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

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==Solution==
 
==Solution==
  
Let the width be <math>w</math> and the length be <math>l</math>. Then, the original perimeter is <math>2(w+1)</math>.  
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Let the width be <math>w</math> and the length be <math>l</math>. Then, the original perimeter is <math>2(w+l)</math>.  
  
 
After the increase, the new width and new length are <math>1.1w</math> and <math>1.1l</math>, so the new perimeter is <math>2(1.1w+1.1l)=2.2(w+l)</math>.
 
After the increase, the new width and new length are <math>1.1w</math> and <math>1.1l</math>, so the new perimeter is <math>2(1.1w+1.1l)=2.2(w+l)</math>.

Revision as of 22:05, 1 June 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 $2(w+l)$.

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
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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