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1962 AHSME Problems/Problem 8: Difference between revisions

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==Problem==
==Problem==


Given the set of n<math></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:  
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==
"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}}

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 $\boxed{D}$

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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