1952 AHSME Problems/Problem 16

Problem

If the base of a rectangle is increased by $10\%$ and the area is unchanged, then the altitude is decreased by:

$\textbf{(A) \ }9\% \qquad \textbf{(B) \ }10\% \qquad \textbf{(C) \ }11\% \qquad \textbf{(D) \ }11\frac{1}{9}\% \qquad \textbf{(E) \ }9\frac{1}{11}\%$

Solution

$b\cdot a=\frac{11}{10}b\cdot xa\implies x=\frac{10}{11}$ . Hence, the altitude is decreased by $1-x=\frac{1}{11}=\boxed{\textbf{(E)}\ 9\frac{1}{11}\%}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1952 AHSME Problems/Problem 16

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