2010 AMC 10B Problems/Problem 14

Revision as of 18:46, 13 July 2015 by Gamjawon (talk | contribs) (Solution)

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:

\[\begin{align*} 100(100x)=99(50)+x& 10000x=99(50)+x& 9999x=99(50)& 101x=50& x=\boxed{\textbf{(B)}\ \frac{50}{101}} \end{align*}\]

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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