CAT 2023 Slot 2 · Quant · Question 19

The region lies between the Y-axis and the line x=5, so only 0\le x\le 5 matters.

Throughout that strip x-5 is at most zero, so \left| x-5 \right|=5-x.

The given equation becomes \left| x-y \right|-\left( 5-x \right)=2, that is \left| x-y \right|=7-x.

A modulus equation splits into two cases.

Case one: x-y=7-x, which rearranges to y=2x-7.

Case two: x-y=-\left( 7-x \right)=x-7, which simplifies to y=7.

So the two boundary lines are the slanted line y=2x-7 and the horizontal line y=7.

Find the four corners by meeting these with the two vertical boundaries.

At x=0: y=2\left( 0 \right)-7=-7 and y=7, giving the corners \left( 0,-7 \right) and \left( 0,7 \right).

At x=5: y=2\left( 5 \right)-7=3 and y=7, giving the corners \left( 5,3 \right) and \left( 5,7 \right).

The two vertical sides are parallel, so the figure is a trapezium lying on its side.

The side at x=0 has length 7-\left( -7 \right)=14, and the side at x=5 has length 7-3=4.

The distance between them is 5.

The area is \frac{1}{2}\left( 14+4 \right)\times 5=\frac{1}{2}\times 18\times 5=45.

Shortcut. Restricting to the strip kills the outer modulus immediately, and the inner one splits into a slanted line and a flat one. Since the two parallel sides are vertical, the trapezium formula applies with the horizontal gap of 5 as the height — no shoelace calculation needed.