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

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== Problem 10 ==
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{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #10]] and [[2010 AMC 10B Problems|2010 AMC 10B #14]]}}
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== Problem ==
 
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>?
 
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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== Solution ==
 
== Solution ==
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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>.
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We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\cdot50</math>. Then, we know that the sum of the series is <math>99\cdot50+x</math>. There are <math>100</math> terms, so we can divide this sum by <math>100</math> and set it equal to <math>100x</math>:
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--[[User:Kestzh|Kestzh]] 03:13, 30 January 2011 (UTC)
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<cmath>\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x</cmath>
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Using difference of squares:
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<cmath>x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}</cmath>
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Thus, the answer is <math>\boxed{\textbf{(B) }\frac{50}{101}}</math>.
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==Video Solution==
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https://youtu.be/vYXz4wStBUU?t=413
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~IceMatrix
  
 
== See also ==
 
== See also ==
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{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}
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{{AMC12 box|year=2010|num-b=9|num-a=11|ab=B}}
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{{stub}}
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{{AMC10 box|year=2010|num-b=13|num-a=15|ab=B}}
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{{MAA Notice}}

Latest revision as of 16:06, 6 July 2023

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

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 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 $\boxed{\textbf{(B) }\frac{50}{101}}$.

Video Solution

https://youtu.be/vYXz4wStBUU?t=413

~IceMatrix

See also

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
2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 10 Problems and Solutions

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