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

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==Solution==
 
==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>.
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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==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=7|num-a=9}}
 
{{AMC8 box|year=2009|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:52, 14 August 2021

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*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

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

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