × c) ! where
(a) 1 (b) 2
b is 1 more than a & c is 2 more than a and a is
the pruduct of the first two positive prime (c) 5 (d) 0
numbers. 27. Find number of zero 1 × 22 × 33........8080
1
esa 'kwU;ksa dh la[;k Kkr dhft,
(a × b × c) ! ¡b,tgk
a ls 11 × 22 × 33........8080 esa 'kwU; dh la[;k Kkr dhft,A
1 vf/d gS vkSj c, a ls 2 vf/d gS vkSj a çFke nks
A
(a) 825 (b) 820
/ukRed vHkkT; la[;kvksa dk xq.kuiQy gSA
(c) 835 (d) 830
(a) 82 (b) 67
28. Which of the following cannot be the number
(c) 21 (d) 80 of zeros at the end of any factorial?
20. Find no. of zero in (2 × 4 × 6 × 8 ×.....170) fuEufyf[kr esa ls dkSu fdlh ! ds var esa 'kwU;ksa dh la[
2 × 4 × 6 × 8 ×.....170 esa 'kwU;ksa dh la[;k Kkr dhft, ugha gks ldrk gS\
(a) 10 (b) 15 (a) 24 (b) 27
(c) 25 (d) 20 (c) 29 (d) 31
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29. What will be the number of zero in the end of 30. If (n + 1)! has 2 more zeros at the end as
(45!)4!? compared to how many two digit values (n+1)
can be assume?
(45!)4! ds var esa 'kwU;ksa dh la[;k D;k gksxh\
(a) 10 vxj (n + 1)! esan! dh rqyuk esa var esa 2 vf/d 'kwU;
(b) 40! gSa] fdrus nks vadksa ds ekuds ekus tk ldrs gSa\
(n+1)
(c) 40 (a) 4 (b) 3
(d) 240 (c) 2 (d) None of these
ANSWER KEY
1.(c) 2.(a) 3.(b) 4.(c) 5.(c) 6.(a) 7.(d) 8.(a) 9.(a) 10.(c)
11.(d) 12.(c) 13.(c) 14.(a) 15.(a) 16.(a) 17.(d) 18.(b) 19.(a) 20.(d)
r
si
21.(c) 22.(b) 23.(c) 24.(d) 25.(c) 26.(d) 27.(d) 28.(c) 29.(d) 30.(b)
an by
n
ja
R s
a th
ty a
di M
A
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SOLUTIONS
1. (c) 5 475 7. (d) 5 649
5 95 5 129
5 19 5 25
3 5 5
No. of zeros 1
95 + 19 + 3 = 117 No. of zeros
2. (a) 1 × 2 × 3 × .......× 108 129 + 25 + 5 + 1 = 160
108! 8. (a) 10 + 100 + 1000 + ....... 10000000
5 108 No. of zero = 1
5 21 9. (a) ATQ,
4 5 241
r
No. of zero = 25 5 48
si
3. (b) 1 × 3 × 5........× 23 × 25 5 9
These are all odd no. 2 is not detected. 1
So, cannot make pair
= no of zero is 0.
an by No. of 5 in factorial 241! = (48 + 9 + 1) = 58
n
4. (c) 5 15870 No. of 5 in 25 = 2
No. of 5 in 35 = 1
5 3174 ja Total no. of 5 = 61
R s
5 634
then, Total no. of zeros = 61
5 126
a th
10. (c) 1532! × 481! × 34!
5 25
5 5 5 1532
5 481
1 5 306 5 34
ty a
5 96
5 61 5 6
No. of zeros 5 19
5 12
di M
= (3174 + 634 + 126 + 25 + 5 + 1) = 3965 1
3
5. (c) 15370! = 1 × 2 × 3 ........ 15370 2
5 15370 Total 5 = 381, Total 5 = 118, Total 5 = 7
5 3074 Total No. of zeroes = Total No. of 5
5 614
= (381 + 118 + 7) = 506
5 122
11. (d) 1² × 2² × 3³ × .......... × 4949
5 24
5 in 55 × 1010 × 1515 × 2020 × 2525 × .... 4545
4
No. of 5 = 5 + 10 + 15 + 20 + 50 + 30 + 35 + 40 + 45
No. of zeros = 250
= 3074 + 614 + 122 + 24 + 4 = 3838 So, no. of zeroes = 250
A
6. (a) 5 15245
12. (c) 11! × 22! × 33! × ..... × 1010!
5 3049 No. of 5 = 5! + 10!
5 609
So, No. of zeroes = (5! + 10!)
