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

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==Solution==
 
==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==
 
==See also==
  
 
{{AHSME box|year=1958|num-b=5|num-a=7}}
 
{{AHSME box|year=1958|num-b=5|num-a=7}}

Revision as of 12:21, 4 June 2011

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 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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