IPMAT Indore 2022 · Quantitative Ability · Question 2

Step 1 — put the inequality in intercept form. Divide through by 6:

\frac{\left| x \right|}{3}+\frac{\left| y \right|}{2}\le 1

Step 2 — identify the shape. With both moduli present, the region is a rhombus centred at the origin with vertices on the axes at

\left( \pm 3, 0 \right) \text{and}\left( 0, \pm 2 \right)

Step 3 — compute the area. The diagonals are 2\times 3=6 and 2\times 2=4, and for a rhombus

\text{Area}=\frac{{d}_{1}{d}_{2}}{2}=\frac{6\times 4}{2}=12

Equivalently: the first-quadrant piece is a right triangle of legs 3 and 2, area 3, and there are four such pieces.

Answer: 12

Shortcut. Memorise the formula. The region \frac{\left| x \right|}{a}+\frac{\left| y \right|}{b}\le 1 is a rhombus of area

\boxed{2ab}

Here a=3, b=2, so the area is 2\left( 3 \right)\left( 2 \right)=12 — one division and one product.

The common error is forgetting the moduli and computing the triangle in the first quadrant only, giving 3. The four-fold symmetry is the whole point of the absolute values.