CAT 2023 Slot 3 · Quant · Question 3

Two linear equations have infinitely many solutions only when one is a multiple of the other.

Comparing with x+y=4, every coefficient of the second equation must be the same multiple of the corresponding coefficient of the first.

So \frac{a+5}{1}=\frac{{b}^{2}-15}{1}=\frac{8b}{4}.

The last ratio is 2b, so both a+5 and {b}^{2}-15 equal 2b.

From the second of these, {b}^{2}-2b-15=0.

Factorising, \left( b-5 \right)\left( b+3 \right)=0, so b=5 or b=-3.

When b=5, a+5=10 gives a=5, and then ab=25.

When b=-3, a+5=-6 gives a=-11, and then ab=\left( -11 \right)\times \left( -3 \right)=33.

The larger of the two is 33 — the second option.

Shortcut. The presence of 25 among the options is the trap: it is the answer from the obvious positive root. Two negatives multiply to a positive, so the negative root is the one worth checking — and it wins.