IPMAT Indore 2026 · Quantitative Ability · Question 21

Write c=\pi -3\approx 0.1416, a small positive number. We need the largest of three quantities to equal c, which means one of them equals c and the other two are at most c.

Step 1 — the candidate \min \left( x,\,2-x \right)=c. This min equals x when x\le 1 and 2-x when x>1:

x=c\approx 0.142. Check the others: x-4\approx -3.86\le c ✓, 2x-8\approx -7.7\le c ✓ — valid

x=2-c\approx 1.858. Then x-4\approx -2.14 ✓, 2x-8\approx -4.28 ✓ — valid

Step 2 — the candidate x-4=c, i.e. x=c+4\approx 4.142. But then

2x-8\approx 0.283>c

so the maximum would exceed c — rejected.

Step 3 — the candidate 2x-8=c, i.e. x=\frac{c+8}{2}\approx 4.071. Then

x-4\approx 0.071\le c ✓, \min \left( x,\,2-x \right)=2-x\approx -2.07\le c ✓

— valid

Step 4 — count.

x\approx 0.142, 1.858, 4.071 \Rightarrow 3\text{values}

Answer: Option 4 — 3

Shortcut. Treat it as three candidate equations and verify each against the other two — that rejection step is the whole question. The \min branch naturally gives two solutions (one on each side of its peak at x=1), and of the two linear branches only the steeper one survives.