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

My Solution: We first sum the first $99$ numbers: $\frac{99(100)}{2}=99\times50=4,950$ . Then, we know that the sum of the series is $4,950+x$ . Since the average is $100x$ , and there are $100$ terms, we also find the sum to equal $10,000x$ . Setting equal - $10,000x=4,\,950+x \Rightarrow 9,999x=4,\,950 \Rightarrow x=\frac{4,950}{9,999} \Rightarrow x= \frac{50}{101}$ . Thus, the answer is $\boxed{\text{B}}$ . --Kestzh 03:13, 30 January 2011 (UTC)

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2010 AMC 12B Problems/Problem 10

Problem 10

Solution

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