5 121
13. (c) 10 × 100 × 1000 × .... × 10000000000
5 24
No. of zeroes
4
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
No. of zeros 14. (a) 5!5! + 20!20! + 50!50! + 100!100!
= 3049 + 609 + 121 + 24 + 4 = 3807 No. of minumum 5 in expression = 5!
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= 5 × 4 × 3 × 2 = 120 No. of zeroes = No. of 5 in 336!
No. of zeros = 120 5 336
15. (a) 3500! 3500! 5 67
=
40 n
(5 2³)n 5 13
2
n = No. of 5 in 3500!
5 3500 Total = 82
5 700 20. (d) (2 × 4 × 6 × 8 × .....× 170)
5 140 285 [1 × 2 × 3 × 4×.......× 85]
5 28 No. of 5's
5 5 5 85
1 5 17
3
Total = (700 + 174) = 874
So, n = 874 No. of zero = (17 + 3) = 20
r
21. (c) (5 × 10 × 15 ......× 200)
16. (a) 5335! 5335! 540 (1 × 2 × 3 .....40)
si
=
63n (3² 7)n
5 40
No. of 7 =
7 5335
an by 5 8
1
n
7 762
Total no. of 5 = 49
7 108
ja
R s
7 15 2 40
2 2 20
a th
Total 7 = 887 2 10
So, n = 887 2 5
2 2
ty a
17. (d) 50! 50!
= 1
(12600)n (2³32 52 7)n
di M
No. of 7 = No. of 2 = (20 + 10 + 5 + 2 + 1) = 38
7 50 No. of zeros = 38
7 7 22. (b) 10 × 20 × 30 × .......× 500
1050 [1 × 2 × 3 × .....× 50]
1
1050 × 50!
Total = 7 + 1 = 8
n=8 5 50
18. (b) No. b/w 66 & 69 are 67 & 68 5 10
we can take one (n) = 67 2
No. of 5 in 67! No. of 5
5 67 then,
A
5 13 No. of zero = (50 + 12) = 62
2 23. (c) 1 × 3 × 5 × 7 × .......× 99 × 128
1 × 3 × 5 × 7 × ......× 99 × 27
Total = 15
1 × 3 × 5 × 7 × .....× 15.....× 25...× 35....× 45....×
No. of zeros = 15 55....× 95 .....× 99 × 27
19. (a) a = 2 × 3 = 6 Hence 5 is used more than 27 times. We will
b = (6 + 1) = 7 take 27 to make a pair.
c = (6 + 2) = 8 Therefore
(a × b × c)! = 336! No. of zero = 7
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24. (d) (48! + 49!) 27. (d) 12 × 22 × 33......8080
(48! + 49 × 48!) No. of 5's 55 × 1010 × 1515 .....8080
48!(1 + 49) (5 + 10 + 15 + 20 + .....80) + 25 + 50 + 75 = 150
48! × 50 n
Sn (a + l ), n = 16
5 48 2
5 9
16
1 (5 + 80) 85 × 8
2
No. of 5's
(680 + 150) = 830 zero
510 × 52
28. (c) Let,
No. of zero = 12
100! = 24 zero
25. (c) (76! – 75!)
105! = 25 zero
(76 × 75! – 75!)
110! = 26 zero
r
75! (76 – 1)
115! = 27 zero
si
75! × 75 120! = 28 zero
5 75 125! = 31 zero
5 15
3
an by (29 zero not present) = 29
29. (d) (45!)4! (1 × 2 × 3× ......× 45)4!
n
No. of 5's = 18 5 45
ja
Total no. of 5 = 518 × 52 5 9
R s
then, 1
a th
No. of zero = 20 No of 5's 9 + 1 = 10
26. (d) (329 – 328 – 327) (731 – 729 – 728) No of zero (210 × 510)4 × 3 × 2 × 1 = 240
327 × 728 (32 – 3 – 1) (73 – 71 – 1) 30. (b) Let
ty a
327 × 728 [9 – 3 – 1] [343 – 7 – 1) n! = 24! +2
di M
3 27
×7 28
[5] × 335] (n + !)! = 25! Increased
3 27 28
×7 (1675) First no. 24!
No. of zero = 0 Second No. 49!
[because it does not have 2] third No. 74!
A
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Arithmetic Progression (lekUrj Js.kh)
(CLASSROOM SHEET)
nth Term of a Sequence/vuqØe dknok¡ in If sequence has n terms then, last term is
also denoted by 'l' sometimes.
nth term of s sequence is an expression in
which if we put n = 1, we get 1st term, ;fn vuqØe esad in gSa] rks vafre in dks dHkh&dHkh
if we put n = 2, we get 2nd term & so on 'l' }kjk Hkh n'kkZ;k tkrk gSA
vuqØe dk nok¡ in ,d ,slh vfHkO;fÙkQ gS ftlesa ;fn
1. For which value of k; the series 2, 3 + k
ge n ¾ 1 j•rs gSa] rks gesa igyk in çkIr gksrk gS] and 6 are in A.P.?
