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

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Latest revision as of 06:07, 3 October 2014

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 $x^2-10x+20$.)

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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