BANK & INSURANCE (IBPS PO MAINS 2025) MOCKTEST 6

Total Questions: 35

31. Find the pattern of the series and answer the question given below.

Series: (K+1), (2^S − 2), (P² − 19), 120(K + 1), C

Note:
A: K = Square root of smallest two-digit perfect square.
B: P = Second smallest two-digit prime number.

C: S = Natural number, Maximum value of 2^S < 40.

Quantity I: 4 × 3rd term of the given series.
Quantity II: C/3
Quantity III: K × S³ + 100.

Correct Answer: (c) Quantity I = Quantity II = Quantity III (d) Quantity I ≤ Quantity II < Quantity III
Solution:

Smallest two-digit perfect square = 16
Square root of smallest two-digit perfect square = 4
K = 4

Second smallest two-digit prime number = 13
P = 13

Natural number, Maximum value of
2^S < 40

2^5 < 40
32 < 40

So, S < 5

Series = (4+1), (2^5−2), (13² − 19), 120(4 + 1), C
Series = 5, 30, 150, 600, C

The pattern of the series:

5, 30, 150, 600, C
×6 ×5 ×4 ×3

C = 600 × 3 = 1800

Quantity I: 4 × 3rd term of the given series
= 4 × 150 = 600

Quantity II: C/3 = 1800/3 = 600

Quantity III: K × S³ + 100 = 4 × 5³ + 100
= 600

So, Quantity I = Quantity II = Quantity III

32. I. (x(x − 5) + 20) / x = 9 − x II. y² + 2√17 y − 5² = √68 y

Correct Answer: (e) If x = y or no relation can be established between x and y
Solution:

x(x − 5) + 20 / x = 9 − x

x² − 5x + 20 = 9x − x²

2x² − 14x + 20 = 0

x² − 7x + 10 = 0

x² − 2x − 5x + 10 = 0

x = 2, 5

II: y² + 2√17y − 5² = √68 y
y² + 2√17 y − 5² = 2√17 y
y² = 25
y = 5, −5

So, no relation can be established between x and y

33. Find the ratio of P to Q. Find the pattern of the series and answer the question given below.

Series A: 0.5, 3, 16, 97, P, 5441
Series B: Q, 0-400, 3.5P+172, 2296, 2100, 1956

Correct Answer: (b) 170 : 819 
Solution:

The pattern of the series A:
0.5, 3, 16, 97, P=680, 5441

×4+1 ×5+1 ×6+1 ×7+1 ×8+1

The pattern of the series B:
Q, Q − 400, 3.5P + 172, 2296, 2100, 1956

Q  Q−400  2552  2296  2100  1956
400  324  256  196  144

Q − 400 − 324 = 2552
Q = 3276

Required ratio = 680 : 3276
= 170 : 819

34. Find the value of √(P − 4). Find the pattern of the series and answer the question given below.

Series A: 0.5, 3, 16, 97, P, 5441
Series B: Q, 0-400, 3.5P+172, 2296, 2100, 1956

Correct Answer: (b) 26 
Solution:

The pattern of the series A:
0.5, 3, 16, 97, P=680, 5441

×4+1 ×5+1 ×6+1 ×7+1 ×8+1

The pattern of the series B:
Q, Q − 400, 3.5P + 172, 2296, 2100, 1956

Q  Q−400  2552  2296  2100  1956
400  324  256  196  144

Q − 400 − 324 = 2552
Q = 3276

Required value = √(P − 4)

= √(680 − 4)
= √676
= 26

35. On a 300 km long racing track, three bikes — A, B, and C take part in a race. Initially, their speeds are 50 km/hr for Bike A, 40 km/hr for Bike B, and 60 km/hr for Bike C.

After the first 2 hours of racing, their speeds are altered as follows:
Bike A adopts a new speed which is 25% more than Bike C’s speed at that moment.
Bike B increases its speed by 10% compared to its original speed.
Bike C reduces its speed to 80% of Bike A’s original speed.

Find the time gap between the bikes that finished first and second.

Correct Answer: (e) 30 minutes
Solution:

In 2 hours,

Total distance covered by bike A = 2×50 = 100 km
Total distance covered by bike B = 2×40 = 80 km
Total distance covered by bike C = 2×60 = 120 km

After 2 hours, now the speeds

Speed of bike B = 40 × 110/100 = 44 km/hr
Speed of bike C = 50 × 80/100 = 40 km/hr
And, speed of bike A = 40 × 125/100 = 50 km/hr

Now, the time taken by bike to cover remaining distance.

Time taken by bike A = (300−100)/50 = 200/50 = 4 hours
Time taken by bike B = (300−80)/44 = 5 hours
Time taken by bike C = (300−120)/40
= 180/40 = 4.5 hours

Total time taken by each bike to cover 300 km,

Time taken by bike A = 2 + 4 = 6 hours
Time taken by bike B = 2 + 5 = 7 hours
Time taken by bike C = 2 + 4.5 = 6.5 hours

We know which two bikes taken least time cover 300 km reached first and second.

So, A and C taken least time respectively to cover 300 km.

So, time gap between the bikes that finished first and second = 6.5 − 6 = 30 minutes