Difference between revisions of "2010 AMC 12B Problems/Problem 10"

Revision as of 19:43, 26 May 2020

The following problem is from both the 2010 AMC 12B #10 and 2010 AMC 10B #14, so both problems redirect to this page.

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: $\frac{99(100)}{2}=99\cdot50$ . Then, we know that the sum of the series is $99\cdot50+x$ . There are $100$ terms, so we can divide this sum by $100$ and set it equal to $100x$ : $\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x$ Using difference of squares: $x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}$ Thus, the answer is $\boxed{\text{B}}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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+{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #10]] and [[2010 AMC 10B Problems|2010 AMC 10B #14]]}}
 == Problem 10 ==
 The average of the numbers <math>1, 2, 3,\cdots, 98, 99,</math> and <math>x</math> is <math>100x</math>. What is <math>x</math>?

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Difference between revisions of "2010 AMC 12B Problems/Problem 10"

Revision as of 19:43, 26 May 2020

Problem 10

Solution

See also