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Difference between revisions of "2010 AMC 12B Problems/Problem 10"
(Made the solution much easier computationally.)
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== Solution ==
 
== Solution ==
−
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>.
+
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>:
 +
<cmath>\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x</cmath>
 +
Using difference of squares:
 +
<cmath>x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}</cmath>
 +
Thus, the answer is <math>\boxed{\text{B}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=9|num-a=11|ab=B}}
 
{{AMC12 box|year=2010|num-b=9|num-a=11|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:29, 18 January 2020

Problem 10

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 $\boxed{\text{B}}$.

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

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