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

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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:
 
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,
 
100(100x)=99(50)+x,
 
10000x=99(50)+x,
 
10000x=99(50)+x,
 
9999x=99(50),
 
9999x=99(50),
 
101x=50,
 
101x=50,
 +
</math>
 +
x=\frac{50}{101}
 
<math>
 
<math>
x=\frac{50}{101}
+
This gives us our answer. </math> \boxed{\mathrm{(B)}= \frac{50}{101}} $
</math>
 
This gives us our answer. <math> \boxed{\mathrm{(B)}= \frac{50}{101}} </math>
 

Revision as of 15:14, 24 January 2011

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: $100(100x)=99(50)+x, 10000x=99(50)+x, 9999x=99(50), 101x=50,$ x=\frac{50}{101} $This gives us our answer.$ \boxed{\mathrm{(B)}= \frac{50}{101}} $