 GeeksforGeeks App
Open App Browser
Continue

# SBI Clerk Prelims Quantitative Aptitude Question Paper 2020

Direction (1-5):-What will come in the place of the question mark ‘?’ in the following question?

Que.1
6, 3 , 3 , 4.5 , 9 , 22.5 , ?
A) 65.5
B) 63.25
C) 67.5
D) 90
E) 45

Explanation :-
6 × .5 = 3
3 × 1 = 3
3 × 1.5 = 4.5
4.5 × 2 = 9
9 × 2.5 = 22.5
22.5 × 3 = 67.5

Que.2
? , 600, 3000, 12000, 36000, 72000
A) 50
B) 100
C) 120
D) 150
E) 200

Explanation:-
100 × 6 = 600
600 × 5 = 3000
3000 × 4 = 12000
12000 × 3 = 36000
36000 × 2 = 72000

Que.3
3, 10, 41, 206,? , 8660
A) 4237
B) 1000
C) 1200
D) 1237
E) 4320

Explanation:-
(3 × 3) + 1 = 10
(10 × 4) + 1 = 41
(41 × 5)+1 = 206
(206 × 6) + 1 = 1237
(1237 × 7) + 1 = 8660

Que.4
107, 105, 102, 97, 90, 79,?
A) 70
B) 66
C) 68
D) 67
E) 62

Explanation:-
107 – 2 = 105
105 – 3 = 102
102 – 5 = 97
97 – 7 = 90
90 – 11 = 79
79 – 13 = 66

Que.5
2, 4, 8, 14, 22, 32, ?
A) 44
B) 65
C) 53
D) 40
E) 46

Explanation:-
2 + 2 = 4
4 + 4 = 8
8 + 6 = 14
14 + 8 = 22
22 + 10 = 32
22 + 12 = 44

Que.6
A can do a work in 20 days and B can do the same work in 20% less time taken by A. A and B start working together, then find in how much time 80% of the total work will get completed.
A) 8 1/9
B) 7 1/9
C) 5 1/9
D) 6 1/9
E) None of these

Explanation:-
20% of 20 = 20 × 1/5 = 4
A = 20 days
B = (20-4) = 16 days
A’s 1 day work = 1/20
B’s 1 day work = 1/16
(A+B)’s 1 day work = 1/20 + 1/16 = 9/80
80% work = 80/100 = 4/5
:. Required time taken

= 80/9 × 4/5 = 64/9 = 7 1/9 days

Que.7
A invested some amount of money in a scheme and B invested Rs. 100 more than the money invested by A in the same scheme at the rate of 10% per annum compounded annually for 2 years. If the total compound Interest is Rs. 189, then find the amount invested by A.
A) Rs. 300
B) Rs. 400
C) Rs. 500
D) Rs. 600
E) None of these

Explanation:-
Let, amount invested by A = x
Amount invested by B = ( x + 100)
Rate of compound interest for 2 years = (10+10+10×10/100) = 21%
According to question,
21% of x + 21% of (x + 100) = 189
=> 21x/100 + 21(x+100)/100 = 189
=> 21x + 21x + 2100 = 18900  ( multiplying both sides by 100)
=> 42x = 18900 – 2100
=> x = 16800/42
:. x = 400
:. The amount invested by A =x =Rs. 400

Que.8
The ratio of age of father to the age of son 5 years ago was 7:2 and the ratio of their age after 10 years will be 2: 1. Find the present age of father(in years).
A) 40
B) 45
C) 50
D) 35
E) None of these

Explanation:-
Age gap between father and son should be same in every case.
Father : Son = 7 : 2 —–(×1)  = 7 : 2
= 2 : 1 —–(×5) = 10 : 5
Here years gap = 3 unit
Now, 3 unit = 5 + 10
1 unit = 5
:. Father’s age 5 years ago = 7 × 5 = 35 years
Thus,  present age of father = 35+5 = 40 years

