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Q.

A train X leaves station A at 3 p.m and reaches station B at, while train Y leaves station B and reaches station at 4:30 p.m, A at 4 p.m. The two trains cross each other at:

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a

3:36 p.m

b

3:30 p.m

c

3:20 p.m

d

3:40 p.m

answer is A.

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Detailed Solution

Concept: We will let the distance between the stations evaluate the speeds of each train, and then we assume the time after they meet. We will equate the distance traveled by each train to the total distance and evaluate the time.
Let the distance between two stations A and B be

d .

speed = \frac{distance}{time},

Here, the time taken by train X is 1.5 hours, so the speed of the train X =

\frac{d}{1.5} = \frac{2 d}{3} km h^{- 1}

and, the time taken by train Y is 1 hours, so the speed of the train Y =

\frac{d}{1} = d km h^{- 1}

Let the train meet

h

hours after 3p.m,
Now,
Distance traveled by train X in

x

hours =

x \times \frac{2 d}{3} km

Distance traveled by train Y in

x

hours =

x \times d  km

For the total distance, as the total distance between the station is

d

,

\Rightarrow d = x \times \frac{2 d}{3} + x \times d

\Rightarrow \frac{2 x}{3} + x = 1

\Rightarrow \frac{5 x}{3} = 1

\Rightarrow x = \frac{3}{5}

The trains meet

\frac{3}{5}

hours after 3p.m.

\frac{3}{5}

hours into minutes =

\frac{3}{5} \times 60 = 36 minutes

That means, both the trains will meet at 3:00 p.m + 36 minutes = 3:36 p.m Hence, the correct answer is option 1.

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