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

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==Problem==
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#redirect [[2010 AMC 12B Problems/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>?
 
 
 
<math>\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}</math>
 
 
 
==Solution==
 
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:
 
 
 
\[\begin{align*}
 
100(100x)=99(50)+x&\\
 
10000x=99(50)+x&\\
 
9999x=99(50)&\\
 
101x=50&\\
 
x=\boxed{\textbf{(B)}\ \frac{50}{101}}&\\
 
\end{align*}\]
 
 
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 

Latest revision as of 20:42, 26 May 2020

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