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: | ||
− | + | \[\begin{align*} | |
− | 100(100x) | + | 100(100x)=99(50)+x& |
− | 10000x | + | 10000x=99(50)+x& |
− | 9999x | + | 9999x=99(50)& |
− | 101x | + | 101x=50& |
− | x | + | x=\boxed{\textbf{(B)}\ \frac{50}{101}} |
+ | \end{align*}\] | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2010|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:46, 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 |
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