After travelling 170 km, a train met with an accident. Its speed then became \(\frac{4}{5}\) of its usual speed, and it reached the station 36 minutes late. Had the accident happened 25 km further on, the train would have reached 24 minutes late. What is the reduced speed of the train (in km/hr)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 25
The delay arises only on the stretch run at the reduced speed. Moving the accident 25 km further on puts those 25 km at the usual speed, and that is what cuts the delay from 36 to 24 minutes.
Let the usual speed be \(v\) km/hr, so the reduced speed is \(\frac{4}{5}v\)
Time for 25 km at the reduced speed \(\displaystyle =\frac{25}{\frac{4}{5}v}=\frac{125}{4v}\) hr
Time for 25 km at the usual speed \(\displaystyle =\frac{25}{v}=\frac{100}{4v}\) hr
Extra time on these 25 km \(\displaystyle =\frac{125}{4v}-\frac{100}{4v}=\frac{25}{4v}\) hr
This equals the drop in delay: \(36-24=12\) min \(\displaystyle =\frac{1}{5}\) hr
\(\displaystyle \frac{25}{4v}=\frac{1}{5}\), so \(4v=125\) and \(v=31.25\) km/hr
Reduced speed \(\displaystyle =\frac{4}{5}\times31.25=25\) km/hr