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Difference between revisions of "1952 AHSME Problems/Problem 34"

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== Solution ==
 
== Solution ==
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<math>\fbox{\frac{10000}{10000} (choice E)}</math>
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Converting the word problem in to an equation, we take x as the initial amount, we get <cmath>x(1+\dfrac{p}{100})(1-\dfrac{p}{100}) = 1</cmath> Using, <math>(a+b)(a-b) = a^{2} - b^{2}</math>. We now simplify it to, <cmath>x(1-\dfrac{p^{2}}{10000})=1</cmath> Simplifying to, <cmath>x(\dfrac{10000-p^{2}}{10000}) = 1</cmath> Dividing both sides by <math>\dfrac{10000-p^{2}}{10000}</math>, we get the value of x. <math>\boxed{\textbf{(E)}\dfrac{10000}{10000-p^{2}}}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 13:14, 16 December 2017

Problem

The price of an article was increased $p\%$. Later the new price was decreased $p\%$. If the last price was one dollar, the original price was:

$\textbf{(A)}\ \frac{1-p^2}{200}\qquad \textbf{(B)}\ \frac{\sqrt{1-p^2}}{100}\qquad \textbf{(C)}\ \text{one dollar}\qquad\\ \textbf{(D)}\ 1-\frac{p^2}{10000-p^2}\qquad \textbf{(E)}\ \frac{10000}{10000-p^2}$

Solution

Converting the word problem in to an equation, we take x as the initial amount, we get \[x(1+\dfrac{p}{100})(1-\dfrac{p}{100}) = 1\] Using, $(a+b)(a-b) = a^{2} - b^{2}$. We now simplify it to, \[x(1-\dfrac{p^{2}}{10000})=1\] Simplifying to, \[x(\dfrac{10000-p^{2}}{10000}) = 1\] Dividing both sides by $\dfrac{10000-p^{2}}{10000}$, we get the value of x. $\boxed{\textbf{(E)}\dfrac{10000}{10000-p^{2}}}$

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 33 		Followed by
Problem 35
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All AHSME Problems and Solutions

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