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 $99/100 = \boxed{\textbf{(B)}\ 99}$ .

Solution 2

If you take the length as $x$ and the width as $y$ then A(OLD)= $xy$ A(NEW)= $1.1x\times.9y$ = $.99xy$ $0.99/1=99\%$

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2009 AMC 8 Problems/Problem 8

Contents

Problem

Solution

Solution 2

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