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 21:07, 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 |