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Difference between revisions of "2010 AMC 10B Problems/Problem 14"

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==Solution==
 
==Solution==
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We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives:
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We must find the average of the numbers from <math> 1 </math> to <math> 99 </math> and <math> x </math> in terms of <math> x </math>. The sum of all these terms is <math> \frac{99(100)}{2}+x=99(50)+x </math>. We must divide this by the total number of terms, which is <math> 100 </math>. We get: <math> \frac{99(50)+x}{100} </math>. This is equal to <math> 100x </math>, as stated in the problem. We have: <math> \frac{99(50)+x}{100}=100x </math>. We can now cross multiply. This gives <math>100(100x)=99(50)+x \rightarrow 10000x=99(50)+x \rightarrow 9999x=99(50) \rightarrow 101x=50 \rightarrow x=\boxed{\text{(B)} \frac{50}{101}}</math>
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<math>\[\begin{align*}
 
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100(100x)=99(50)+x&\\
 
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10000x=99(50)+x&\\
 
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9999x=99(50)&\\
 
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101x=50&\\
 
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x=\boxed{\textbf{(B)}\ \frac{50}{101}}&\\
 
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\end{align*}\]</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2010|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:17, 16 July 2016

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 must find the average of the numbers from $1$ to $99$ and $x$ in terms of $x$. The sum of all these terms is $\frac{99(100)}{2}+x=99(50)+x$. We must divide this by the total number of terms, which is $100$. We get: $\frac{99(50)+x}{100}$. This is equal to $100x$, as stated in the problem. We have: $\frac{99(50)+x}{100}=100x$. We can now cross multiply. This gives $100(100x)=99(50)+x \rightarrow 10000x=99(50)+x \rightarrow 9999x=99(50) \rightarrow 101x=50 \rightarrow x=\boxed{\text{(B)} \frac{50}{101}}$

See Also

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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