IPMAT Indore 2022 · Quantitative Ability · Question 29

Step 1 — use the Bipin information to get his speed. When Ashok finishes, Bipin is 150 m short, and he covers that 150 m in the 15 seconds by which he loses. So

\text{Bipin’s speed}=\frac{150}{15}=10 m/s

Step 2 — find Bipin’s total time.

{t}_{B}=\frac{400}{10}=40 s

Step 3 — find Ashok’s and Chandan’s times. Ashok beats Bipin by 15 s:

{t}_{A}=40-15=25 s

Ashok beats Chandan by 25 s:

{t}_{C}=25+25=50 s

Step 4 — find Chandan’s speed.

\text{Chandan’s speed}=\frac{400}{50}=8 m/s

Step 5 — compare Bipin and Chandan. Bipin finishes at t=40 s. In 40 seconds Chandan covers

8\times 40=320 m

so he is 400-320=80 m behind.

Answer: Option 1 — 80 metres

Shortcut. Convert “beats by x metres and y seconds” into a speed immediately: the loser’s speed is \frac{x}{y}. Here Bipin runs at \frac{150}{15}=10 m/s, which unlocks every other time in the question.

From there the times fall out as 40, 25 and 50 seconds, and the final gap is

400-8\times 40=80 m

The trap is option (c) 150 m — reusing the margin given in the question. The gaps are not additive or transferable between pairs of runners; you must go through the speeds.