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)= $1.1x\times.9y$ = $.99xy$ $0.99/1=99\%$

Video Solution

https://youtu.be/4io4bjzgQKI

~savannahsolver

Easier to understand Video Solution

https://www.youtube.com/watch?v=cZUT7lYu474&t=4s

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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

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

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