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 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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