# 1956 AHSME Problems/Problem 16

## Problem 16

The sum of three numbers is $98$. The ratio of the first to the second is $\frac {2}{3}$, and the ratio of the second to the third is $\frac {5}{8}$. The second number is: $\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$

## 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

## See Also

 1956 AHSME (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 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 All AHSME Problems and Solutions

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