How to Solve Linear Races and Head Start Problems
Solve linear race aptitude problems with distance and time head starts using the distance-speed-time method, with worked examples.
Expected Interview Answer
In a linear race, "A gives B a head start of x meters" means B starts x meters ahead, while "A gives B a start of t seconds" means B starts running t seconds earlier; both cases are solved by tracking each runner’s effective distance covered by the time the race ends.
A distance head start reduces the ground the trailing runner must cover, so if A runs distance D while B only needs to run D−x to reach the same finish line, they finish together exactly when D/speedA = (D−x)/speedB. A time head start means B has already been running for t seconds before A starts, covering speedB×t extra distance before A even begins, which must be added to B’s position. "A beats B by x meters" means when A finishes distance D, B has covered only D−x. Converting every head-start statement into an equation of covered distance versus time is the universal technique.
- One equation — distance = speed × time — resolves every head-start variant
- Distance and time head starts are handled by the same effective-distance idea
- "Beats by x meters/seconds" translates directly into a solvable equation
AI Mentor Explanation
If a fast bowler gives a slower bowler a head start by letting them bowl 2 overs before joining, that time head start means the slower bowler has already delivered speedB × 2-overs-worth of balls before the fast bowler even starts — that extra output must be added when comparing totals at any later point. A distance head start instead would be like starting the slower bowler’s over count already 5 overs ahead, directly reducing how many more overs they need to bowl to reach the same target over count as the fast bowler.
Worked example
A covers
- D/8 seconds
B covers
- (D−40)/6 seconds
Solve
- 6D = 8D − 320
- D = 160 m
Step-by-Step Explanation
Step 1
Classify the head start
Determine if it is a distance head start (meters) or a time head start (seconds).
Step 2
Convert to effective distance
Distance head start reduces required distance; time head start adds pre-race distance at the runner’s own speed.
Step 3
Set up the finish-time equation
Use distance = speed × time for each runner, equating finish times if they tie.
Step 4
Solve for the unknown
Isolate the race distance, head start, or speed as required by the question.
What Interviewer Expects
- Clear distinction between distance and time head starts
- Correct equation setup using distance = speed × time
- Correct interpretation of “beats by x meters/seconds”
- Careful handling of units (meters vs seconds) throughout
Common Mistakes
- Treating a time head start as if it were a distance head start directly
- Adding the head start to the wrong runner’s distance
- Forgetting to convert a time head start into distance using the runner’s own speed
- Confusing “beats by x meters” with “beats by x seconds” and mixing formulas
Best Answer (HR Friendly)
“I first figure out whether the head start is given in distance or time. A distance head start simply means the trailing runner has less ground left to cover, so I subtract it from the total distance. A time head start means that runner has already been moving before the race clock starts, so I convert that time into distance at their own speed and add it to their position. From there it is just distance equals speed times time for each runner, and I solve for whatever is unknown.”
Follow-up Questions
- How do you find the winning margin in meters at the finish line?
- How does a “dead heat” (tie) condition change the equation?
- How would you handle a race with three runners and two different head starts?
- How does "A beats B by x seconds" translate differently from "A beats B by x meters"?
MCQ Practice
1. A runs at 10 m/s, B runs at 8 m/s. In a 200m race, A gives B a head start of how many meters so they finish together?
A finishes 200m in 20s. In 20s, B covers 8×20 = 160m, so B needs 200−160 = 40m head start.
2. A and B run a 100m race. A finishes in 10s, B finishes in 12.5s. By how many meters does A beat B?
B’s speed = 100/12.5 = 8 m/s. In A’s 10s, B covers 80m, so A beats B by 100−80 = 20m.
3. A gives B a head start of 5 seconds in a race where A runs at 6 m/s and B runs at 4 m/s. How much distance has B covered before A starts?
B’s head start distance = speedB × time = 4 × 5 = 20m.
Flash Cards
Distance head start meaning? — Trailing runner starts x meters ahead, covering only D−x to finish.
Time head start meaning? — Trailing runner starts t seconds earlier, gaining speed×t extra distance before the other begins.
"A beats B by x meters" means? — When A finishes distance D, B has covered only D−x.
Universal race-problem equation? — Distance = Speed × Time, applied to each runner separately.