2009 AMC 8 Problems/Problem 8

Problem

The length of a rectangle is increased by $10\%$ percent and the width is decreased by $10\%$ percent. What percent of the old area is the new area?


$\textbf{(A)}\  90  \qquad \textbf{(B)}\   99  \qquad \textbf{(C)}\   100  \qquad \textbf{(D)}\   101  \qquad \textbf{(E)}\   110$

Solution

In a rectangle with dimensions $10 \times 10$, the new rectangle would have dimensions $11 \times 9$. The ratio of the new area to the old area is $99/100 = \boxed{\textbf{(B)}\ 99}$.

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AJHSME/AMC 8 Problems and Solutions

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