Difference between revisions of "1962 AHSME Problems/Problem 8"
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==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> | 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 | |
| 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. 
