Math shift wise

the ages of these two are

average age of the remaining excluded, the Q, side of square-100
three years less than the playens is Sol.2.(d) ATQ. Area of square - 100 x 100- I0000
ihe whole team What is average age of Sum - 288 and HHCE- I6 The ratio of area of the rhombus
uf the
the average age So the numbers can be l6x and loy. S065
whole team? 28
square - 10000 40
(a) 24 years (b) 28 years xy should be -
16
(c) 26 years (d) 25 years Sol.7.(b)
I8 can be factored as (1, I8)and (2, 9).
2pairs are possible, so 2 pairs of such
934. How many straight lincs does a numbers can be formed.
cuboid have?
26 cm 28 cm
(a) 10 (b) 12 (c) l6 (d) 24 Sol.3.(d) Total no. of students - 120
There are 5 subjects and the no. of
Q.35. Observe the graph and answer tlhe students distributed is equal. 20 cm
questions below. . no,of students in cach subject 26+20+20
The total annual carnings ofa family of 120
24
Semiperimeter 2
36cm
four members is Rs 12 Lakhs. Thc bar 5
Area - Vs (S - a)(S - b)(S - c)
graph shows the percentage of There are two languages Hindi and
/36(36 26)(36 - 26)(36-20)
contributionof cach family member. English,
so, no, of students study languages -36 x 10 x 10 x 16 240 cm'2
Percentage of totalearning of family
-2x 24 48 Sol.8.(c) Sum of weight of P,Q, R
-3 58174 kg
Sol.4.(a) S,-30 k/h - .5min late Sum of weight of P and Q 2 x 54
20 S,- 40 km/h -. 3 min carly -108 kg
Sum of weight of Q and R
Percent earned Duc to increase in speed, Vinod saves 8 2x 48-96 kg
minutes, Weight of Q= (108 + 96) - 174
0
Distance is cqual in both cases, >204 - 174 30 kg.
Father Mother Son Daughter S,xT,-S,xT,
What is the difference in the salary of 30 3 Sol.9.(a) The prime factorisation of
the highest and the lowest carning S, T, 40 4 35280 = 2×2×2x2x3×3x5x7x7
members? IfT,- 4 and T,- 3 Except for 5, all factors make a perfect
(a) Rs. I,560 (b) Rs. ,5600 square.
Diff (time) = 1 ---- 8 min.
(c) Rs. ll,60,000 (d)Rs. I,56,000 ::TT,=4x8= 32 min,=
32
-hours
60 Sol.10.(a) Let the originalnumber = x
80 115
32 New number is = xX
Answer key : Distance = SxT= 30 ×
60
16km 100 100
23
25X
1.(a) 2.(d) 3.(d) 4.(a)
Sol.5.(d) 23
S.(d) 6.(b) 7.(b) 8.(c) A ATQ, x-EX 64
6 cm
9.(a) 10.(a) 11.(a) 12.(c) 26 cm
6 cm 25 X=64
13.(a) 14.(d) 15.(a) 16.(c)
10 cm D 20 cm . X=
64x25 = 800
17.(d) 18.(c) 19.(a) 20.(b) 2
4 cm
21.(a) 22.(b) 23.(b) 24.(c)
C Sol.11.(a) tan 15°= tan(45°. 30°)
25.(c) 26.(d) 27.(a) 28.(a) B4cm 20 cm
24 cm tan 45°-tan30°
29.(d) 30.(c) 31.(c) 32.(a) Radius of circle = OD = BE = 4cm 1+tan 45°xtan 30°

33.(a) 34.(b) 35.(d) N3-1

Sol.6.(b) A rhombus has one of its 1+ 1x V3+1
Solution : diagonal 65% of the other.
cot15°V3+1
So, if one diagonal is 100then the other
is 65. V3-1
Sol.1.(a) HCF of (24, 56, 72) = 8 V3-1 y3+1
Total number of trees = 24+56+72 =152 Area of rhombus = d,d, tan 15°+ cot 1so
V3+1 V3-1
152 =19
Number of rows = 8
1
X100 x 65 => 50 x 65 2 x(3+1-4
2
 Pinnacle Math shift wise

