Train speed increase for return journey

Let the normal speed be $s$ and scheduled time be $t$ minutes.

When speed becomes $\frac{s}{3}$, time taken becomes $3t$. Given $3t - t = 30 \implies 2t = 30 \implies t = 15$ minutes.

On the return journey:

• Total scheduled time available $= 15$ minutes.
• The train travels for 5 minutes at usual speed $s$. Remaining scheduled distance corresponds to $15 - 5 = 10$ minutes of travel at usual speed.
• The train stops for 4 minutes, leaving $15 - 5 - 4 = 6$ minutes to complete the remaining distance.

Let the new speed be $s'$. To cover the remaining distance: $s' \times 6 = s \times 10 \implies s' = \frac{10}{6}s = \frac{5}{3}s$

Percentage increase in speed: $\left(\frac{5/3 - 1}{1}\right) \times 100\% = \frac{2}{3} \times 100\% = 66.67\% \approx 67\%$