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

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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:  
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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"
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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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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 7|num-a=9}}
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[[Category:Introductory Algebra Problems]]
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{{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 (ProblemsAnswer KeyResources)
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

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