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

Revision as of 22:13, 29 January 2011

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
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

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 == Solution ==
+My Solution: We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\times50=4,950</math>. Then, we know that the sum of the series is <math>4,950+x</math>. Since the average is <math>100x</math>, and there are <math>100</math> terms, we also find the sum to equal <math>10,000x</math>. Setting equal - <math>10,000x=4,\,950+x \Rightarrow 9,999x=4,\,950 \Rightarrow x=\frac{4,950}{9,999} \Rightarrow x= \frac{50}{101}</math>. Thus, the answer is <math>\boxed{\text{B}}</math>.
+--[[User:Kestzh|Kestzh]] 03:13, 30 January 2011 (UTC)
 == See also ==
 {{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}
 {{stub}}

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

Revision as of 22:13, 29 January 2011

Problem 10

Solution

See also