Que.9
A woman spends 10% of her salary in household expenses, 10% of the remaining in mutual funds and 50% in transportation expenses. If the amount invested in mutual funds is Rs. 450. Then find the amount spend in transportation.
A) Rs. 2500
B) Rs. 1850
C) Rs. 2750
D) Rs. 2250
E) None of these

Explanation:-
Remaining salary after spending in household expenses= 100-10 = 90
Spend in mutual fund = 90 × 10/100 = 9
Spend in transportation expenses= 90× 50/100 = 45
According to question,
9 unit = 450
1 unit = 50
:. 45 unit = 50×45 = 2250
:.The amount spend in transport = Rs.2250

Que.10
Train A of length 280 m can cross a platform of length 360 m in 32 seconds and the ratio of the speed of train B to train A is 3: 2. Find the length of train B, if train B cross a man in 30 seconds.
A) 500 m
B) 600 m
C) 700 m
D) 900 m
E) None of these

Explanation:-
Total distance travelled by train A = (280+360) = 640 m
Time = 32 sec
Speed of train A = 640/32 m/s = 20 m/s
Here, ratio of speed of train A and B = 3:2
2 unit = 20 m/s
1 unit = 10 m/s
3 unit = 30 m/s
So, speed of train B = 30 m/s
Train B crosses a man which means the train crosses its own length.
So, the distance travelled by train B = 30×30 = 900 m
:. The length of the train B = 900 m.

Direction(11 -20):- What value should come in place of the question mark ‘?’ in the following question?

Que.11
(√64) × 150% -88 ÷ 22 = ?
A) 6
B) 8
C) 3
D) 2
E) 12

Explanation:-
(√64) × 150% -88 ÷ 22 =?
=> 8 × 3/2 – 88 × 1/22 =?
=> 12 – 4 =?
:. ? = 8

Que.12
√121 + ³√216 ÷ 2 × 15 =?
A) 42
B) 84
C) 56
D) 45
E) 34

Explanation:-
√121 + ³√216 ÷ 2 × 15 =?
=> 11 + 6 × 1/2 × 15 =?
=> 11 + 45 =?
:. ? = 56

Que.13
(12×25 – 11×2³)/(12² – 140) =?
A) 212
B) 106
C) 84
D) -84
E) 53

Explanation:-
(12×25 – 11×2³)/(12² – 140) =?
=> ( 300 – 88)/( 144 – 140) =?
=> 212/4 =?
:. ? = 53

Que.14
48 – 55 ÷ 5 × 11 + √289 =?
A) 15
B) 7
C) 55
D) -56
E) -44

Explanation:-
48 – 55 ÷ 5 × 11 + √289 =?
=> 48 – 55 × 1/5 × 11 + 17 =?
=> 48 – 121 + 17 =?
:. ? = – 56

Que.15
[{ 22 + (7+3)÷5}×2^8]/4³ =?
A) 106
B) 96
C) 88
D) 132
E) 68

Explanation:-
[{ 22 + (7+3)÷5}×2^8]/4³ =?
=> [ (22+ 10 × 1/5) × 256]/4³ =?
=> 24 × 256 × 1/64 =?
=> 24×4 =?
:. ? = 96

Que.16
(√484 – 525 ÷ 35)/(³√64 ×7-21) =?
A) 3
B) -1
C) -2
D) 1
E) 2

Explanation:-
(√484 – 525 ÷ 35)/(³√64 ×7-21) =?
=> (22 – 525 × 1/35)/(4×7-21) =?
=> (22 – 15)/(28-21) =?
=> 7/7 =?
:. ? = 1

Que.17
³√1728 – ²√49 × ³√125 +? = √484
A) 84
B) 125
C) 45
D) 44
E) 121

Explanation:-
³√1728 – ²√49 × ³√125 +? = √484
=> 12 – 7 × 5 +? = 22
=> 12 – 35 +? =22
=> ? = 22 +23
:. ? = 45

