# 1959 AHSME Problems/Problem 4

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## Problem 4

If $78$ is divided into three parts which are proportional to $1, \frac13, \frac16,$ the middle part is: $\textbf{(A)}\ 9\frac13 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 17\frac13 \qquad\textbf{(D)}\ 18\frac13\qquad\textbf{(E)}\ 26$

## Solution

Let the part proportional to $1$ equal $x$. Then, the parts are $x, \frac{1}{3}x$, and $\frac{1}{6}x$. The sum of these parts should be $78$, so $\frac{9}{6}x=1\frac{1}{2}x=78$. Solving for $x$, $x=52$. The middle part is $\frac{1}{3}x$, so the answer is $17\frac{1}{3}$, or $\boxed{\textbf{C}}$

## See also

 1959 AHSC (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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