Difference between revisions of "2010 AMC 10B Problems/Problem 14"

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==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}}
fuck all
 

Revision as of 21:07, 25 April 2012

Problem

The average of the numbers $1, 2, 3,\cdots, 98, 99,$ and $x$ is $100x$. What is $x$?

$\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}$

Solution

We must find the average of the numbers from $1$ to $99$ and $x$ in terms of $x$. The sum of all these terms is $\frac{99(100)}{2}+x=99(50)+x$. We must divide this by the total number of terms, which is $100$. We get: $\frac{99(50)+x}{100}$. This is equal to $100x$, as stated in the problem. We have: $\frac{99(50)+x}{100}=100x$. 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 (ProblemsAnswer KeyResources)
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