1959 AHSME Problems/Problem 4

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 $\boxed{\textbf{E}}$


See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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