CAT 2024 Slot 3 · Quant · Question 3

Expanding the left side, {\left( a+b\sqrt{3}\right)}^{2}={a}^{2}+2ab\sqrt{3}+3{b}^{2}.

Grouping, this is \left({a}^{2}+3{b}^{2}\right)+2ab\sqrt{3}.

Since a and b are natural numbers, {a}^{2}+3{b}^{2} is rational and 2ab is rational, so the rational and irrational parts can be matched separately.

Matching the irrational parts, 2ab=30, so ab=15.

Matching the rational parts, {a}^{2}+3{b}^{2}=52.

The natural-number factor pairs with product 15 are \left( 1,15 \right), \left( 3,5 \right), \left( 5,3 \right) and \left( 15,1 \right).

Testing a=5, b=3: {a}^{2}+3{b}^{2}=25+27=52, which fits exactly.

The other pairs give 676, 84 and 228, none of which is 52.

So a+b=5+3=8, the third option.

Shortcut. Write 52 as a sum {a}^{2}+3{b}^{2} where ab=15 — only the balanced pair 5\times 3 keeps both terms small enough. A quick check confirms {\left( 5+3\sqrt{3}\right)}^{2}=52+30\sqrt{3}.