1958 AHSME Problems/Problem 6: Difference between revisions

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

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

@@ Line 6: / Line 6: @@
 ==Solution==
+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>.
-{{solution}}
 ==See also==
 {{AHSME box|year=1958|num-b=5|num-a=7}}

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1958 AHSME Problems/Problem 6: Difference between revisions

Revision as of 12:21, 4 June 2011

Problem

Solution

See also