The remainder in each case should be 5

Sol.12.(c) Let the efficiency of man = M. Sol.18.(c) so the required numbers are
boy = B 17-4 x (5.4 +9) + 6x 1.9 437 and 485.
20M +15B =10days and 25M +10B =17 - 4 X 0.6 t6 X 1.9 Sum = 437+485 =922
=9 days = 17-2.4 +11.4
Work in both case is same, = 17+9= 26 Sol.24.(c) In 18 eggs, 3 are rotten, so
(20 M+ 15 B)x10 =(25 M+10B)>x9 good eggs =15
200M +150B = 225M + 90B Sol.19.(a) Ratio of good eggs to rotten eggs =5:1
’25M= 60B 5

M 60
25 + 10 -(, x} x x Good eggs = 690×=
6
575
B 25
M:B= 12:5
10-25)
Sol.25.(c) x + =7
12 x + 5
’25+10 -
3
Sol.13.(a) 343
Pass percentage of schoolB=32
Ratio of boys and girls = 3:5
10 +-s
Pass % of boys =32x =12
67
10
7
10
-56-5=1 »+* +}343
Pass % of school C = 24 t+3x7=343
Ratio of boys to girls = 1:2 Sol.20.(b)remainder
9 =4
12n
’ x t =343 -21 =322
Pass % of boys =24x 3 8 ’ remainder = 12x4= 48
9
Required ratio =12:8 = 3:2 When we divide 48 further by 9, then the 14641
Sol.26.(d) The sum becomes 10000
remainder will be 3.
Sol.14.(d) CP of the car times of itself.

=1,20,000 x
100
-=1,00,000 Sol.21.(a) The ratio of incomes of Amar IfP= 10000then A = 14641
120 and Komal = 5:4
Now, when profit = 30%, 4 years
.. income of Amar = 5x and income of 10000 14641
130 Komal = 4x
SP =1,00,000 × 100
=1,30,000
Ratio of expenditure = 2:1
Sol.15.(a) Given,
:. expenditure of Amar = 2y and V10000: V14641
expenditure of Komal = y ’ 10: 11
diameter of lead sphere =6 cm Given, saving of each =6,000 1
diameter of beaker =9 cm Rate = 10
x 100 =10%
ATQ,
Increased height of water level = 32 cm
5x - 2y 6000 ---(1)
Let the no. of lead spheres = n, 4x - y= 6000 ----(2) Sol.27.(a) Difference in the number of
Volume of the cylinder On solving equation (1) and (2) male and female,
=nx volume of lead sphere X= 2000 For company A, 250 - 150 =100
4 B, 350 - 250 = 100
: Income of Amar = 5 x 2000 =10,000
nRH =nx C, 200 - 150 = 50
’x32 =nXx 3x 3x 3 Sol.22.(b) We know, sec´0 - tan e = 1
D. 350 - 100 = 250

’n= 18 25x.25-=1 Sol.28.(a) Let the CP of one LED TV =

Sol.16.(c) 15 -6.3 + 7+3 x 1.3 -2

- 100x and CP of other LED TV = 41000 -
100x
= 15 -0.9 +3.9 -2 SP of lst TV = 100x + 20x = 120x and
= 15 +3.9 - 2.9 85
SP of other TV = (41000 - 100x)X
=15 +1= 16 100

10
= 34850 - 85x
Sol.17.(d) ya=8 Hence, 25 5 ATQ,
Squaring both sides, 120x =34850 - 85x

X=64 and, x = 4096 205x= 34850

Sol.23.(b) LCM of (8, 12, 16) = 48
’X= 170
Given, xty = 4112 Ist number that is multiple of 48
SP of both TV sets
’ 4096+ y=4112 between 400 and 500 is = 432
=2 X120 × 170 =40800
And the 2nd number = 480
Dy= 16 yy=4 Loss = 41000 - 40800 = 200

