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>
  
==Solution==
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==Solution #1==
  
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>.
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<math>\boxed{\text{B}}</math>
 
<math>\boxed{\text{B}}</math>
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==Solution #2 (Quick Fakesolve)==
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Assume WLOG that the rectangle is a square with length <math>10</math> and width <math>10</math>.  Thus, the square has a perimeter of <math>40</math>.  Increasing the length and width by <math>10\%</math> will increase the dimensions of the square to 11x11.  Thus, the new square's perimeter is <math>44</math>, and because <math>44</math> is <math>110\%</math> of <math>40</math>, our answer is <math>\boxed{\text{(B) } 10\%}</math>.
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==Video Solution==
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https://youtu.be/jAy9O53u-nQ
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~savannahsolver
  
 
==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]]
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{{MAA Notice}}

Latest revision as of 09:04, 22 January 2023

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 #1

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*}

$\boxed{\text{B}}$

Solution #2 (Quick Fakesolve)

Assume WLOG that the rectangle is a square with length $10$ and width $10$. Thus, the square has a perimeter of $40$. Increasing the length and width by $10\%$ will increase the dimensions of the square to 11x11. Thus, the new square's perimeter is $44$, and because $44$ is $110\%$ of $40$, our answer is $\boxed{\text{(B) } 10\%}$.

Video Solution

https://youtu.be/jAy9O53u-nQ

~savannahsolver

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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