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

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==Solution==
 
==Solution==
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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 old area to the new area is <math>99/100 = \boxed{\textbf{(B)}\ 99}</math>.
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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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===Solution 2===
 +
If you take the length as <math>x</math> and the width as <math>y</math> then
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A(OLD)= <math>xy</math>
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A(NEW)= <math>1.1x\times.9y</math> = <math>.99xy</math>
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<math>0.99/1=99\%</math>
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==Video Solution==
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https://youtu.be/4io4bjzgQKI
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 +
~savannahsolver
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 +
==Easier to understand Video Solution==
 +
https://www.youtube.com/watch?v=cZUT7lYu474&t=4s
  
 
==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}}

Latest revision as of 20:39, 26 December 2023

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

See Also

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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