;fn ge n ¾ 2 j•rs gSa] rks gesa nwljk in çkIr gksrk k ds fdl eku ds fy,_ J`a•yk 2] 3 $ k vkSj 6
gS bR;kfn lekUrj Js.kh esa gSa\
r
Note-01 (a) 4 (b) 3
si
For the sequence 2, 4, 6, 8, ..... (c) 1 (d) 2
Øe 2] 4] 6] 8] ------ ds fy, 2. Twentieth term of an A.P. is 101 and first
Observe for an = 2n
an = 2n ds
an by
fy, è;ku nsa
term is 6. What will be the common
difference?
n
n = 1 gives 1st term, n = 2 gives 2nd term. fdlh lekUrj Js.kh dk chloka in 101 gS vkSj igyk
in 6 gSA lkekU; varj D;k gksxk\
ja
i.e., it's an expression for nth even
R s
number. MP Excise Constable 23/02/2023 (Shift-03)
igyk in nsrk gS]n = 2 nwljk in nsrk gSA
a th
n=1
(a) 7 (b) 4
vFkkZr~ n;goha le la[;k dk O;atd gSA (c) 5 (d) 6
Note-02
3. What is the first term of an AP series
ty a
We all know odd numbers are given by 2n having total 25 terms whose last term is
–1 or 2n + 1 , but do you realize that if I 314 and the common difference is 12?
di M
ask for nth odd number then it would be
just 2n – 1. dqy 25 inksa okyh lekUrj Js.kh dk igyk in D;k gS
ge lHkh tkurs gSa fd fo"ke la[;k,a
2n –1 or 2n + ftldk vafre in 314 gS vkSj lkoZ varj 12 gS\
1 }kjk O;Dr dh tkrh gSa] ysfdu D;k vkidks irk gS MP Excise Constable 21/02/2023 (Shift-03)
fd ;fn ge noha fo"ke la[;k dh ckr djs rks og (a) 28 (b) 26
fliQZ2n – 1 gksxhA (c) 24 (d) 24
(Putting n = 1 should give first odd number th
4. If 5 term of an arithmetic progression
n = 2 should give second odd number etc.)
(AP) series is 16 and 9th term is 22, then
( n ¾ 1 j•us ij igyh fo"ke la[;k] n ¾ 2 j•us find the seventh (7th) term of the series.
ij nwljh fo"ke la[;k vkfn vkuh pkfg,) ;fn ,d lekUrj Js.kh (AP) J`a•yk dk 5oka in 16
Arithmetic Progression/lekUrj Js.kh
A
gS vkSj 9oka in 22 gS] rks J`a•yk dk lkroka (7oka)
Sequence of the terms: a, a + d,a + in Kkr dhft,A
2d,.......+ a + (n – 1) d is called an A.P. (a) 25 (b) 22
inksa dk vuqØe%
a, a + d, a + 2d,.......+ a + (n – 1)d (c) 19 (d) None of these
dks lekUrj Js.kh dgk tkrk gS
5. Find the n term of /dk nok¡ in Kkr
th
dhft, %
nthterm /n ok¡ in = a + (n – 1) d
5, 2, –1, .....
where a 1st term/igyk in]
(a) 5 – 2n (b) 8 – 3n
d Common difference/lkoZUrj (c) 7 – 3n (d) 4 – n
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6. What is the 11th term for the given Note-03
arithmetic progression? Common terms of two APs with common
nh xbZlekUrj Js.khds fy, 11ok¡ in D;k gS\ differences d1 and d2, are also in AP with
common difference = LCM {d1, d2}
2, 6, 10, 14, 18,........
lkoZUrjd1 vkSj d2 okyh nks lekUrj Jsf.k;ksa ds
(a) 38 (b) 44
mHk;fu"B in Hkh lekUrj Js.kh esa gksrs gS] ftudk lkoZUrj
(c) 42 (d) 46 = LCM {d1, d2}
7. What will be the 10th term of the 13. Find 10th common term in: 3, 7, 11, 15,
arithmetic progression 2, 7, 12, ____ ? 19, .... & 1, 6, 11, 16, 21, ....
