Difference between revisions of "2009 AMC 8 Problems/Problem 8"

Revision as of 13:52, 14 August 2021

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?

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

In a rectangle with dimensions 10*10 the new rectangle would have dimensions 11*9 The ratio of the new area to the old area is 99/100 =99

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

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 ==Solution==
-In a rectangle with dimensions <math>10 \times 10</math>, the new rectangle would have dimensions <math>11 \times 9</math>. The ratio of the new area to the old area is <math>99/100 = \boxed{\textbf{(B)}\ 99}</math>.
+In a rectangle with dimensions 10*10 the new rectangle would have dimensions 11*9 The ratio of the new area to the old area is 99/100 =99
 ==See Also==
 {{AMC8 box|year=2009|num-b=7|num-a=9}}
 {{MAA Notice}}