Test Index

IBPS PO Mains 5 Nov 2023 Paper

Question Numbers: 151-152

There are four cities A, B, C, and D which are located at different places such that city B is the east of city A, city C is to the south of city B, and city D is to the west of city C and south of city A. Distance between cities B and C is

(20p + 100)

km and distance between cities C and D is

(p^{2} + 100)

km. Total distance from city A to city D (via cities B and C) is 2700 km.

Train Y takes a total 25 hours to cover the distance between A and D (via B and C).

Table given below shows the time taken by three trains X, Y, and Z to cross a pole and a platform P1.

Trains	Time taken (seconds) to cross pole	Time taken (seconds) to cross platform P1
X	10	$p$
Y	—	$q$
Z	—	$(p + 6)$

Question : 152 of 155

Marks: +1, -0

Find the time taken by train Z to cross the pole if the speed of train X is 20% more than the speed of train Z?

15 secs
16 secs
12 secs
13 secs
10 secs

Validate

Solution:

Concept:

When a train crosses a pole, the distance covered equals the length of the train.

When a train crosses a platform, the distance covered equals train length plus platform length.

Explanation:

Let the speed of train Z be

S_z

Since train X is 20% faster, its speed is

S_x = 1.2 S_z

From the given data, train X crosses a pole in 10 seconds and a 400 m platform in 30 seconds.

So,

L_x = 10 S_x

and

\frac{L_x + 400}{S_x} = 30

Substituting

L_x = 10 S_x

gives

\frac{10 S_x + 400}{S_x} = 30

This gives

10 S_x + 400 = 30 S_x

, so

S_x = 20~\mathrm{m/s}

Therefore,

S_z = \frac{20}{1.2} = \frac{100}{6}~\mathrm{m/s}

Train Z crosses the 400 m platform in 36 seconds:

\frac{L_z + 400}{100/6} = 36

So,

L_z + 400 = 600

, giving

L_z = 200

Time taken by train Z to cross the pole is

t = \frac{L_z}{S_z} = \frac{200}{100/6} = 12

seconds.

Answer:

The time taken by train Z to cross the pole is 12 seconds, i.e. option C.

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