lekUrj Js.kh 2] 7] 12] ------ dk 10ok¡ in D;k gksxk\ 10oka mHk;fu"B in Kkr djsa%
(a) 45 (b) 43 3] 7] 11] 15] 19] ---- vkSj 1] 6] 11] 16] 21] ----
(c) 97 (d) 47 (a) 191 (b) 180
8. In a certain A.P., 5 times of 5th term is (c) 195 (d) 201
r
equal to 8 times the 8th term, then find its Note-04
13th term.
si
In an AP sum of the terms equidistant
,d fuf'pr A.P. esa] 5osa in dk 5 xquk 8osa in ds 8 from beginning & end is same
an by
xquk ds cjkcj gS] rks bldk 13okaKkr
in djsaA lekUrj Js.kh esa vkjaHk vkSj var ls leku nwjh okys
(a) 0 (b) 1 inksa dk ;ksx leku gksrk gS
n
(c) 2 (d) 4 14. Find sum of first 6 terms of an AP if 3rd
ja
9. How many terms are there in the A.P. 3, and 4th terms are 5 & 8.
R s
7, 11, ....., 407? fdlh lekUrj Js.kh ds igys 6 inksa dk ;ksx Kkr dhft,
;fn rhljk vkSj pkSFkk in 5 vkSj 8 gSaA
a th
lekUrj Js.kh 3] 7] 11] ------] 407 esa fdrus in gSa\
(a) 96 (b) 102 (a) 36 (b) 38
(c) 100 (d) 108 (c) 39 (d) 45
ty a
15. Find sum of first five terms of an AP if 3rd
10. Find the tn for an arithmetic progression
term is 4.
where t3 = 22, t17 = – 20.
di M
fdlh lekUrj Js.kh ds igys ikap inksa dk ;ksx Kkr
lekUrj Js.kh ds fy, tn Kkr dhft, tgk¡ t3 = 22,
t17 = – 20 djsa ;fn rhljk in 4 gSA
(a) 24 (b) 16
(a) 3n – 32 (b) 3n – 31
(c) 20 (d) None
(c) –3n + 31 (d) –3n + 32
Note-05
11. In a A.P. if a2 + a5 – a3 = 10 and a2 + a9 =
17 then find a20 Assuming terms in A.P/lekUrj Js.kh esa inksa dks
,d A.P. esa ;fn a2 + a5 – a3 = 10 vkSj2 + a9 = ekuuk
17 gS rks
a20 Kkr dhft, (a) 3 terms/3 in
(b) 4 terms/4 in
A
(a) 6 (b) –6
(c) 8 (d) –10 (c) 5 terms/5 in
th th
12. If the p term of an A.P. is q and the q 16. The sum of three numbers in A.P. is – 3
term is p, then find its rth term. and their product is 8. Find the numbers.
;fn fdlh AP dk pok¡ in q gS vkSjqok¡ in p gS] lekUrj Js.kh esa 3 la[;kvksa dk ;ksxiQy
– 3 rFkk
rks bldk r ok¡ in Kkr dhft,A mudk xq.kuiQy 8 gSA la[;k, Kkr dhft,A
(a) p – q + r (b) p + q + r (a) 2, –1, –4 (b) 2, 1, 4
(c) p – q – r (d) p + q – r (c) –2, –1, –4 (d) 2, 1, –4
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Sum of an A.P./fdlh lekUrj Js.kh dk ;ksx 22. If 1st term of an A.P. is 5, the last term is
45 & sum is 400 then find number of
Sn = a1 + a2 + .... + an where ai's are in AP.
terms.
Sn =
n
(2a +(n– 1)d) ;fn fdlh lekUrj Js.kh dk igyk in 5 gS] vafre
2 in 45 gS vkSj ;ksx 400 gS rks inksa dh la[;k Kkr
or dhft,A
n (a) 15 (b) 14
(a + 1) where l = a + (n – 1)d
2 (c) 18 (d) 16
17. Find sum of first n odd numbers. 23. Find sum of natural numbers between 100
& 1000 which are multiples of 5.
çFken fo"ke la[;kvksa dk ;ksx Kkr dhft,A
100 vkSj 1000 ds chp çkÑfrd la[;kvksa dk ;ksx
(a) 3n² (b) n²
Kkr dhft, tks 5 ds xq.kt gSaA
n2 3n2
(c) (d) (a) 94540 (b) 98450
r
2 2
(c) 95320 (d) 96430
si
18. Find the sum of 23 terms of the A.P. 5, 9,
24. Find sum of natural numbers between 100
13, 17 .......
and 1000 are multiples of 5 and 7
;ksx Kkr dhft,A
an by
lekUrj Js.kh5, 9, 13, 17 ....... ds 23 inksa dk simultaneously.
n
100 vkSj 1000 ds chp çkÑfrd la[;kvksa dk ;ksx
(a) 1172 (b) 1127 Kkr dhft, tks 5 vkSj 7 dh ,d lkFk xq.kt gksA
(c) 1217
ja (d) None of these
R s
(a) 15204 (b) 13102
19. Find the sum upto 151 term of the (c) 11105 (d) 14105
a th
sequence 243, 256, 269, ...
