2015/16 AHSMC Part 1 Problem 4

If $\ell$ and $w$ denote the length and width of the rectangle then its area is $A = \ell \cdot w$ while the area of the increased rectangle is $$\Bigl(\ell + \frac{20\ell}{100}\Bigr) \cdot \Bigl(w + \frac{30w}{100}\Bigr) = \ell \cdot w \cdot \frac{156}{100} = A \cdot \frac{156}{100} = A + A \cdot \frac{56}{100}.$$ Thus the area of the rectangle is increased by $56\%$. The answer is (D).