1964 AHSME Problems/Problem 23

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Problem

Two numbers are such that their difference, their sum, and their product are to one another as $1:7:24$. The product of the two numbers is:

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96$

Solution

Set the two numbers as $x$ and $y$. Therefore, $x+y=7(x-y),  xy=24(x-y)$, and $24(x+y)=7xy$. Simplifying the first equation gives $y=\frac{3}{4}x$. Substituting for $y$ in the second equation gives $\frac{3}{4}x^2=6x.$ Solving yields $x=8$ or $x=0$. Substituting $x=0$ back into the first equation yields $1=-7$ which is false, so $x=0$ is not valid and $x=8$. Substituting into $y=\frac{3}{4}x$ gives $y=6$ and $xy=\boxed{\textbf{(D) } 48}$.

See More

1964 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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