25. Find the sum of the first 157 terms of an
vuqØe 243] 256] 269] ---- ds 151 in rd dk A.P. whose first term and third term are
;ksxiQy Kkr dhft,A 160 and 170 respectively.
ty a
UPSI 13/11/2021 (Shift-02) ,d lekarj Js.kh ftldk igyk vkSj rhljk in Øe'k%
160 vkSj 170 gS] blds igys 157 inksa dk ;ksxiQy
di M
(a) 183916 (b) 183917
(c) 183918 (d) 183915 Kkr dhft,A
20. What is the sum of first 200 terms of the UPSI 12/11/2021 (Shift-03)
given series? (a) 84350 (b) 85350
nh xbZ J`a•yk ds igys 200 inksa dk ;ksx D;k gS\ (c) 83350 (d) 86350
1 + 5 + 6 + 10 + 11 + 15 + 16 + 20 + .... 26. A man saves ` 1550 in January 2020 and
SSC CHSL 10/03/2023 (Shift-01) increases his saving by ` 75 every month
over the previous month. What is the
(a) 49400 (b) 48300
annual saving of the man for the year
(c) 50100 (d) 49600
2020? (in `)
A
21. How many terms of series – 9, – 6, – 3, ...
,d vkneh tuojh 2020 esa 1550 :- cpkrk gS vkSj
must be taken so that the sum of all the
terms is 45? gj eghus fiNys eghus dh rqyuk esa viuh cpr esa 75
J`a•yk ds fdrus in & 9] & 6] & 3] --- ysus gksaxs dh o`f¼ djrk gSA ml vkneh us o"kZ 2020 esa druh
rkfd lHkh inksa dk ;ksx 45 gks\ okf"kZd cpr dh\ (` ess)
UPSI 24/11/2021 (Shift-02)
SSC CHSL 10/03/2023 (Shift-01)
(a) 9 (b) 11 (a) 23250 (b) 23350
(c) 10 (d) 8 (c) 23550 (d) 23450
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27. How many positive integers less than 100 29. If the 10th term of an AP is 21 and 17 term
have a remainder of 3 when divided by 7? is 8 more than the 13th term, then find
the AP.
100 ls de fdrus /ukRed iw.kkZadksa dks 7 ls foHkkftr
;fn fdlh lekUrj Js.kh dk 10ok¡ in 21 gS vkSj 17 ok¡
djus ij 'ks"kiQy 3 jgrk gS\
in 13osa in ls 8 vf/d gS] rks lekUrj Js.kh Kkr dhft,A
MP Jail Prahari 11/12/2020 (Shift-01)
MPPEG GROUP 2 31/01/2021 (Shift-01)
(a) 13 (b) 15 (a) 7, 10, 13, 16 .... (b) 3, 8, 11, 14 ....
(c) 12 (d) 14 (c) 5, 7, 9, 11 .... (d) 3, 5, 7, 9 ....
28. If the ratio of the sum to n terms of two 30. If sum of first 2n terms of the AP 2, 5, 8,
A.P.'s is (5n + 3) : (3n + 4), then the ratio ... is equal to the sum of first n terms of
of their 17th terms is 57, 59, 61, ..... Then find n
;fn nks AP ds n inksa ds ;ksx dk vuqikr
(5n + 3) ;fn lekUrj Js.kh 2] 5] 8] --- ds çFke n2 inksa dk
;ksx 57] 59] 61] --- ds çFken inksa ds ;ksx ds
: (3n + 4) gS] rks muds 17osa inksa dk vuqikr gS
cjkcj gS] rksn Kkr dhft,
r
(a) 172 : 99 (b) 168 : 103
(a) 10 (b) 12
si
(c) 175 : 99 (d) 171 : 103
(c) 11 (d) 13
an by Answer Key
n
1.(c) 2.(c) 3.(b) 4.(c) 5.(b) 6.(c) 7.(d) 8.(a) 9.(b) 10.(c)
ja
R s
11.(b) 12.(d) 13.(a) 14.(c) 15.(c) 16.(a) 17.(b) 18.(b) 19.(c) 20.(c)
a th
21.(c) 22.(d) 23.(b) 24.(d) 25.(d) 26.(c) 27.(a) 28.(b) 29.(d) 30.(b)
ty a
di M
A
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Geometric Progression (xq.kksÙkj Js.kh)
(CLASSROOM SHEET)
Geometric Progression/xq.kksÙkj Js.kh 1 1 1
4. Which term of the G.P. 2, 1, , ,... is ?
Sequence of the form: a, ar, ar , ..., ar2 n –1
is 2 4 128
called GP.
