1952 AMC 12 Problem 34

Question

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:

\dfrac{1-p^2}{200}

\dfrac{\sqrt{1-p^2}}{100}

One dollar

1-\dfrac{p^2}{10{,}000-p^2}

\dfrac{10{,}000}{10{,}000-p^2}

Represent the increase and decrease by multiplicative factors

If the original price is

P,

then

P(1+\frac{p}{100})(1-\frac{p}{100})=1

If the original price is

P

dollars, then after both changes its price is

\begin{aligned} &P\left(1+\frac p{100}\right) \left(1-\frac p{100}\right)\\ &\qquad=P\left(1-\frac{p^2}{10{,}000}\right). \end{aligned}

Setting this equal to

1

gives

P=\frac{10{,}000}{10{,}000-p^2}.

Thus, the correct answer is E.

Problem 34 in other years