Difference between revisions of "1993 AJHSME Problems/Problem 21"

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==Solution==
 
==Solution==
 
If a rectangle had dimensions <math>10 \times 10</math> and area <math>100</math>, then its new dimensions would be <math>12 \times 15</math> and area <math>180</math>. The area is increased by <math>180-100=80</math> or <math>80/100 = \boxed{\text{(D)}\ 80\%}</math>.
 
If a rectangle had dimensions <math>10 \times 10</math> and area <math>100</math>, then its new dimensions would be <math>12 \times 15</math> and area <math>180</math>. The area is increased by <math>180-100=80</math> or <math>80/100 = \boxed{\text{(D)}\ 80\%}</math>.
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==Alternate Solution==
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Let the dimensions of the rectangle be <math>x \times y</math>. This rectangle has area <math>xy</math>. The new dimensions would be <math>1.2x \times 1.5y</math>, so the area is <math>=1.8xy</math>, which is <math>\boxed{80\%}</math> more than the original area.
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1993|num-b=20|num-a=22}}
 
{{AJHSME box|year=1993|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 11:32, 10 May 2015

Problem

If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%$, then the area is increased by

$\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\%$

Solution

If a rectangle had dimensions $10 \times 10$ and area $100$, then its new dimensions would be $12 \times 15$ and area $180$. The area is increased by $180-100=80$ or $80/100 = \boxed{\text{(D)}\ 80\%}$.

Alternate Solution

Let the dimensions of the rectangle be $x \times y$. This rectangle has area $xy$. The new dimensions would be $1.2x \times 1.5y$, so the area is $=1.8xy$, which is $\boxed{80\%}$ more than the original area.

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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