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

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,
$$\eqalign{ & = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3} \cr & \frac{{50}}{3} - S = \frac{{120}}{{18}} \cr & \Rightarrow \frac{{50 - 3S}}{3} = \frac{{20}}{3} \cr & \Rightarrow 50 - 3S = 20 \cr & \Rightarrow 3S = 50 - 20 \cr & \Rightarrow 3S = 30 \cr & \therefore S = 10\,\,{\text{m/sec}} \cr} $$
Or, Speed of the 2^nd train = $${\text{10}} \times \frac{{18}}{5}$$  = 36 kmph.