Question

A train leaves the station at 11 a.m. traveling at the rate of 40 miles per hour. A faster train leaves the same station at 1 p.m. that afternoon and travels in the same direction on a parallel track at a rate of 60 miles per hour. At what time will the faster train overtake the slower one

Answer 1

The faster train overtake the slower on 5 p.m.

How to solve such time related questions?

The key concept for solving such questions is the relationship between speed, distance and time.

The relationship between them is Distance = Speed x Time.

Let the number of hours after 1pm at which faster train will overtake slower train be x.

At overtaking point,

Distance travelled by slower train = Distance travelled by faster train

Given :

• Speed of slower train = 40 miles per hour
• Speed of faster train = 60 miles per hour
• Time taken by faster train = x
• Time taken by slower train = x + 2.

On equating distances we get

40 x ( x + 20) = 60 x (x)

40x + 80 = 60x

20x = 80

x = 4.

∴The faster train overtake the slower on 1 + 4 = 5 p.m.

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