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SBI Clerk Prelims Quantitative Aptitude Question Paper 2019

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  • Difficulty Level : Basic
  • Last Updated : 14 Jun, 2022
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Directions (1 – 5): What will come in the place of question mark (?) in the following number series:

Que.1
11, ?, 16, 21, 29, 41 
(a) 12
(b) 14
(c) 15
(d) 13
(e) 11

Answer : D
Explanation:-
11    (13)   16     21    29    41
    \/       \/     \/     \/       \/
    2        3      5      8      12
         \/     \/      \/     \/
          1      2       3      4

Que.2
1800, ?, 60, 15, 5, 2.5 
(a) 300
(b) 600
(c) 120
(d) 240
(e) 360

Answer:-A
Explanation:-
2.5 × 2 = 5
5 × 3    = 15
15 × 4  = 60 
60 × 5  = 300
300 × 6 = 1800

Que.3
4, 3, 4, 9, 32, ? 
(a) 75
(b) 155
(c) 125
(d) 175
(e) 165

Answer :-B
Explanation:-
(4 × 1 ) – 1 = 3
(3 × 2) – 2 = 4
(4 × 3) – 3 = 9
(9 × 4) – 4 = 32
(32 × 5) – 5 = 155

Que.4
?, 100, 150, 375, 1312.5 
(a) 50
(b) 100
(c) 75
(d) 25
(e) 200

Answer :-E
Explanation:-
200 × 0.5 = 100
100 × 1.5 = 150
150 × 2.5 = 375
375 × 3.5 = 1312.5

Que.5
0, 6, 24, 60, ?, 210 
(a) 130
(b) 170
(c) 90
(d) 120
(e) 150

Answer : D
Explanation:-
1³ – 1 = 0
2³ – 2 = 6
3³ – 3 = 24
4³ – 4 = 60
5³ – 5 = 120
6³ – 6 = 212

Directions (6 – 10): Study the bar chart given below and answer the following questions.
 

The bar chart shows the number of books read by 4 different persons (A, B, C & D) in 2005 and 2006.

 

Solution (6-10):-
A read in 2005 = 72
A read in 2006 = 90
B read in 2005 = 60
B read in 2006 = 78
C read in 2005 = 90
C read in 2005 = 72
D read in 2005 = 48
D read in 2006 = 70

Que.6
Find average number of books read by A, C & D in 2005.
(a) 64
(b) 70
(c) 75
(d) 60
(e) 56
Answer :-B
Explanation:-
Average number of books read by (A+C+D) together in 2005
= (72+90+48)/3 = 70

Que.7
Find the ratio of books read by B & C together in 2005 to books read by A & D together in 2006.
(a) 15 : 16
(b) 5 : 6
(c) 1 : 5
(d) 4 : 7
(e) 2 : 3

Answer :-A
Explanation:-
Books read by (B+C) together in 2005 = (60+90) = 150
Books read by (A+D) together in 2006 = (90+70) = 160
:. Required ratio = 150 : 160 = 15 : 16

Que.8
Books read by A & D together in 2005 are what percent more than books read by C in 2006? 
(a) 46 ⅔%
(b) 54 ¼ %
(c) 25 ⅔ %
(d) 33⅓ %
(e) 66 ⅔ %

Answer :- E
Explanation:-
Books read by (A+D) together in 2005 = (72+48) = 120
Books read by C in 2006 = 72
:. Required more percent = (120-72)/72 × 100% = 66(⅔)

Que.9
Books read by A & C together in 2005 are how much more or less than books read by B & D together in 2006?
(a) 24
(b) 14
(c) 18
(d) 22
(e) 28

Answer :-B
Explanation:-
Books read by (A+C) together in 2005 = (72+90) = 162
Books read by (B+D) together in 2006 = (78+70) = 148
:. Required more books read = (162-148) = 14

Que.10
Books read by B & C together in 2006 are what percent of books read by B in 2005? 
(a) 100%
(b) 120%
(c) 250%
(d) 200%
(e) 160%

Answer :-C
Explanation:-
Books read by B & C together in 2006= (78+72) = 150
Books read by B in 2005 = 60
:. Required percent = 150/60 × 100% = 250%
 

Directions (11 – 20): – What will come in place of the (?) question mark in the following questions?

