1964 AHSME Problems/Problem 23

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 $xy=\boxed{\textbf{(D) } 48}$ .

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1964 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1964 AHSME Problems/Problem 23

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