IPMAT Indore 2026 · Quantitative Ability · Question 45

Step 1 — the two distances. Let the circle have radius r.

\text{Circle lap}=2\pi r

The largest inscribed square has the circle’s diameter as its diagonal:

\text{diagonal}=2r \Rightarrow \text{side}=\frac{2r}{\sqrt{2}}=r\sqrt{2}\Rightarrow \text{perimeter}=4r\sqrt{2}

Step 2 — the two times.

{t}_{\text{walk}}=\frac{2\pi r}{v},{t}_{\text{run}}=\frac{4r\sqrt{2}}{3v}

Step 3 — take the ratio. The r and v both cancel:

\frac{{t}_{\text{walk}}}{{t}_{\text{run}}}=\frac{2\pi r}{v}\times \frac{3v}{4r\sqrt{2}}=\frac{6\pi}{4\sqrt{2}}=\frac{3\pi}{2\sqrt{2}}

Answer: Option 4 — \frac{3\pi}{2\sqrt{2}}

Shortcut. Set r=1 immediately — the ratio cannot depend on it. Then it is 2\pi against \frac{4\sqrt{2}}{3}, and the only thing to remember is that the inscribed square’s diagonal is the diameter, so its side is \sqrt{2}, not 2.