1956 AHSME Problems/Problem 16

Solution

Let the $3$ numbers be $a$ $b$ and $c$ . We see that $a+b+c = 98$ and $\frac{a}{b} = \frac{2}{3} \Rrightarrow 3a = 2b$ $\frac{b}{c} = \frac{5}{8} \Rrightarrow 8b = 5c$ Writing $a$ and $c$ in terms of $b$ we have $a = \frac{2}{3} b$ and $c = \frac{8}{5} b$ . Substituting in the sum, we have $\frac{2}{3} b + b + \frac{8}{5} b = 98$ $\frac{49}{15} b = 98$ $b = 98 \cdot \frac{15}{49} \Rrightarrow b = 30$ $\boxed{C}$

~JustinLee2017

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1956 AHSME Problems/Problem 16

Solution