How to Solve Speed Problems Involving Stoppage Time
Solve journey stoppage-time aptitude problems with the moving-speed vs effective-speed ratio formula, worked example and practice MCQs.
Expected Interview Answer
When a journey includes stoppage time, the speed excluding stoppages (moving speed) is always higher than the speed including stoppages (effective speed), and the gap between them reveals exactly how many minutes per hour the vehicle spends stopped.
The key relation is: stoppage time per hour = (Speed excluding stops − Speed including stops) / Speed excluding stops, all multiplied by 60 to convert into minutes. This works because in one hour of "including stoppage" travel, the vehicle only actually moves for a fraction of that hour, and that fraction covers the effective distance at the faster moving speed. So if a train’s speed without stops is 60 km/h but its effective (including stops) speed over a route is 50 km/h, it must have been stationary for the time it would have taken to cover that missing 10km at 60 km/h — 10 minutes out of every hour. The same principle inverts to find the effective speed if given the moving speed and stoppage minutes per hour.
- A single ratio formula finds stoppage minutes per hour directly
- Explains intuitively why effective speed is always lower with stops
- Generalizes to any two known speeds (with/without stoppage) plus one unknown
- Reinforces the same distance-time reasoning used in every speed-time-distance problem
AI Mentor Explanation
A bowler’s raw delivery pace when actually bowling a ball is like the speed excluding stoppages, but the over-rate the umpires track includes time spent walking back to the mark between deliveries, which is like the effective speed including stoppages. If a bowler could theoretically deliver a full over’s worth of balls in half the actual time an over takes on the clock, half the over’s duration was stoppage — the walk-back time. The gap between raw delivery pace and the clocked over-rate reveals exactly how much time per over is lost to non-bowling pauses, mirroring the stoppage-time formula precisely.
Worked example
Speed excluding stops
- 60 km/h
Speed including stops
- 50 km/h
Stoppage per hour
- (60-50)/60 × 60 = 10 min
Step-by-Step Explanation
Step 1
Identify both speeds
Determine speed excluding stoppages and speed including stoppages.
Step 2
Apply the stoppage ratio
Stoppage min/hr = (S_excl − S_incl)/S_excl × 60.
Step 3
Interpret the result
The result is minutes stopped per every 60 minutes of scheduled travel.
Step 4
Invert if needed
Given stoppage minutes and one speed, solve algebraically for the other speed.
What Interviewer Expects
- Correct stoppage-time ratio formula and its derivation
- Understanding why effective speed is always ≤ moving speed
- Ability to invert the formula to solve for an unknown speed
- Clear reasoning about what “per hour” means in the stoppage context
Common Mistakes
- Dividing by the effective (including-stoppage) speed instead of the excluding-stoppage speed
- Forgetting to multiply the ratio by 60 to convert to minutes
- Assuming stoppage time is a fixed absolute value rather than a per-hour rate
- Mixing up which of the two speeds is larger before applying the formula
Best Answer (HR Friendly)
“The moving speed, without any stops, is always faster than the effective speed once stoppages are factored in — the gap between the two tells you exactly how many minutes per hour are lost to stopping. The formula is the difference between the two speeds, divided by the moving speed, times 60 to get minutes. It is a clean, one-line ratio once you’re clear on which speed is the numerator and which is the denominator.”
Follow-up Questions
- How would you find the effective speed given the moving speed and stoppage minutes?
- How does this stoppage concept relate to average speed over a mixed journey?
- What if stoppage time varies between different legs of the same journey?
- How would you find total distance covered in a day given both speeds and total hours?
MCQ Practice
1. A bus travels at 50 km/h excluding stoppages and 40 km/h including stoppages. Stoppage time per hour is?
(50−40)/50 × 60 = 12 minutes.
2. A train’s speed excluding stoppages is 75 km/h and it stops for 8 minutes per hour. Its speed including stoppages is?
Moving 52/60 of the hour: 75 × 52/60 = 65 km/h.
3. If a vehicle never stops, what is its stoppage time per hour?
No stoppage means speed excluding and including stoppages are equal, giving 0 stoppage minutes.
Flash Cards
Stoppage time per hour formula? — (Speed excl. stoppage − Speed incl. stoppage) / Speed excl. stoppage × 60.
Which speed is always larger? — Speed excluding stoppages is always ≥ speed including stoppages.
What does the formula output represent? — Minutes stopped per every 60 minutes of scheduled travel.
How to find effective speed from moving speed and stoppage minutes? — Effective speed = Moving speed × (60 − stoppage minutes) / 60.