# 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 $\boxed{\textbf{(A)}}$.

## See Also

 1961 AHSC (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

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