Difference between revisions of "2010 AMC 10B Problems/Problem 14"
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We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives: | We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives: | ||
<math> | <math> | ||
| + | \linebreak | ||
100(100x)=99(50)+x | 100(100x)=99(50)+x | ||
| + | \linebreak | ||
10000x=99(50)+x | 10000x=99(50)+x | ||
| + | \linebreak | ||
9999x=99(50) | 9999x=99(50) | ||
| + | \linebreak | ||
101x=50 | 101x=50 | ||
| + | \linebreak | ||
x=\frac{50}{101} | x=\frac{50}{101} | ||
| + | \linebreak | ||
</math> | </math> | ||
This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math> | This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math> | ||
Revision as of 23:06, 8 August 2010
We must find the average of the numbers from 
to 
and 
in terms of . The sum of all these terms is 
. We must divide this by the total number of terms, which is 
. We get: 
. This is equal to 
, as stated in the problem. We have: 
. We can now cross multiply. This gives:
This gives us our answer. 
