IPMAT Indore 2026 · Quantitative Ability · Question 16

Step 1 — note the geometry. A and B are diametrically opposite on a 14 km track, so the arc from A to B is 7 km either way.

Step 2 — when is each runner at B?

Rekha starts at A, so she reaches B after 7 km, then every full lap after that:

\text{Rekha’s distance at B}=7+14k, k=0,1,2,\ldots

Sajal starts at B, so he returns to B after each complete lap:

\text{Sajal’s distance at B}=14m, m=1,2,\ldots

Step 3 — they must be there at the same time. With speeds 5v and 2v:

\frac{7+14k}{5v}=\frac{14m}{2v}\Rightarrow \frac{7+14k}{5}=7m \Rightarrow 1+2k=5m

Step 4 — find the smallest solution. Try m=1: 1+2k=5\Rightarrow k=2. ✓

\text{Rekha’s distance}=7+14\left( 2 \right)=35\text{km}

Check: time =35/\left( 5v \right)=7/v; Sajal covers 2v\times 7/v=14 km = one full lap, so he is back at B. ✓

Answer: Option 1 — 35

Shortcut. The condition is not “they meet” but “they meet at B”, so write down separately when each is at B and match the times. Since their speeds are 5:2, in the time Rekha covers 35 km Sajal covers 14 — exactly one lap. Checking the ratio against the options gets you there fast.