# Difference between revisions of "1993 AJHSME Problems/Problem 21"

## Problem

If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$, then the area is increased by

$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$

## Solution

If a rectangle had dimensions $10 \times 10$ and area $100$, then its new dimensions would be $12 \times 15$ and area $180$. The area is increased by $180-100=80$ or $80/100 = \boxed{\text{(D)}\ 80\%}$.

## Alternate Solution

Let the dimensions of the rectangle be $x \times y$. This rectangle has area $xy$. The new dimensions would be $1.2x \times 1.5y$, so the area is $=1.8xy$, which is $\boxed{80\%}$ more than the original area.