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 Pinnacle
0.6. The value so
Math shit wiy
Sol.29.(d) We know, Sum and the obtained
a +b + the
3abc =
(a+btc) [(a + b +
c)»-3(ab+bc+ca)]
a+ b + c -3 × 15 =
numbers 3.03 and 2.05. is
(a)60.06
(c) 6.06
(b)600.6
di f erence
We can see in the above
figure that there (d) 0.606
x[(14)-3 x47]
14
are 12 straight lines in a cuboid. 0.7. Which of the
’a+ b + c= 14 x(196 - 141) ++ 45
’ a+ b + c= (14 x
55 )+ 45 ’ 770
Sol.35.(d) The total annual earning of a
rational number?
fol owing is
+ 45 = 815 family = 12 lakhs
Highest earning is of father and lowest is
R+4,/12. 96, V125 and.G
(a) y900
of Daughter, (b) V125
Sol.30.(c) Let the rate of interest = R% difference = 33 - 20 = 13% (c) V3 + 4 (d) 12.96
SI obtained from
Dubey
7200xR×3
100 = 216R
= 12,00,000 X 13_1.56,000
100
Q.8. If the denominator
of ara
SI obtained fromn Raghav number is of the form
8400×RX4
100 =336R RRB NTPC2019 TIER-I m are non-negative
will be the decimal
S,where 1
integers, then
Total SI = 216R+336R= 552R 05-01-2021 (Morning) number? expansion
Interest obtained is Rs 4,968 (a) Non-terminating and
So, 552R = 4968
’R= 4968 =9%
552
Q.1. If x=2 and y =5, evaluate 5xy- y
.(a) 20 (b) 25 (c) 40 (d) 0
(b) Non-terminating but recuring
(c) Terminating
|non-recutta
(d) Can't be determined
Sol.31.(c) The greatest number of 5 Q.2. A bus passes two persons moving
digits = 99999 in the direction of the moving bus at a 0.9. What is the total surface area nfs
When we divide 99999 by 468, we get speed of 3 km/h and 5 km/h, visible faces in the given figure?
315 as remainder. respectively. The bus passes the first 20 cm
5 cm
So, the required 5-digit number which is person in 10s and the second person in
8 cm
divisible by 468 is 11s. The speed of the bus is 15 cm
= 99999 - 315 = 99684 (a) 27 km/h (b)25 km/h
(c) 24 km/h (d) 28 km/h
7 cm
Sol.32.(a) Q.3. The sum of the deviations about the
In this case time is equal for both cars= (a) 384 cm (b)905 Cm
mean is always 2
(a) positive (c) 580 Cm (d) 1325 cm
Distance = 4500 km
(b) zero
Let speeds of both cars= x andy km/h
(c) the range Q.10. In a frequency distribution,
For opposite direction, (d) the total standard deviation mid value of a class is 12 and its widths
Relative speed =x+ ykm/h 6. The lower limit of the class is
4500
Relative speed = 9
= 500 km/h. 0.4. What is the least number, which (a) 12 (b)9 (c) 6 (d) 18
when divided by 12, 21 and 35, leaves
the same remainder 6? Q.11. The ratio of two weights, "
Sol.33.(a) Age of captain = 35 years
(a) 420 (b) 576 (c) 414 (d) 426 and 108g, is
Age of wicket keeper = 40 years
Let the average age of whole team X (a) 240 :1 (b) 270:I
Q.5. In the given figure, LABD = 55° (c) 250 :1 (d) 300:1
years
and LACD = 30°. If 2BAC = y°
ATQ, and
11x =9(x-3) + 35+40 non-reflex ZBDC = x°, then what is the Q.12. A student required 20% mars
value ofx- y? Secured1M
’ l l x =9x - 27+ 75 pass in psychology. He'she What
’ 2x = 48 marks and failed by 20 marks.
’ X=24 years. the passing marks?
(a) 40 (b) 60 (c) 20 (d) 50
Sol.34.(b) of20
loss
C Q.13. AshopkeeperrincursSaRs48OOIo
for
(a) 95 (b) 105 (c) 85 (d) 15 afler selling a machine 20%, atwha
order to gain a profit of

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