Difference between revisions of "1962 AHSME Problems/Problem 8"

Latest revision as of 21:14, 3 October 2014

Problem

Given the set of $n$ numbers; $n > 1$ , of which one is $1 - \frac {1}{n}$ and all the others are $1$ . The arithmetic mean of the $n$ numbers is:

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ n-\frac{1}{n}\qquad\textbf{(C)}\ n-\frac{1}{n^2}\qquad\textbf{(D)}\ 1-\frac{1}{n^2}\qquad\textbf{(E)}\ 1-\frac{1}{n}-\frac{1}{n^2}$

Solution

Just take $\frac{1(n-1)+(1-\frac{1}{n})}{n}$ . You get $\frac{n-1+1-\frac{1}{n}}{n}$ , which is just $\frac{n-\frac{1}{n}}{n}$ , which is just $\boxed{D}$

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 ==Solution==
-"Unsolved"
+Just take <math>\frac{1(n-1)+(1-\frac{1}{n})}{n}</math>. You get <math>\frac{n-1+1-\frac{1}{n}}{n}</math>, which is just <math>\frac{n-\frac{1}{n}}{n}</math>, which is just <math>\boxed{D}</math>
+==See Also==
+{{AHSME 40p box|year=1962|before=Problem 7|num-a=9}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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Difference between revisions of "1962 AHSME Problems/Problem 8"

Latest revision as of 21:14, 3 October 2014

Problem

Solution

See Also