CAT 2019

Slot 2

QA

Question & Solution

Each interior angle in an n-sided polygon = $\frac{(n-2)180}{n}$

It is given that each interior angle of B is $\frac{3}{2}$ times each interior angle of A and $b = 2a$

$\frac{(b-2)180}{b} = \frac{3}{2} \times \frac{(a-2)180}{a}$

$2(b-2)a = 3(a-2)b$

$2(ab-2a) = 3(ab-2b)$

$2ab - 4a = 3ab - 6b$

$ab - 6b + 4a = 0$

$a(2a) - 6(2a) + 4a = 0$

$2a^2 - 12a + 4a = 0$

$2a^2 - 8a = 0$

$a(2a-8) = 0$

$a$ cannot be zero so $2a = 8$

$a=4$, $b = 4 \times 2 = 8$

$a+b = 12$

Each interior angle of a regular polygon with 12 sides = $\frac{(12-2) \times 180}{12}$

= $150^\circ$