1962 AHSME Problems/Problem 8: Difference between revisions

Revision as of 06:12, 11 December 2013

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 $\boxed{D}$

Revision as of 21:53, 10 November 2013 view source Fadebekun (talk \| contribs) 300 edits →Solution ← Older edit		Revision as of 06:12, 11 December 2013 view source DANCH (talk \| contribs) 90 edits No edit summary Newer edit →
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	==Solution==		==Solution==
	{{~~solution~~}}		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>

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1962 AHSME Problems/Problem 8: Difference between revisions

Revision as of 06:12, 11 December 2013

Problem

Solution