2010 AMC 10B Problems/Problem 14
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 | |
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