Skip to main contentSkip to solution

A train travelled at one-thirds of its usual speed, and hence reached the destination 3030 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 55 minutes but then stopped for 44 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

Solution

✅ Correct Option: 4

A train has two journeys to analyze:

Forward journey: Travels at 13\tfrac{1}{3} usual speed, arrives 30 minutes late

Return journey: Travels normally for 5 minutes, stops 4 minutes, then needs to speed up

We need to find the percentage increase in speed required for the return journey.


Let's define our variables:

Usual speed = S

Usual time for the journey = T

Distance = D = S × T

Forward Journey Analysis:

Speed used = S3\tfrac{S}{3} (one-third of usual speed)

Time taken = DS/3=3DS=3T\frac{D}{S/3} = \frac{3D}{S} = 3T

When speed is reduced to 13\tfrac{1}{3}, the time increases by a factor of 3 since Speed × Time = Distance (constant).

Finding T:

Delay = Time taken - Usual time = 3T−T=2T=303T - T = 2T = 30 minutes

Therefore: T=15T = 15 minutes

The usual journey time is 15 minutes.


Initial Conditions:

Total scheduled time = T=15T = 15 minutes

Distance to cover = D=S×15D = S × 15

Journey Breakdown:

First 5 minutes: Travels at usual speed S

Distance covered = S×5=5SS × 5 = 5S

Remaining distance = D−5S=15S−5S=10SD - 5S = 15S - 5S = 10S

Time Analysis:

Stop time: 4 minutes

Remaining time = 15−5−4=615 - 5 - 4 = 6 minutes


Speed Calculation:

Distance remaining = 10S10S

Time remaining = 6 minutes

Required speed = 10S6=5S3\frac{10S}{6} = \frac{5S}{3}

Speed Increase:

Usual speed = SS

New speed = 5S3\frac{5S}{3}

Increase = 5S3−S=5S3−3S3=2S3\frac{5S}{3} - S = \frac{5S}{3} - \frac{3S}{3} = \frac{2S}{3}

Percentage Increase:

Percentage = IncreaseUsual speed×100%\frac{\text{Increase}}{\text{Usual speed}} × 100\%

Percentage = 2S/3S×100%=23×100%=66.67%\frac{2S/3}{S} × 100\% = \frac{2}{3} × 100\% = 66.67\%


The train must increase its usual speed by 66.67% (approximately 67%) to reach the destination on time.

In time-speed-distance problems, when speed changes by a factor, time changes by the reciprocal factor (assuming distance stays constant).

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question