Difference between revisions of "1951 AHSME Problems/Problem 8"

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[[Without loss of generality]], let the price of the article be 100. Thus, the new price is <math>.9\cdot  100= 90</math>. Then, to restore it to the original price, we solve <math>90x=100</math>. We find <math>x=1\dfrac{1}{9}</math>, thus, the percent increase is <math>1\dfrac{1}{9}-1=\dfrac{1}{9}=\boxed{\textbf{(C) \ } 11\frac{1}{9}\%}</math>.
 
[[Without loss of generality]], let the price of the article be 100. Thus, the new price is <math>.9\cdot  100= 90</math>. Then, to restore it to the original price, we solve <math>90x=100</math>. We find <math>x=1\dfrac{1}{9}</math>, thus, the percent increase is <math>1\dfrac{1}{9}-1=\dfrac{1}{9}=\boxed{\textbf{(C) \ } 11\frac{1}{9}\%}</math>.
  
== See also ==
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== See Also ==
{{AHSME box|year=1951|num-b=7|num-a=9}}  
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{{AHSME 50p box|year=1951|num-b=7|num-a=9}}  
  
[[Category:Introductory Geometry Problems]]
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:20, 5 July 2013

Problem

The price of an article is cut $10 \%.$ To restore it to its former value, the new price must be increased by:

$\textbf{(A) \ } 10 \% \qquad\textbf{(B) \ } 9 \% \qquad \textbf{(C) \ } 11\frac{1}{9} \% \qquad\textbf{(D) \ } 11 \% \qquad\textbf{(E) \ } \text{none of these answers}$

Solution

Without loss of generality, let the price of the article be 100. Thus, the new price is $.9\cdot  100= 90$. Then, to restore it to the original price, we solve $90x=100$. We find $x=1\dfrac{1}{9}$, thus, the percent increase is $1\dfrac{1}{9}-1=\dfrac{1}{9}=\boxed{\textbf{(C) \ } 11\frac{1}{9}\%}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AHSME Problems and Solutions

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