Difference between revisions of "1958 AHSME Problems/Problem 6"

Latest revision as of 05:09, 3 October 2014

Problem

The arithmetic mean between $\frac {x + a}{x}$ and $\frac {x - a}{x}$ , when $x \not = 0$ , is:

$\textbf{(A)}\ {2}\text{, if }{a \not = 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a = 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x$

Solution

We have $\frac{1}{2}\cdot \left(\frac{x + a}{x} + \frac{x - a}{x}\right) = \frac{2}{2} = \boxed{\text{(B) }{1}}$ .

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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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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 ==See also==
-{{AHSME box|year=1958|num-b=5|num-a=7}}
+{{AHSME 50p box|year=1958|num-b=5|num-a=7}}
 {{MAA Notice}}

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Difference between revisions of "1958 AHSME Problems/Problem 6"

Latest revision as of 05:09, 3 October 2014

Problem

Solution

See also