A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to

Let the usual speed of the train be v and the scheduled time be t minutes.

When the train travels at speed $\frac{v}{3}$, the time taken is inversely proportional to the speed. Since speed becomes $\frac{1}{3}$, the time taken becomes 3t.
We are given that the delay is 30 minutes:
3t - t = 30 ⇒ 2t = 30 ⇒ t = 15 minutes.
So the scheduled duration for the journey is 15 minutes.

On the return journey:
The train travels at usual speed for 5 minutes (remaining scheduled time = 10 minutes).
It stops for 4 minutes (remaining scheduled time = 10 - 4 = 6 minutes).
The remaining distance is what would normally be covered in 10 minutes at the usual speed v.

Let the new speed be v'. The train must cover this remaining distance in 6 minutes:
v' × 6 = v × 10
$v^{\prime }=\frac{5}{3}v$

The fractional increase in speed is:
$\frac{v^{\prime }-v}{v}=\frac{2}{3}\approx 66.67\%$

The nearest integer percentage is 67%.