Difference between revisions of "2010 AMC 10B Problems/Problem 14"
(→See Also) |
|||
| Line 16: | Line 16: | ||
==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}} | ||
| − | |||
Revision as of 20:07, 25 April 2012
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}} (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 | ||
