BANK & INSURANCE (SBI CLERK PRELIMS 2025) MOCKTEST 4

Total Questions: 50

21. 40% of total students who appeared in exam B are girls and 80 girls did not pass the exams out of total girls appeared in exam B. Find the difference between number of girls who passed the exam B and number of boys who did not pass the exam B?

Correct Answer: (b) 90 
Solution:

Total girls appeared in exam B = 450 × 40/100 = 180
Total boys appeared in exam B = 450 − 180 = 270
Total passed girls in exam B = 180 − 80 = 100
Total passed boys in exam B = 180 − 100 = 80

So, boys who did not pass the exam B = 270 − 80 = 190

Required difference = 190 − 100 = 90

22. Four years ago, the ratio of ages of Sonam and Niharika was 5:7. 12 years hence, the ratio between the ratio of ages of Sonam and Niharika will be 9:11. The present age of Sonam and Niharika are X years and Y years respectively

Quantity I: X − 4Y/8
Quantity II: 0.2Y + 1.5X

Correct Answer: (b) Quantity I < Quantity II
Solution:

Let the present ages of Sonam and Niharika be 5x + 4 and 7x + 4 years, respectively.

Given,
(5x+4+12)/(7x+4+12) = 9/11

55x + 176 = 63x + 144
32 = 8x
4 = x

X = 5x + 4
X = 5(4) + 4 = 24

Y = 7x + 4
Y = 7(4) + 4 = 32

Quantity I: X − 4Y/8 = 24 − (4×32)/8
= 24 − 16
= 8

Quantity II: 0.2Y + 1.5X
= 0.2×32 + 1.5×24
= 6.4 + 36
= 42.4

So, Quantity I < Quantity II

23. Given, ‘a’ and ‘b’ are two distinct positive integers such that 4b + 2a = 24 and (6b − a)/7 = 4.

Quantity I: value of 2a
Quantity II: value of b

Correct Answer: (b) Quantity I < Quantity II
Solution:

4b + 2a = 24 ...(i)

And (6b − a)/7 = 4
6b − a = 28
6b − 28 = a

a value put in (i)
4b + 2(6b − 28) = 24
4b + 12b − 56 = 24
16b = 80
b = 5

6b − 28 = a
30 − 28 = a
2 = a

Quantity I: 2a = 4
Quantity II: b = 5

So, Quantity I < Quantity II

24. The average of three numbers P, Q, and R is 16, and another number S is 9 more than the average of Q and R. Find the value of 2S + P.

Correct Answer: (c) 66
Solution:

Given, P + Q + R = 48 ...(i)

And
S − 9 = (Q + R)/2

2S − 18 = Q + R

Q + R value put in (i)

P + 2S − 18 = 48
∴ 2S + P = 66

25. I: 3x² − 16x + 5 = 0 II: 2y² − 7y + 3 = 0

Correct Answer: ((e) x = y or relation cannot be determined
Solution:

I: 3x² − 16x + 5 = 0
3x² − 15x − x + 5 = 0
3x(x − 5) − 1(x − 5) = 0
(x − 5)(3x − 1) = 0
x = 5, 1/3

II: 2y² − 7y + 3 = 0
2y² − 6y − y + 3 = 0
2y(y − 3) − 1(y − 3) = 0
(2y − 1)(y − 3) = 0
y = 1/2, 3

So, relation cannot be determined.

26. 50% of the solid sphere is melted to form cubes of sides (64/336)th of the radius of the sphere. If the curved surface area of the sphere is 5544 sq. cm. Find the maximum number of cubes that can be formed.

Correct Answer: (e) 303
Solution:

Let the radius of the sphere be R cm

Given,
4 × 22/7 × r × r = 5544

r × r = 441
r = 21

Side of the cube = 21 × 64/336 = 4 cm

Let n number of cubes formed

ATQ,
1/2 × 4/3 × 22/7 × 21 × 21 × 21 = n × 4 × 4 × 4

n = 303.1875

So, maximum number of cubes formed = 303

27. I: 0.5x² + √144x² + √169x² + 3.5x = ³√343 + 22

II: 9y² + 16y(0.5 × √36) − √289 = 0

Correct Answer: (e) x = y or relation cannot be determined
Solution:

I: 0.5x² + √144x² + √169x² + 3.5x = ∛343 + 22

x²/2 + 12x + 13x + 3.5x = 29
x²/2 + 28.5x = 29

x² + 57x − 58 = 0
x² + 58x − x − 58 = 0
x(x + 58) − 1(x + 58) = 0
(x + 58)(x − 1) = 0

x = −58, 1

II: 9y² + 16y(0.5 × √36) − √289 = 0
9y² + 48y − 17 = 0
9y² + 51y − 3y − 17 = 0
3y(3y + 17) − 1(3y + 17) = 0
(3y − 1)(3y + 17) = 0

y = 1/3, −17/3

So, relation cannot be determined.

28. I: 3x² − 5x − 2 = 0

II: 2y² − 3y + 1 = 0

Correct Answer: (e) x = y or relation cannot be determined
Solution:

I: 3x² − 5x − 2 = 0
3x² + x − 6x − 2 = 0
(x − 2)(3x + 1)

x = 2, −1/3

II: 2y² − 3y + 1 = 0
2y² − 2y − y + 1 = 0
2y(y − 1) − 1(y − 1) = 0
(2y − 1)(y − 1) = 0

y = 1/2, 1

So, relation cannot be determined

29. Vessel P contains 56 liters of a mixture of milk and water in the ratio of 3:1, and vessel Q contains 40% water and the rest milk. 25% of mixture Q and 60% of mixture P are taken out and poured into empty vessel R. If the ratio of water to milk in vessel R is 52:141, respectively, then how much milk was taken out from vessel Q (in liters)?

Correct Answer: (a) 3 
Solution:

In vessel P,
Milk = 56 × 3/4 = 42 litres
Water = 56 − 42 = 14 litres

Let the quantity of mixtures in vessel Q be 20x litres

Milk = 20x × 60/100 = 12x litres
Water = 20x − 12x = 8x litres

Quantity of mixture taken out from vessel Q
= 25/100 × 20x = 5x litres

Quantity of mixtures taken out from vessel P
= 60/100 × 56 = 33.6 litres

ATQ,
[(42×60)/100 + (12x x 25/100)]/[(14*60/100) + (8x x 25/100) = 141/52​

⇒ 52(25.2 + 3x) = 141(84 + 2x)

⇒ 1310.4 + 156x = 1184.4 + 282x

⇒ x = 1

Required answer = 12x × 23/100 − 3x
= 3x = 3 litres

30. a/b/c is a mixed fraction and the whole number P is prime number. The denominator is Q, which is a multiple of 2. The product of the whole number and the denominator is 18. If the product of the denominator and numerator is 24, then find the value of b.

Correct Answer: (b) 4 
Solution:

Q − c = multiple of 2
P = a = prime number

Given, P × Q = 18
Possible values of P = 2, 3, 5, 7, 9, 11, 13 & 17
Only P = 3 and Q = 6 is satisfy the condition

a  b/c = 3  b/6

Also given, product of the denominator and numerator is 24, i.e.

b = 24/6 = 4