Que.11
(17.28 ÷?)/(3.6×0.2) = 200
(a) 120 
(b) 1.20
(c) 12
(d) 0.12 
(e) None of these

Answer : D
Explanation:-
(17.28 ÷?)/(3.6×0.2) = 200
=> (17.28 ÷?)/.72 = 200
=> 17.28 × 1/? = 200 × 0.72
=> 17.28 = 144 ×?
=>? = 17.28/144
:. ? = 0.12

Que.12
486 ÷ ? × 7392 ÷ 66 = 1008
(a) 54
(b) 55
(c) 52
(d) 53
(e) 51 

Answer :-A
Explanation:-
486 ÷ ? × 7392 ÷ 66 = 1008
=> 486 × 1/? × 7392 × 1/66 = 1008
=> 486 × 112 = 1008 ×?
=>? = (486 × 112 )/1008
:. ? = 54

Que.13
14 (2/7) of 4200 ÷ √576 =(?)½
(a) 125
(b) 225
(c) 25
(d) 50
(e) 625

Answer : D
Explanation:-
14 (2/7) of 4200 ÷ √576 =(?)½
=> 100/7 × 4200 × 1/24 = (?)½
=> 2500 = (?)½
=>? = √2500
:. ? = 50

Que.14
2/7 × 5/6 × 3/8 × ? = 90
(a) 1208
(b) 1108
(c) 1008
(d) 1128
(e) 1348

Answer :-C
Explanation:-
2/7 × 5/6 × 3/8 × ? = 90
=> 5/56 ×? = 90
=>? = 90×56/5
:. ? = 1008

Que.15
(0.05 × 6.25) ÷ 2.5 =?
(a) 12.55
(b) 0.125
(c) 0.115
(d) 1.25
(e) None of these

Answer :-B
Explanation:-
(0.05 × 6.25) ÷ 2.5 =?
=> 0.3125/2.5 =?
:. ? = 0.125

Que.16
1496 ÷ 17 = ?% of 220
(a) 25
(b) 40
(c) 50
(d) 75
(e) None of these

Answer :-B
Explanation:-
1496 ÷ 17 = ?% of 220
=> 88 = ?/100 × 220
=> ? = (88 ×10)/22
:. ? = 40

Que.17
(36% of 180) ÷ 0.4 = ?
(a) 160 
(b) 164
(c) 166
(d) 162
(e) 180

Answer : D
Explanation:-
(36% of 180) ÷ 0.4 =?
=> (36/100 × 180) ÷ 0.4 =?
=> (36 × 9)/(5×0.4) =?
:. ? = 162

Que.18
0.08% of 55500 – 16.4 = ?
(a) 26.6
(b) 28
(c) 29.2
(d) 30.4
(e) 32

Answer :-B
Explanation:-
0.08% of 55500 – 16.4 =?
=> 0.08/100 × 55500 – 16.4 =?
=> 44.4 – 16.4 =?
:. ? = 28

Que.19
35% of 150 × 16 = ? – 22
(a) 865
(b) 932
(c) 864
(d) 862
(e) None of these

Answer : D
Explanation:-
35% of 150 × 16 = ? – 22
=> 35/100 × 150 × 16 = ? – 22
=> 35 × 3 × 8 + 22 =?
=>? = 840 + 22
:. ? = 862

Que.20
(3080 + 6160) ÷ ? = 330 
(a) 26 
(b) 22
(c) 28
(d) 29 
(e) 18

Answer :-C
Explanation:-
(3080 + 6160) ÷ ? = 330 
=> 9240/? = 330
=>? = 9240/330
:. ? = 28

Que.21
The difference between the compound interest received in the first year and second year at 20% per annum at CI is Rs 1200 then find the sum?
(a) Rs 25,000
(b) Rs 36,000
(c) Rs 35,000
(d) Rs 24,000
(e) Rs 30,000

Answer :-E
Explanation:-
CI rate for First year = 20%
CI rate for second years = 20 + 20 × 20/100 = 24%
Rate difference = (24-20)% = 4%
4% = 1200
=> 1% = 300
:. 100% = 30000
:. Required sum = 30,000

Que.22
Find the total distance covered by boat in each upstream and downstream in 7 hours if the speed of boat in still water and the speed of the current is 21 km/h and 3 km/h respectively?
(a) 280 km
(b) 294 km
(c) 315 km
(d) 301 km
(e) 322 km

