IPMAT Indore 2026 · Quantitative Ability · Question 15

Step 1 — deal with the denominator first. For the square root to be defined and non-zero we need

100-{n}^{3}>0 \Rightarrow{n}^{3}<100

Since {4}^{3}=64<100 but {5}^{3}=125>100, this gives n\le 4. Note the denominator is then positive.

Step 2 — so the sign is decided by the numerator.

|n+6|\ge |n-3|

Step 3 — square both sides (both are non-negative, so this is safe):

{\left( n+6 \right)}^{2}\ge{\left( n-3 \right)}^{2}

{n}^{2}+12n+36\ge{n}^{2}-6n+9

18n\ge -27 \Rightarrow n\ge -1.5

For integers, n\ge -1.

Step 4 — combine both conditions.

-1\le n\le 4 \Rightarrow n\in \{-1,0,1,2,3,4\}

6\text{values}

Answer: 6

Shortcut. Check the denominator before touching the modulus — it is what caps n at 4, and it is positive throughout, so the inequality reduces to the numerator alone. Then \left| n+6 \right|\ge \left| n-3 \right| just says n is closer to -6… no, farther from -6 than from 3 is false; squaring is the safe one-line route.