Difference between revisions of "1958 AHSME Problems/Problem 13"

Problem

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

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

Solution

$x+y=10$

$xy=20$

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

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