Difference between revisions of "1985 AJHSME Problems/Problem 19"

(New page: ==Problem== If the length and width of a rectangle are each increased by <math>10\% </math>, then the perimeter of the rectangle is increased by <math>\text{(A)}\ 1\% \qquad \text{(B)}\ ...)
 
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==Problem==
 
==Problem==
  
If the length and width of a rectangle are each increased by <math>10\% </math>, then the perimeter of the rectangle is increased by
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If the [[length]] and width of a [[rectangle]] are each increased by <math>10\% </math>, then the [[perimeter]] of the rectangle is increased by
  
 
<math>\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\% </math>
 
<math>\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\% </math>
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==See Also==
 
==See Also==
  
[[1985 AJHSME Problems]]
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{{AJHSME box|year=1985|num-b=18|num-a=20}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Introductory Algebra Problems]]

Revision as of 12:29, 16 May 2009

Problem

If the length and width of a rectangle are each increased by $10\%$, then the perimeter of the rectangle is increased by

$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$

Solution

Let the width be $w$ and the length be $l$. Then, the original perimeter is $2(w+1)$.

After the increase, the new width and new length are $1.1w$ and $1.1l$, so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$.

Therefore, the percent change is \begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= \frac{.2}{2} \\ &= 10\% \\ \end{align*}

$\boxed{\text{B}}$

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions
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