1958 AHSME Problems/Problem 13

Problem

The sum of two numbers is $10$ ; their product is $20$ . The sum of their reciprocals is:

$\textbf{(A)}\ \frac{1}{10}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$

$x+y=10$

$xy=20$

$\frac1x+\frac1y=\frac{y}{xy}+\frac{x}{xy}=\frac{x+y}{xy}=\frac{10}{20}=\boxed{\frac12\textbf{ (B)}}$

(Which is a lot easier than finding the roots of $x^2-10x+20$ .)

1958 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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