Question 4.

Let the total distance be ‘D’ km and the speed of the train be ’s’ kmph. The time taken to cover D at speed d is ’t’ hours. Based on the information: on equating the distance, we get

$s \times t = \frac{s}{3} \times (t + \frac{1}{2})$

On solving we acquire the value of $t = \frac{1}{4}$ or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed ’s’ {distance traveled in terms of s = $\frac{s}{12}$}. Remaining distance in terms of ’s’ = $\frac{s}{4} - \frac{s}{12}$ = $\frac{s}{6}$ 

The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in $15 - 9 = 6$ mins or {$\frac{1}{10}$ hours}. Thus, let ‘x’ kmph be the new value of speed. Based on the above, we get $\frac{s}{6x} = \frac{1}{10}$ $or$ $x = \frac{10s}{6}$

Since the increase in speed needs to be calculated: $\frac{(\frac{10s}{6} - s)}{s} \times 100 = \frac{200}{3} \approx 67$% increase.

Hence, Option B is the correct answer.