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1. A train travelling with a speed of 60 kmph catches another train travelling in the same direction and then leaves it 120 m behind in 18 seconds. The speed of the second train is

A. 26 kmph

B. 35 kmph

C. 36 kmph

D. 63 kmph

Solution:

Let speed of the 2^nd train is S m/sec.
And,
60 kmph

= \frac{60 \times 5}{18} = \frac{50}{3}

m/sec.
As trains are traveling in same distance, Then Relative distance,

\begin{aligned} = \frac{60 \times 5}{18} = \frac{50}{3} \\ \frac{50}{3} - S = \frac{120}{18} \\ \Rightarrow \frac{50 - 3 S}{3} = \frac{20}{3} \\ \Rightarrow 50 - 3 S = 20 \\ \Rightarrow 3 S = 50 - 20 \\ \Rightarrow 3 S = 30 \\ ∴ S = 10 m/sec \end{aligned}

Or, Speed of the 2^nd train =

10 \times \frac{18}{5}

= 36 kmph.

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