CAT 2022 Slot 2 · Quant · Question 4

The constraint fixes a once b is chosen: a=6-2b.

Both variables are non-negative, so b\ge 0 and 6-2b\ge 0.

The second condition gives b\le 3, so b ranges over 0\le b\le 3.

Substituting into the target, a+b=\left( 6-2b \right)+b=6-b.

This decreases steadily as b increases, so the extremes sit at the ends of the range.

At b=0, a+b=6 — the maximum, with a=6.

At b=3, a+b=3 — the minimum, with a=0.

The average of the two is \frac{6+3}{2}=4.5 — the second option.

Shortcut. Writing the target in one variable turns it into the linear expression 6-b, and a linear function on an interval always peaks at an endpoint. So only b=0 and b=3 need checking.