# 1993 AJHSME Problems/Problem 21

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## 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.

 1993 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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