1952 AHSME Problems/Problem 34

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. $\boxed{\textbf{(E)}\dfrac{10000}{10000-p^{2}}}$

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 33	Followed by Problem 35
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1952 AHSME Problems/Problem 34

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