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

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==Solution==
 
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We have <math> \frac{1}{2}\cdot \left(\frac{x + a}{x} + \frac{x - a}{x}\right) = \frac{2}{2} = \boxed{\text{(B) }{1}}</math>.
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==See also==
 
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{{AHSME box|year=1958|num-b=4|num-a=6}}
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{{AHSME 50p box|year=1958|num-b=5|num-a=7}}
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{{MAA Notice}}

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

See also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AHSME Problems and Solutions

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