Danny weighs $40 \mathrm{~kg}$. Steven weighs $20 \%$ more than Danny. Steven's weight is
(A) $50 \mathrm{~kg}$
(B) $60 \mathrm{~kg}$
(C) $48 \mathrm{~kg}$
(D) $32 \mathrm{~kg}$
(E) $72 \mathrm{~kg}$
Solution 1

Since Danny weighs $40 \mathrm{~kg}$, then $20 \%$ of his weight is $\frac{20}{100} \times 40=\frac{1}{5} \times 40=8 \mathrm{~kg}$.

Since Steven weighs $20 \%$ more than Danny, his weight is $40+8=48 \mathrm{~kg}$.

## Solution 2

Since Steven weighs $20 \%$ more than Danny, then Steven's weight is $120 \%$ of Danny's weight.

Since Danny's weight is $40 \mathrm{~kg}$, then Steven's weight is $\frac{120}{100} \times 40=\frac{6}{5} \times 40=48 \mathrm{~kg}$.