Atul gives Vishu a head-start of 20 seconds in a 900 m race and beats him by 135 m. While running the same race again Atul gives a start of 189 m and beats him by 8 seconds. In how much time can Vishu complete the full race of 900 m?
- (a)2 minutes 50 seconds
- (b)3 minutes 20 seconds
- (c)3 minutes 10 seconds
- (d)3 minutes 30 seconds
Answer
Why
Correct — B. Turn each margin into the distance Vishu actually ran. Let V be his time for 900 m and A be Atul's.
Race 1 — Vishu starts 20 s early and is 135 m short when Atul finishes:
he covers 900 − 135 = 765 m, which takes 765/900 × V = 0.85V
he has been running for A + 20 seconds, so A + 20 = 0.85V
Race 2 — Vishu starts 189 m up the track and finishes 8 s later:
he runs only 900 − 189 = 711 m, taking 711/900 × V = 0.79V
both set off together, so A + 8 = 0.79V
Subtract: 0.85V − 20 = 0.79V − 8
0.06V = 12 → V = 200 s = 3 minutes 20 seconds → option (b)
Check: A = 150 s, so Atul runs at 6 m/s and Vishu at 4.5 m/s, which fits both races.
Why the others are wrong
- (a)2 minutes 50 seconds — 170 s is how long Vishu ran in the first race, not his time for 900 m — Atul's 150 s plus the 20 s head start. In those 170 s he covered only 765 m.
- (c)3 minutes 10 seconds — 190 s fits neither race. It makes Atul's 900 m 141.5 s by the first race and 142.1 s by the second, and one runner cannot have two times.
- (d)3 minutes 30 seconds — 210 s is too slow. It demands 158.5 s from Atul in the first race and 157.9 s in the second, so no single pair of speeds satisfies both.
Concept
A 'start' can be given in metres or in seconds, and the two behave differently.
A start of 189 m shortens the receiver's race to 711 m, with both runners setting off together. A start of 20 seconds leaves the race at its full 900 m but lets him run for 20 seconds before his rival's clock begins.
The margins split the same way. 'Beats him by 135 m' fixes a position at a moment — when Atul crosses the line Vishu is at 765 m. 'Beats him by 8 seconds' fixes a time — Vishu reaches his own finish 8 seconds after Atul reaches his.
Both equations carry Atul's time A, and A never has to be found to answer the question — the subtraction removes it.
Working it out afterwards is still worth ten seconds: A = 150 s reproduces the 135 m and the 8 s exactly, which is the only proof the reading of the stem was right.
Key facts
- A head start of 20 seconds means Vishu has run for (A + 20) seconds when Atul finishes in A seconds.
- 'Beats him by 135 m' in a 900 m race means the loser stands at 765 m at that moment.
- A start of 189 m cuts the receiver's own race to 711 m.
- The consistent solution is Vishu 900 m in 200 s at 4.5 m/s and Atul in 150 s at 6 m/s.
Study next
Common traps
- Keeping Vishu's second race at 900 m after the 189 m start, which forces a negative time
- Treating the 20 second head start as a distance and subtracting metres from 900
- Answering 170 s, the time Vishu spent running in the first race rather than his 900 m time
SSC mixes a time start with a distance start inside one stem, so a single conversion habit will not carry you through both halves of the question.
The same rate bookkeeping appears at Quant Q.17 of this shift, where one pipe fills a tank one and a half times as fast as the other.
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