Que.18
[{(99×11)÷3}×(4)½ × ³√729]/27 =?
A) 524
B) 254
C) 121
D) 242
E) 423

Explanation:-
[{(99×11)÷3}×(4)½ × ³√729] /27 =?
=> [ 99 × 11 × 1/3 × 2 × 9 ] / 27 =?
=> 99 × 11 × 6 × 1/27 =?
=>? = 11 × 11 × 2
:. ? = 242

Que.19
(77×11÷7+2) ÷ ³√27 =?
A) 36
B) 41
C) 47
D) 51
E) 66

Explanation:-
(77×11÷7+2) ÷ ³√27 =?
=> ( 77 × 11 × 1/7 + 2) ÷ 3 =?
=> ( 121 + 2) ÷ 3 =?
=> 123 × 1/3 =?
:. ? = 41

Que.20
20% of 1800 + 60% of 900 =?
A) 1200
B) 900
C) 1000
D) 800
E) 700

Explanation:-
20% of 1800 + 60% of 900 =?
=> 1/5 × 1800 + 3/5× 900 =?
=> 360 + 540 =?
:. ? = 900

Direction( 21 – 25):-In the given question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer –
A) x > y
B) x < y
C) x ≥ y
D)  x ≤ y
E)  x = y or the relation cannot establish.

Que.21
I. x² – 5x + 6 = 0
II. y = ³√8

Explanation:-
x² – 5x + 6 = 0
=> x² – 3x – 2x + 6 = 0
=> (x -3)(x -2) = 0
:. X = 3,2

y = ³√8
:. y = 2
:.x ≥ y

Que.22
I. x² – 6x + 8 = 0
II. y² – 8y + 15 = 0

Explanation:-
x² – 6x + 8 = 0
=> x² – 4x – 2x + 8 = 0
=> (x – 4)(x – 2) = 0
:. x = 4, 2
y² – 8y + 15 = 0
=> y² – 5y – 3y + 15 = 0
=> (y – 5)(y – 3) = 0
:. y = 5, 3
Relation can not be established.

Que.23
I. x² – 5x – 24 = 0
II. y² – 3y – 40 = 0

Explanation:-
x² – 5x – 24 = 0
=> x² – 8x + 3x – 24 = 0
=> (x – 8)( x + 3) = 0
:. x = 8, -3

y² – 3y – 40 = 0
=> y² – 8y + 5y – 40 = 0
=> ( y – 8)( y + 5) = 0
:. y = 8, -5
Relation can not be established.

Que.24
I. x² + 9x + 14 = 0
II. 2y² + 7y + 6 = 0

Explanation:-
x² + 9x + 14 = 0
=> x² + 7x + 2x + 14 = 0
=> (x + 7)(x + 2) = 0
:. x = -7,-2

2y² + 7y + 6 = 0
=> 2y² + 4y + 3y + 6 = 0
=> ( 2y + 3)(y + 2)= 0
:. y = -3/2, -2
x ≤ y

Que.25
I. x² – 9x + 20 = 0
II. y² – 3y – 4 = 0

Explanation:-
x² – 9x + 20 = 0
=> x² – 5x – 4x + 20 = 0
=> ( x – 5)( x -4) = 0
:. x = 5, 4
y² – 3y – 4 = 0
=> y² – 4y + y – 4 = 0
=> (y – 4)( y+1) = 0
:. y = 4, -1
x ≥ y

Que.26
The average age of three persons A, B and C is 25 and the average age of A and B is 24.Find the age of C.
A) 29 years
B) 25 years
C) 24 years
D) 27 years
E) 30 years

Explanation:-
Total age of A, B and c together = 25×3 = 75
Total age of A and B together = 24 × 2 = 48
:. The age of C = 75 – 48 = 27 years.

