IPMAT Indore 2026 · Quantitative Ability · Question 31
IPMAT Indore 2026 previous year question on Coordinate Geometry (Basics), with the answer and a step-by-step solution.
- A12
- B13
- C36
- D37
Answer & solution
Answer: Option D
We are counting lattice points inside or on a circle of radius \sqrt{10}\approx 3.16.
Step 1 — note the range. Since {x}^{2}\le 10, we have -3\le x\le 3, and the same for y.
Step 2 — count row by row. For each x, find how many y satisfy {y}^{2}\le 10-{x}^{2}:
x | 10-{x}^{2} | allowed y | count |
|---|---|---|---|
0 | 10 | -3\ldots 3 | 7 |
\pm 1 | 9 | -3\ldots 3 | 7 each |
\pm 2 | 6 | -2\ldots 2 | 5 each |
\pm 3 | 1 | -1\ldots 1 | 3 each |
Step 3 — add up.
7+2\left( 7 \right)+2\left( 5 \right)+2\left( 3 \right)=7+14+10+6=37
Answer: Option 4 — 37
Shortcut. Count by columns and exploit symmetry — only x=0,1,2,3 need any thought, and everything but x=0 doubles. The trap is forgetting that x=\pm 3 still works, since 9\le 10 leaves room for y\in \{-1,0,1\}.
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