CAT 2022 Slot 1 · Quant · Question 15

Each equation has a common factor that is worth taking out.

The first is a\left( a+b+1 \right)=14.

The second is b\left( a+b+1 \right)=28.

Both share the bracket \left( a+b+1 \right), so dividing one by the other removes it.

That gives \frac{b}{a}=\frac{28}{14}=2, so b=2a.

Substituting into the first equation, a\left( a+2a+1 \right)=14, that is a\left( 3a+1 \right)=14.

Expanding, 3{a}^{2}+a-14=0.

Factorising, \left( 3a+7 \right)\left( a-2 \right)=0, so a=2 or a=-\frac{7}{3}.

Only a=2 is a natural number, and then b=4.

Checking: 4+8+2=14 and 16+8+4=28, both correct.

So 2a+b=4+4=8 — the fourth option.

Shortcut. Adding the two equations gives \left( a+b \right)\left( a+b+1 \right)=42=6\times 7, so a+b=6 straight away. With b=2a from the ratio, a=2 and b=4 follow without any quadratic.