Difference between revisions of "2010 AMC 10B Problems/Problem 14"
| Line 6: | Line 6: | ||
101x=50, | 101x=50, | ||
</math> | </math> | ||
| + | <math> | ||
x=\frac{50}{101} | x=\frac{50}{101} | ||
| − | <math> | + | </math> |
| − | This gives us our answer. < | + | This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math> |
Revision as of 14:14, 24 January 2011
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. 
