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 $\boxed{\textbf{E}}$

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1959 AHSME Problems/Problem 4

Problem 4

Solution

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