IPMAT Indore 2026 · Quantitative Ability · Question 28

Step 1 — compute k after the replacement.

{k}_{\text{new}}=\frac{{\left( x+y \right)}^{2}+{\left( x-y \right)}^{2}}{\left( x+y \right)|x-y|}=\frac{2{x}^{2}+2{y}^{2}}{|{x}^{2}-{y}^{2}|}

using {\left( x+y \right)}^{2}+{\left( x-y \right)}^{2}=2\left({x}^{2}+{y}^{2}\right) and \left( x+y \right)\left| x-y \right|=\left|{x}^{2}-{y}^{2}\right|.

Step 2 — set it equal to the original k.

\frac{2\left({x}^{2}+{y}^{2}\right)}{|{x}^{2}-{y}^{2}|}=\frac{{x}^{2}+{y}^{2}}{xy}

Cancel the common factor {x}^{2}+{y}^{2} (which is positive):

\frac{2}{|{x}^{2}-{y}^{2}|}=\frac{1}{xy}\Rightarrow |{x}^{2}-{y}^{2}|=2xy

Step 3 — convert to a single variable. Put t=\frac{x}{y}>0 and divide by {y}^{2}:

|{t}^{2}-1|=2t

Taking t>1: {t}^{2}-2t-1=0\Rightarrow t=1+\sqrt{2}.

Step 4 — evaluate k. Note k=\frac{{x}^{2}+{y}^{2}}{xy}=t+\frac{1}{t}:

k=\left( 1+\sqrt{2}\right)+\frac{1}{1+\sqrt{2}}=\left( 1+\sqrt{2}\right)+\left( \sqrt{2}-1 \right)=2\sqrt{2}

Answer: Option 4 — 2\sqrt{2}

Shortcut. Spot early that k=t+\frac{1}{t} where t=x/y — then the whole problem lives in one variable. The identity \frac{1}{1+\sqrt{2}}=\sqrt{2}-1 makes the last line instant. Note also k\ge 2 always, which kills options (a) and (b) on sight.