CAT 2024 Slot 2 · Quant · Question 2

A modulus behaves differently on either side of zero, so the sign of x and the sign of y have to be settled first.

Suppose x<0. Then \left| x \right|=-x and the first equation collapses to -x+x+y=15, giving y=15.

With y=15>0, the second equation becomes x+15-15=20, so x=20 — a contradiction, because x was assumed negative.

So x>0, and the first equation reads x+x+y=15, that is 2x+y=15.

Now suppose y>0. Then \left| y \right|=y and the second equation collapses to x=20, which forces y=15-40=-25 — again a contradiction.

So y<0, \left| y \right|=-y, and the second equation becomes x-2y=20.

Solving 2x+y=15 and x-2y=20 together gives x=10 and y=-5.

Both signs check out, and x-y=10-\left( -5 \right)=15, which is the fourth option.

Shortcut. Notice that \left| x \right|+x is 2x when x>0 and 0 when x<0, while \left| y \right|-y is 0 when y>0 and -2y when y<0. If either vanishes, one equation loses an unknown and the two immediately clash, so only the combination x>0, y<0 can survive — the other three cases die on sight.