Que.27
The average of 11 numbers is 33. If the average of the first six numbers and the last six numbers are 25.5 and 40.5 respectively, then find the sixth number.
A) 34
B) 33
C) 35
D) 36
E) None of these

Explanation:-
Total of 11 numbers = 33×11 = 363
Total of first six numbers = 25.5 × 6 = 153
Total of last six numbers = 40.5 × 6 = 243
sum of first six and last six numbers = 153+243 = 396
:. The sixth number = 396 – 363 = 33

Que.28
The ratio of milk and water in a mixture 5 : 4. If 45 litres of the mixture is taken out and replaced by the same amount of water then the ratio of milk and water becomes 5:7.Find the amount of milk in the original mixture.
A) 100 litres
B) 80 litres
C) 90 litres
D) 75 litres
E) none of these

Explanation:-
After 45 litres of mixture taken out,
Milk taken out = 45 × 5/9 = 25 litres
Water taken out = 45 × 4/9 = 20 litres
Let, Milk in original mixture = 5x & water = 4x
According to the question,
(5x – 25)/( 4x – 20 + 45) = 5/7
=> 35x – 175 = 20x + 125
=> 15x = 300
:. x = 20
Thus, Milk in the original mixture = 5×20 = 100 litres

Que.29
The average cost price of two shirts A and B is Rs.400 and the profit per cent on these shirts are 10% and 20% respectively. If the total selling price of these shirts is Rs.928, then find the cost price of shirt B.
A) Rs.360
B) Rs.320
C) Rs.480
D) Rs.420
E) none of these

Explanation:-
Total cost price of two shirts = 400×2 = 800
Overall Profit % = (928-800)/800 ×100% = 128/8 % = 16%
Applying alligation method,
10%          20%
\       /
\ /
16%
/       \
/            \
4       :       6
=  2       :       3
:. Cost price of shirt B = 3/5 × 800 =Rs. 480

Que.30
The distance covered by a boat in 4 hours 48 minutes is 7800 km downstream. If the ratio of the speed of stream and the speed of the boat in still water is 2:3.Find the upstream speed of the boat.
A) 280 km/hr
B) 270 km/hr
C) 375 km/hr
D) 325 km/hr
E) none of these

Explanation:-
4 hours 48 mins =( 4 + 48/60) hours = 24/5 hours
Downstream speed = 7800/(24/5) km/hr =( 7800 × 5)/24 = 1625 km/hr
Let, Speed of stream = 2x and speed of boat = 3x
Upstream speed = 3x – 2x = x
Downstream speed = 2x + 3x = 5x
So,
5x = 1625
:. x = 325 km/hr
:. The upstream speed of boat = 325 km/hr

Directions ( 31 – 35):-Read the following table carefully and answer the questions given below.
The following table shows the number of males and females living in 5 different societies A, B, C, D and E.

Que.31
Find the total number of persons living in society B and D together.
A) 1200
B) 1250
C) 1350
D) 1300
E) 1100

Explanation:-
Total persons living in society B = (350+500) = 850
Total persons living in society D = (300+150)  = 450
:. Total number of persons living in society B and D together
= ( 850 + 450) = 1300

Que.32
Find the ratio between the number of males in society C to the number of females in society E.
A) 3:5
B) 5:4
C) 4:9
D) 5:3
E) 2:3

Explanation:-
Required ratio = 250 : 200 = 5 : 4

Que.33
The number of males living in society D is what per cent of the number of females living in society B?
A) 20%
B) 25%
C) 75%
D) 66.66%
E) 60%

Explanation:-
Required percent = 300/500 × 100 % = 60%

Que.34
What is the difference between the total number of persons living in society A and the total number of persons living in society C?
A) 150
B) 100
C) 200
D) 50
E) None of these

Explanation:-
Total persons living in A =( 200+250) = 450
Total persons living in C =( 250+300) = 550
:. Required difference = ( 550 – 450) = 100

Que.35
The number of females living in society D is how much per cent less than the number of males living in society A?
A) 20%
B) 33.33%
C) 25%
D) 10%
E) 40%