Difference between revisions of "2010 AMC 10B Problems/Problem 14"
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\[\begin{align*} | \[\begin{align*} | ||
| − | 100(100x)=99(50)+x& | + | 100(100x)=99(50)+x&\\ |
| − | 10000x=99(50)+x& | + | 10000x=99(50)+x&\\ |
| − | 9999x=99(50)& | + | 9999x=99(50)&\\ |
| − | 101x=50& | + | 101x=50&\\ |
| − | x=\boxed{\textbf{(B)}\ \frac{50}{101}} | + | x=\boxed{\textbf{(B)}\ \frac{50}{101}}&\\ |
\end{align*}\] | \end{align*}\] | ||
Revision as of 17:50, 13 July 2015
Problem
The average of the numbers 
and 
is 
. What is ?
Solution
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:
\[\begin{align*} 100(100x)=99(50)+x&\\ 10000x=99(50)+x&\\ 9999x=99(50)&\\ 101x=50&\\ x=\boxed{\textbf{(B)}\ \frac{50}{101}}&\\ \end{align*}\]
See Also
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
