1961 AHSME Problems/Problem 18

Problem

The yearly changes in the population census of a town for four consecutive years are, respectively, 25% increase, 25% increase, 25% decrease, 25% decrease. The net change over the four years, to the nearest percent, is:

$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 12$

Solution

A 25% increase means the new population is $\frac{5}{4}$ of the original population. A 25% decrease means the new population is $\frac{3}{4}$ of the original population.

Thus, after four years, the population is $1 \cdot \frac{5}{4} \cdot \frac{5}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{225}{256}$ times the original population.

Thus, the net change is -12%, so the answer is $\boxed{\textbf{(A)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1961 AHSME Problems/Problem 18

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