1 1 1
a, ar, ar2, ..., ar n –1 izdkj ds vuqØe dks xq.kksÙkj xq.kksÙkj Js.kh 2] ,1] dk dkSu lk in gS\
2 4 128
Js.kh dgrs gSA
(a) 8 (b) 9
arn–1 is nth term/nok¡ in
(c) 7 (d) 10
a 1st term/ izFke in
r Common ratio 'CR'/lokZuqikr 5. If x, 2x + 2, 3x + 3 are first three terms of
r
a GP then find its 4th term.
Ex. nth term of 8, 4, 2, ..., is 24 – n
si
Ex. 6th term of 4, 12, 36,..., is 4(3)5 ;fn x, 2x + 2, 3x + 3 fdlh GP ds igys rhu
Note:- in gSa rks bldk pkSFkk in Kkr dhft,A
(i) a,b,c AP 2b = a + c
a,b,c GP b2 = ac
an by –27
n
(a) (b) –27
(ii) No term of GP can be 0. 2
Ex. If the 2nd and 5th terms of a GP are 24 and
ja
(c) 27 (d) None
R s
81, respectively, then find the GP.
6. Three positive numbers form an increasing
;fn fdlh GP ds nwljs vkSj 5osa in Øe'k% 24 vkSj GP. If the middle term in this GP is
a th
81 gSa] rks
GP Kkr djsaA doubled, then new numbers are in AP, then
1. If the first term is 216 and the common common ratio of GP is:
5 rhu /ukRed la[;k,¡ ,d c<+rh gqbZ xq.kksÙkj Js.kh cukrh
ty a
ratio is , what will be the 4th term of
6
gSaA ;fn bl xq.kksÙkjesaJs.kh
eè; in nksxquk dj fn;k
the G.P.?
tk,] rks ubZ la[;k,¡ lekarj Js.kh esa vk tkrh gSa] rks
di M
;fn igyk in 216
5
gS vkSj lkoZ vuqikr gS] rks xq.kksÙkj Js.kh dk lokZuqikr gS%
6
xq.kksÙkj Js.kh dk pkSFkk in D;k gksxk\ (a) 2 3 (b) 3 2
UPSI 12/11/2021 (Shift-03)
(c) 2 – 3 (d) 2 3
(a) 129 (b) 123
(c) 127 (d) 125 7. Find product of first 5 terms of a GP if a3 = 2
2. Find the 11th term in the series. ;fn a3 = 2 gS rks xq.kksÙkj Js.kh ds igys 5 inksa dk
J`a•yk esa 11ok¡ in Kkr dhft,A xq.kuiQy Kkr djsa
6, 12, 24, ___________ ?
(a) 16 (b) 32
(a) 6812 (b) 6720
A
(c) 6144 (d) 6446 (c) 64 (d) None
3. How many terms are there in G.P. 6, 18, 8. If 4th and 5th term of a G.P. are 2 and 8
54, ........, 39366? respectively, then the product of the first
xq.kksÙkj Js.kh
(G.P.) 6, 18, 54, ........, 39366 esa 8 terms is
fdrus in gSa\ ;fn fdlh xq.kksÙkj Js.kh dk pkSFkk vkSj ik¡pok¡ in
UPSI 14/11/2021 (Shift-02) Øe'k% 2 vkSj 8 gSa] rks igys 8 inksa dk xq.kuiQy gS
(a) 7 (b) 11 (a) 88 (b) 168
(c) 13 (d) 9 (c) 48 (d) 64
Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs1
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Sum of Geometric Progression 13. Find the sum up to n terms of the
sequence:
xq.kksÙkj Js.kh dk ;ksx
vuqØe dsn inksa rd dk ;ksx Kkr dhft,%
Sn =a
1 – r ;
n
r 1
1– r 0.7, 0.77, 0.777, ...
