Question

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

• A. 58
• B. 67
• C. 61
• D. 50

Answer 1

The Correct Option is B

The correct answer is (B) : \(67\)

Let the usual speed of the train be s and time taken at that speed be \('t'\).

Given by travelling at \(s/3\), it reached \(30\) min late. 

Hence the usual time:

Distance travelled = \(s×t\)

Distance travelled in the first \(5\) min = \(\frac{s×t}{3}.D\)

Distance to be travelled in the last \(6\) min = \(\frac{2st }{ 3}\)

Required speed to cover that distance on time = \(\frac{\frac{2st}{3}}{\frac{2t}{5}}\) i.e., \(\frac{5s'}{3}\)

Hence the percentage increase in its speed = \(\bigg(\frac{2}{3}\bigg)×100\) i.e., \(66\frac{2}{3}\%\) or \(67\%\)

Answer 2

Let's say the train's regular speed is 's' and the usual time it takes is 't'.
When it travels at \(\frac{s}{3}\), it's 30 minutes late.
So, the usual time:
Distance covered = s × t
Distance covered in the first 5 minutes = s ×\(\frac{t}{3}\)
Distance to cover in the last 6 minutes = 2s \(\frac{t}{3}\)
To be on time, the speed required for that distance is 5 \(\frac{s}{3}\)
So, the percentage increase in speed is \((\frac{(\frac{5}{3}) }{(\frac{2}{3}) })\)× 100, which simplifies to 67%.