Question

A train X leaves station A at 3 p.m. and reaches station B at 4.30 p.m, while another train A leaves station B at 3.00 p.m. and reaches station A at 4.00 p.m. These two trains cross each other at
a)3.36 p.m.
b)3.30 p.m.
c)3.20 p.m.
d)3.40 p.m.

Answer 1

Hint: To solve this question, we use the formula for speed s = $\dfrac{{\text{d}}}{{\text{t}}}$,which means speed equals distance divided by time. And similarly, to solve for time use the formula for time, t = $\dfrac{{\text{d}}}{{\text{s}}}$ which means time equals distance divided by speed.

Complete step-by-step answer:

As given in question,

Train X takes =1.5 hrs. to go from A to B

Train Y takes =1 hr. to go from B to A

Let Distance between AB=3 kms

Speed of train X =2 kmph

Speed of train Y =3 kmph

And we know the formula,

                    t = $\dfrac{{\text{d}}}{{\text{s}}}$

where, t- time, d- distance and s- speed

Time taken by trains to meet = $\dfrac{{\text{3}}}{{\left( {{\text{2 + 3}}} \right)}}$= 0.6 hrs.

i.e. 36 minutes after 3 hours

Therefore, these two trains cross each other at 3:36 p.m.

So, option (a) is the correct answer.

Note- Speed and time are inversely proportional (when distance is constant)

                                    ⟹ speed ∝ $\dfrac{{\text{1}}}{{{\text{time}}}}$(when distance is
constant)

That means If the ratio of the speeds of A and B is a: b, then, the ratio of the time taken by

them to cover the same distance is

$\dfrac{{\text{1}}}{{\text{a}}}{\text{:}}\dfrac{{\text{1}}}{{\text{b}}}$= b : a