(a) 7n–1
a
S = ; |r| < 1
1– r 7
(b) 9n – 1 10 – n
n number of term 81
9. Find the sum of 3 + 32 + 33 + ... + 38.
7n
3 + 32 + 33 + ... + 38 dk ;ksx Kkr dhft,A (c)
81
(a) 6561 (b) 6560
(c) 9840 (d) 3280 7n 1
10. Find the sum of the G.P. (d)
10n
fuEu xq.kksÙkj Js.kh dk ;ksxiQy Kkr dhft,A
r
Note:
si
2 2 2 2
, , , , ..... to n terms. Assuming terms in G.P./xq.kksÙkj Js.kh esa inksa
5 25 125 625
dks ekuuk
an by
UPSI 12/11/2021 (Shift-02)
(a) 3 terms
n
n n
1 1 2 1 (b) 4 terms
(a) 2 1 –
5 (b) 5 1 –
5
(c) 5 terms
ja
R s
n n 14. Sum of first three terms of a GP is 19 &
4 1 5 1
1 – 1 – their product is 216. Find the three
a th
(c) 5 (d) 4
5 5 numbers.
11. Find the sum of the following series: ,d xq.kksÙkj Js.kh ds igys rhu inksa dk ;ksx 19 gS
fuEufyf•r J`a•yk dk ;ksx Kkr dhft,% vkSj mudk xq.kuiQy 216 gSA rhuksa la[;k,¡ Kkr dhft,A
ty a
1 1 1 1 1 1 (a) 6, 8, 12 (b) 6, 9, 12
di M
+ ...
2 32 23 34 25 36 (c) 2, 4, 6 (d) 4, 6, 9
18 19 15. Ram gives his son Rs. 100 on one day, Rs.
(a) (b) 50 on the second day. Rs 25 on third day
25 24
and so on,. What will be total amount
18 18 given by Ram to his son starting from the
(c) (d) first day, if he lives forever?
23 25
12. Evaluate (a) Sn = 4 + 44 + 444 + ... n terms jke vius csVs dks ,d fnu esa 100 #- nwljs fnu 50
ewY;kadu djsa
(a) Sn = 4 + 44 + 444 + ... n in #- rhljs fnu 25 #i;s vkSj blh rjgA ;fn jke lnSo
thfor jgs] rks igys fnu ls 'kq: djds jke }kjk vius
4 10
(a) (10 n – 1) – n csVs dks nh xbZ dqy jkf'k fdruh gksxh\
9 9
A
(a) 200 (b) 120
7 (c) 250 (d) 100
(b) 9n – 1 10 – n
81
16. After striking the floor, a ball rebounds to
7n 4
(c) th of the height from which it has
81 5
fallen. Find the total distance that it
7n 1
(d) travels coming to rest if it has been gently
10n
dropped from a height of 120 meters.
Aditya Ranjan (Excise Inspector) Selected gSSelection fnyk,axs2
Join Telegram- Maths by Aditya Ranjan Geometric Progression
4 20. Sum of first 63 terms of a G.P. is equal to
iQ'kZ ls Vdjkus ds ckn] ,d xsan ml ÅapkbZoka
ds the sum of the first 61 terms in the same
5
G.P. When second term is –653, what is
Hkkx rd mNyrh gS tgka ls og fxjh gSA ;fn bls 120 the sum of 50 terms in the G.P.?
ehVj dh ÅapkbZ ls /hjs ls fxjk;k x;k gks rks vkjke
xq.kksÙkj Js.kh ds çFke 63 inksa dk ;ksx leku xq.kksÙk
dh fLFkfr esa vkus rd r; dh xbZ dqy nwjh Kkr dhft,A
(a) 540 mtrs (b) 960 mtrs
Js.kh ds çFke 61 inksa ds ;ksx ds cjkcj gSA ;fn mlh
(c) 1080 mtrs (d) 1120 mtrs xq.kksÙkj Js.kh esa nwljk in &653 gS] rks 50 inksa dk
17. A square is drawn by joining the mid ;ksx crk,
points of the sides of a square. A third UPSI 15/11/2021 (Shift-03)
square is drawn inside the second square (a) 0 (b) 4
in the same way and the process is (c) 2 (d) 6
continued infinitely. If the side of the 21. Sum of first 13 items of a G.P. is equal
sqaure is 10 cm, then the sum of areas of to the sum of the first 11 terms in the
all the squares so formed is same G.P. Sum of the first 15 terms is
,d oxZ dh Hkqtkvksa ds eè; fcanqvksa dks feykdj ,d 1200, what is the 21st term in the same
oxZ cuk;k tkrk gSA nwljs oxZ ds vanj Hkh blh çdkjG.P.?
