2010 AMC 12B Problems/Problem 10

Problem 10

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 first sum the first $99$ numbers: $\frac{99(100)}{2}=99\times50=4,950$ . With 99 terms, the average is $\frac{99\times50}{99}$ , which equals 50. Since the average is $100x$ , we set set $100x$ equal to $50$ and solve for $x$ = $\frac{1}{2}$ . Thus, the answer is $\boxed{\text{C}}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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2010 AMC 12B Problems/Problem 10

Problem 10

Solution

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