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.

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 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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