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

(Solution)
Line 7: Line 7:
 
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:
  
<cmath>\begin{align*}
+
\[\begin{align*}
100(100x)&=99(50)+x\\
+
100(100x)=99(50)+x&
10000x&=99(50)+x\\
+
10000x=99(50)+x&
9999x&=99(50)\\
+
9999x=99(50)&
101x&=50\\
+
101x=50&
x&=\boxed{\textbf{(B)}\ \frac{50}{101}}</cmath>
+
x=\boxed{\textbf{(B)}\ \frac{50}{101}}
 +
\end{align*}\]
  
 
==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 18:46, 13 July 2015

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:

\[\begin{align*} 100(100x)=99(50)+x& 10000x=99(50)+x& 9999x=99(50)& 101x=50& x=\boxed{\textbf{(B)}\ \frac{50}{101}} \end{align*}\]

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS