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*} | + | <math>\[\begin{align*} |
100(100x)=99(50)+x&\\ | 100(100x)=99(50)+x&\\ | ||
10000x=99(50)+x&\\ | 10000x=99(50)+x&\\ | ||
| Line 13: | Line 13: | ||
101x=50&\\ | 101x=50&\\ | ||
x=\boxed{\textbf{(B)}\ \frac{50}{101}}&\\ | x=\boxed{\textbf{(B)}\ \frac{50}{101}}&\\ | ||
| − | \end{align*}\] | + | \end{align*}\]</math> |
==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 21:34, 30 December 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:
$\[\begin{align*} 100(100x)=99(50)+x&\\ 10000x=99(50)+x&\\ 9999x=99(50)&\\ 101x=50&\\ x=\boxed{\textbf{(B)}\ \frac{50}{101}}&\\ \end{align*}\]$ (Error compiling LaTeX. Unknown error_msg)
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. 
