How to Solve Races and Relative Start-Distance Problems
Solve race aptitude problems involving distance and time head starts, with a worked example, MCQs, and dead-heat problem strategy.
Expected Interview Answer
In race problems, 'A gives B a start of x meters’ means B only needs to cover (track length − x) while A covers the full track length in the same time, so the ratio of their speeds equals the ratio of distances each actually covers in that shared time.
If A finishes a race of length L while B, given a start of x meters, only had to run (L − x) in that same time, then the speed ratio A:B equals L : (L − x), since both runners take the same time to finish. A separate but related concept is 'A gives B a start of t seconds,' meaning B starts running t seconds before A, so B’s total running time exceeds A’s by t seconds when they finish together — this is a time head-start, not a distance head-start, and must be modeled with a time offset instead of a distance offset in the equations. The two forms combine in 'dead heat' problems, where the goal is to find the start (distance or time) that makes both racers finish simultaneously, solved by equating each racer’s finishing time.
- Converts a start distance directly into a clean speed ratio using L : (L−x)
- Distinguishes distance head-starts from time head-starts, which need different equations
- The dead-heat condition (equal finishing time) unifies both start types into one solvable equation
- Extends naturally to multi-runner race comparisons using chained speed ratios
AI Mentor Explanation
If a senior batter completes 30 runs’ worth of running between wickets while a junior batter, given a head start of 5 runs’ worth of ground already covered, only needs to cover 25 in the same time, their running-speed ratio is 30:25, or 6:5 — derived purely from the distances each actually covered in an identical time window. Race problems use exactly this logic: a start of x meters out of a total L means the ratio of speeds is L : (L−x).
Worked example
A's time
- 500m ÷ 10 m/s = 50 s
B's distance & time
- 400m (500−100 start)
- Time = 50+20 = 70 s
B's speed
- 400 ÷ 70 ≈ 5.71 m/s
Step-by-Step Explanation
Step 1
Identify the type of start
Distance head start reduces the distance the trailing runner covers; time head start shifts when they begin running.
Step 2
Compute actual distance covered
Trailing runner with distance start x covers (L − x) while the leader covers L.
Step 3
Set up the shared time or finish condition
Use equal finishing time, or a stated time gap, to relate the two runners' equations.
Step 4
Solve for the unknown speed, time, or start
Isolate the requested variable using speed = distance/time for each runner.
What Interviewer Expects
- Clear distinction between a distance head start and a time head start
- Correct reduction of the trailing runner's distance to (L − x)
- Accurate handling of finish-time differences stated in seconds
- Ability to set up and solve the dead-heat (equal finish) condition
Common Mistakes
- Treating a distance head start and a time head start with the same equation
- Forgetting the trailing runner still runs the reduced distance (L − x), not the full L
- Sign errors when a finish-time gap is added versus subtracted
- Not converting all given quantities to consistent units before solving
Best Answer (HR Friendly)
“I first figure out whether the start is a distance head start or a time head start, since they need different setups. For a distance start, the trailing runner only covers the track length minus the start, so I compare that to the leader’s full distance in the same time to get the speed ratio. If there’s a finish-time gap given in seconds, I add that directly to the trailing runner’s total time before computing their speed.”
Follow-up Questions
- How would you solve for the start distance that produces a dead heat between two runners?
- How does a time head start problem differ algebraically from a distance head start problem?
- How would you extend this to a three-runner race with two different head starts?
- How does the speed ratio A:B translate into "A can give B a start of how many meters" in a different race length?
MCQ Practice
1. In a 200m race, A beats B by 40m. What is the ratio of their speeds?
In the time A covers 200m, B covers only 160m (200−40), so speed ratio A:B = 200:160 = 5:4.
2. A can run 1 km in 5 minutes; B can run 1 km in 6 minutes. In a 1 km race, by how many meters does A beat B?
When A finishes in 5 min, B has run (5/6)×1000 ≈ 833.3m, so A beats B by 1000−833.3 ≈ 166.7 m.
3. "A gives B a start of 10 seconds" means?
A time head start means the trailing runner (B) begins running earlier, gaining a 10-second time advantage.
Flash Cards
What does a distance start of x meters mean? — The trailing runner only needs to cover (track length − x) while the leader covers the full length in the same time.
How is speed ratio derived from a distance start? — Speed ratio (leader:trailer) = full length L : (L − x).
What does a time start of t seconds mean? — The trailing runner begins running t seconds before the leader, a time offset rather than a distance offset.
What is a dead heat? — A race outcome where both runners finish at exactly the same time, given the head start conditions.