Difference between revisions of "1962 AHSME Problems/Problem 8"
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==Problem== | ==Problem== | ||
− | Given the set of | + | Given the set of <math>n</math> numbers; <math>n > 1</math>, of which one is <math>1 - \frac {1}{n}</math> and all the others are <math>1</math>. The arithmetic mean of the <math>n</math> numbers is: |
<math> \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} </math> | <math> \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} </math> | ||
==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> | |
+ | |||
+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1962|before=Problem 7|num-a=9}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:14, 3 October 2014
Problem
Given the set of numbers; , of which one is and all the others are . The arithmetic mean of the numbers is:
Solution
Just take . You get , which is just , which is just
See Also
1962 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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All AHSME Problems and Solutions |
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