Difference between revisions of "2010 AMC 10B Problems/Problem 14"
(Created page with 'We must find the average of the numbers from 1 to 99 and x in terms of x. The sum of all these terms is 99(100)/2+x=99(50)+x. We must divide this by the total number of terms, wh…') |
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| − | We must find the average of the numbers from 1 to 99 and x in terms of x. The sum of all these terms is 99(100) | + | We must find the average of the numbers from 1 to 99 and x in terms of x. The sum of all these terms is \frac{99(100)}{2}+x=99(50)+x. We must divide this by the total number of terms, which is 100. We get: \frac{99(50)+x}{100}. This is equal to 100x, as stated in the problem. We have: \frac{99(50)+x}{100}=100x. We can now cross multiply. This gives: |
100(100x)=99(50)+x | 100(100x)=99(50)+x | ||
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9999x=99(50) | 9999x=99(50) | ||
101x=50 | 101x=50 | ||
| − | x=50 | + | x=\frac{50}{101} |
| − | This gives us our answer. <math> \boxed{\mathrm{(B)}= 50 | + | This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math> |
Revision as of 23:58, 8 August 2010
We must find the average of the numbers from 1 to 99 and x in terms of x. The sum of all these terms is \frac{99(100)}{2}+x=99(50)+x. We must divide this by the total number of terms, which is 100. We get: \frac{99(50)+x}{100}. This is equal to 100x, as stated in the problem. We have: \frac{99(50)+x}{100}=100x. We can now cross multiply. This gives:
100(100x)=99(50)+x 10000x=99(50)+x 9999x=99(50) 101x=50 x=\frac{50}{101}
This gives us our answer. 