Answer :-B
Explanation:-
Upstream speed = ( 21 – 3) = 18 km/hr
Downstream speed = ( 21 + 3) = 24 km/hr
:. Required total distance covered

= ( 24 × 7 + 18 × 7) Km
 = 294 km

Que.23
The ratio of income of A to that of B is 5:9. If the expenditure of A is 3/8th of his income and expenditure of B is 4/9th of his income and the sum of their saving is Rs 1950 then find the difference between their income?
(a) Rs 900
(b) Rs 1000
(c) Rs 880
(d) Rs 960
(e) Rs 920

Answer : D
Explanation:-
Let, 
Income of A = 5x
Income of B = 9x
Savings of A = (5x – 5x * 3/8) = 25x/8
Savings of B = ( 9x – 9x * 4/9) = 5x
According to question,
(25x/8 + 5x) = 1950
=> 65x/8 = 1950
=> x = (1950*8)/65
:. x = 240
:. Difference between their income 

= (9x-5x) =4 * 240 = 960

Que.24
A alone can do a work in 12 days while A and B together can do that work in 7.5 days. Find the time taken by C alone to do that work if C takes 3 days more than that of B alone to do that work?
(a) 33 days
(b) 30 days
(c) 23 days
(d) 27 days
(e) 28 days

Answer:-C
Explanation:-
A’s 1 day work = 1/12
(A+B)’s 1 day work = 1/7.5 = 2/15
B’s 1 day work = (2/15 – 1/12) = 1/20
B alone can do the work in  20 days
So, C alone can do the work in (20+3) = 23 days.

Que.25
The ratio of base and perpendicular side of a right-angled triangle is 3:4 and its base is equal to the side of a square having area 81 cm². Find the perimeter of the triangle?
(a) 30 cm
(b) 36 cm
(c) 33 cm
(d) 42 cm
(e) 40 cm

Answer:-B
Explanation:-
Let , Base and perpendicular of right angle triangle be 3x & 4x respectively.
Area of square = 81 cm²
:. Side of area = √81 = 9 cm = base of the triangle
Now, 3x = 9
:. x = 3
Base = 9 cm & perpendicular = 4x = 4*3 = 12 cm
Hypotenuse = √(9²+12²) = √225 = 15 cm
:. Perimeter of the triangle = ( 9 + 12 + 15) cm = 36 cm

Directions (26 – 30): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
A) x > y
B) x < y
C) x ≥ y
D)  x ≤ y
E)  x = y or the relation cannot establish.

Que.26
I. x² – 13x + 40 = 0
II. 2y² – y – 15 = 0

Answer :-A
Explanation:-
x² – 13x + 40 = 0
=> x² – 8x – 5x + 40 = 0
=> (x – 8)(x – 5) = 0
:. x = 8, 5
2y² – y – 15 = 0
=> 2y² – 6y + 5y – 15 = 0
=> (2y +5)(y – 3) = 0
:. y = 3, -5/2
x > y

Que.27
I. 5x² + 17x + 6 = 0
II. 2y² + 11y + 12 = 0

Answer :-E
Explanation:-
5x² + 17x + 6 = 0
=> 5x² + 15x + 2x + 6 = 0
=> (5x + 2)(x + 3) = 0
:. x = -3, -2/5

2y² + 11y + 12 = 0
=> 2y² + 8y +3y + 12 = 0
=> (2y + 3)(y + 4) = 0
:. y = -4, -3/2
Relationship can not establish.

Que.28
I. 7x² – 19x + 10 = 0
II. 8y² + 2y – 3 = 0

Answer :-A
Explanation:-
7x² – 19x + 10 = 0
=> 7x² – 14x – 5x + 10 = 0
=> (7x -5)( x – 2) = 0
:. x = 2, 5/7

8y² + 2y – 3 = 0
=> 8y² + 6y – 4y – 3 = 0
=> ( 2y – 1)(4y + 3) = 0
:. y = 1/2 , -3/4
x > y

Que.29
I. x² – 8x + 15 = 0
II. y2 – 3y + 2 = 0

Answer :-A
Explanation:-
x² – 8x + 15 = 0
=> x² – 5x – 3x + 15 = 0
=> (x – 5)( x – 3) = 0
:. x = 5, 3
y2 – 3y + 2 = 0
=> y² – 2y – y + 2 = 0
=> (y – 2)(y -1) = 0
:. y = 2, 1
x > y

