CAT 2022 Slot 2 · Quant · Question 3

Let polygon A have n sides, so polygon B has 2n sides.

The interior angle of a regular polygon with k sides is \frac{180\left( k-2 \right)}{k} degrees.

So A's interior angle is \frac{180\left( n-2 \right)}{n}.

And B's is \frac{180\left( 2n-2 \right)}{2n}.

Their ratio is \frac{3}{4}, so \frac{180\left( n-2 \right)}{n}\div \frac{180\left( 2n-2 \right)}{2n}=\frac{3}{4}.

The 180s cancel, and dividing by a fraction means multiplying by its reciprocal: \frac{n-2}{n}\times \frac{2n}{2n-2}=\frac{3}{4}.

The n cancels too, leaving \frac{2\left( n-2 \right)}{2n-2}=\frac{n-2}{n-1}=\frac{3}{4}.

Cross-multiplying, 4\left( n-2 \right)=3\left( n-1 \right), so 4n-8=3n-3.

Therefore n=5, and B has 2n=10 sides.

Checking: a pentagon's interior angle is {108}^{∘} and a decagon's is {144}^{∘}, and 108:144=3:4.

Shortcut. The ratio of interior angles collapses to \frac{n-2}{n-1} once the doubling is substituted, so the whole question is one small fraction equation. Recognising \frac{3}{4} as \frac{5-2}{5-1} gives n=5 on sight.