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

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png