r
rhljk oxZ cuk;k tkrk gS vkSj ;g çfØ;k vuar dky xq.kksÙkj Js.kh ds inksa dk ;ksx mlh xq.kksÙkj Js.kh
si
rd tkjh jgrh gSA ;fn oxZ dh Hkqtk 10 lseh gS] rks izFke11 inksa ds ;ksx ds cjkcj gsA izFke
15 inksa
bl çdkj cus lHkh oxks± ds {ks=kiQyksa dk ;ksx gS dk ;ksx 1200 gS] mlh xq.kksÙkj Js.kh
an by
21oka
esa in
(a) 100 cm2 (b) 200 cm2
D;k gS\
n
2
(c) 250 cm (d) None of these
UPSI 24/11/2021 (Shift-01)
18. An equilateral triangle by joining the
(a) 1300 (b) 1400
ja
midpoints of a given equilateral triangle of
R s
side 18cm. A thord triangle is drawn inside (c) 1200 (d) 1100
the second in the same way. The process Note:
a th
is constructed indefinitely. The sum of the For two positive numbers a and b.
areas of all such triangle in cm2 is
ab
18 lseh Hkqtk okys fdlh fn, x, leckgq f=kHkqt ds Arithmetic mean lekUrj ekè; (A.M.)=
2
ty a
eè; fcanqvksa dks feykdj ,d leckgq f=kHkqt cuk;k
tkrk gSA nwljs ds vanj Hkh blh çdkj rhljk f=kHkqtGeometric mean xq.kksÙkj ekè;(G.M.)= ab
di M
cuk;k tkrk gSA çfØ;k vfuf'pr dky ds fy, cukbZ 2ab
xbZ gSA ,sls lHkh f=kHkqtksa ds {ks=kiQyksa
2
esa dk ;ksx lseh+ mean gjkRed ekè;(H.M.)= a b
Harmonic
gS
AM × HM = GM²
(a) 324 3 (b) 108 3 22. The geometric mean of two numbers is 8
(c) 54 3 (d) 200 3 and their harmonic mean is 6.4, the
numbers are
19. The mid points of an equilateral triangle
nks la[;kvksa dk T;kferh; ekè; 8 gS vkSj mudk gkekZsfud
are joined. Forming a second triangle, a
third triangle is formed joining the ekè; 6-4 gS] la[;k,¡ gSa
midpoints of the second and the process (a) 2, 8 (b) 4, 16
continued infinitely. If the area of the first (c) 6, 16 (d) 8, 16
A
traingle is 24 units, then find the sum of 23. The harmonic mean and the geometric
areas of all triangles. mean of two numbers are 10 and 12
,d leckgq f=kHkqt ds eè; fcanq tqM+s gq, gSaA nwljs f=kHkqt
respectively. What is their arithmetic
mean?
ds eè; fcUnqvksa dks feykdj ,d rhljk f=kHkqt curk
gS vkSj ;g çfØ;k vuUr dky rd pyrh jgrh gSA nks la[;kvksa ds gjkRed ekè; vkSj xq.kksÙkj ekè; Øe'k%
;fn çFke f=kHkqt dk {ks=kiQy 24 bdkbZ gS] rks 10 vksj 12 gSA mudk lekarj ekè; D;k gS\
lHkh
f=kHkqtksa ds {ks=kiQyksa dk ;ksx Kkr dhft,A (a)
25
(b)
3 20
(a) 26 (b) 28
(c) 32 (d) 36 (c) 11 (d) 14.4
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Join Telegram- Maths by Aditya Ranjan Geometric Progression
24. For two observations, the sum is S and 25. If arithmetic ad geometric mean of x and
product is P. What is the harmonic mean y is 8 and 3 7 respectively, then the
of these two observations?
value of x³ + y³ is?
nks voyksduksa ds fy,] ;ksx
S gS vkSj xq.kuiQy
P gSA
bu nks voyksduksa dk gkeksZfud ekè; D;k gS\ ;fn x vkSjy dk lekarj vkSj xq.kksÙkj ekè; Øe'k%
8
CDS 02/02/2020 vkSj3 7 gS] rksx³ + y³ dk eku D;k gS\
2S S Selection post phase IX 01/08/2022 (Shift-04)
(a) (b)
P (2P)
(a) 1072 (b) 945
2P P (c) 559 (d) 855
(c) (d)
S (2S)
Answer Key
r
1.(d) 2.(c) 3.(d) 4.(b) 5.(a) 6.(d) 7.(b) 8.(c) 9.(c) 10.(a)
si
an by
11.(b) 12.(a) 13.(b) 14.(d) 15.(a) 16.(c) 17.(b) 18.(b) 19.(c) 20.(a)
n
21.(c) 22.(b) 23.(d) 24.(c) 25.(a)
ja
R s
a th
ty a
di M
A
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