Que.30
I. 3x² –7x + 4 = 0 
II. 2y² – 9y + 10 = 0 

Answer :-B
Explanation:-
3x² –7x + 4 = 0 
=> 3x² – 3x – 4x + 4 = 0
=> (3x – 4)( x – 1) = 0
:. x = 1, 4/3

2y² – 9y + 10 = 0 
=> 2y² – 4y – 5y + 10 = 0
=> (2y – 5)( y – 2) = 0
=> y = 2, 5/2
x < y

Que.31
A person travels half of the distance at the speed of x km/h and the remaining half of the distance at 4x km/h. Find the value of ‘x’ if the average speed is 36.8 km/h?
(a) 21
(b) 25
(c) 24
(d) 23
(e) 20

Answer : D
Explanation:-
Average speed = 2*x*4x/(x+4x) = 8x²/5x = 8x/5
According to question,
8x/5 = 36.8
=> x = (36.8 * 5)/8
:. x = 23

Que.32
A, B and C invested in a ratio of 7: 8: 5 in a business. They got an annual profit of Rs. 136800. If A and C withdrew their amount at the end of 3 months and 7 months respectively. Then find the difference between A and C’s share of profit?
(a) Rs. 12,600
(b) Rs. 11,500
(c) Rs. 13,500
(d) Rs. 10,500
(e) Rs. 13,000

Answer :-A
Explanation:-
Let, Investment of A ,B and C be 7x,8x and 5x respectively.
Profit ratio =( 7x * 3) : ( 8x * 12) : ( 5x * 7) 
                   = 21x : 96x : 35x
:. Required difference = 136800 * (35x-21x)/152x
 = 136800 * 14/152 = 12600

Que.33
The retailer sold one article at 33 ⅓%  profit and another at 100% profit. Find his overall profit percentage if the selling price of both articles is the same?
(a) 60%
(b) 55%
(c) 66⅔%
(d) 75%
(e) 56⅔%

Answer :-A
Explanation:-
Let,
Cost article of two articles be X and Y respectively.
Selling price of one article = X * 400/300 = 4X/3
Selling price of another article = Y * 200/100 = 2Y
According to question,
4X/3 = 2Y
:. X/3 = Y/2
Now,overall Profit = (4X/3 + 2Y) – (X+Y) =( X/3 + Y)
Overall profit percent 

= (X/3 + Y)/(X+Y) * 100%
= (Y/2 + Y)/(3Y/2 +Y) * 100% = 60%

Another method:-
 CP     SP
100     400/3      ———-(×3)
100      200        ———–(×2)
[Since selling price of two articles is same]
CP       SP
300      400
200      400
Total Cost price(CP) = (300+200) =500
Total Selling price(SP) = (400+400) =800
Profit = (800-500) = 300
Overall profit percent = 300/500 × 100 % = 60%

Que.34
A mixture has milk and water in a ratio 4: 1. When 50% of the mixture is taken out and replaced by 24 litres of water then the ratio of milk to water in the mixture becomes 1: 1. Find the initial quantity of the mixture.
(a) 80 litres
(b) 45 litres
(c) 70 litres
(d) 60 litres
(e) 75 litres

Answer :-A
Explanation:-
Let, Initial mixture = 5x
Initial Milk = 4x
Initial Water = x 
Mixture taken out = 5x/2 = 2.5x
Milk taken out = 2x
Water taken out = x/2
Milk remaining = 4x – 2x = 2x
Water remaining = x – x/2 = x/2
According to question,
2x/(x/2 + 24) = 1/1
=> 2x = 0.5x + 24
=> 3/2x = 24
:. x = 16
:. Initial mixture = 5x = 5 × 16 = 80 litres.

Que.35
4 years ago, ratio of Shivam’s age to Deepak’s age was 2: 3 and ratio of Shivam’s age 4 years ago to Deepak’s age 5 years hence is 8: 15. Find present age of Shivam.
(a) 32 years
(b) 28 years
(c) 40 years
(d) 24 years
(e) 36 years

Answer :-B
Explanation:-
Let,
Present age of Shivam = X years
Present age of Deepak = Y years
According to question,
(X – 4)/(Y – 4) = 2/3.           ————-(1)
(X – 4)/(Y + 5) = 8/15.        ————-(2) 
Solving equations (1) & (2),we get
X = 28
Y = 40
:. Present age of Shivam = 28 years.


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