Difference between revisions of "2010 AMC 12B Problems/Problem 10"

Revision as of 14:13, 3 February 2011

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: $\frac{99(100)}{2}=99\times50=4,950$ . With 99 terms, the average is $\frac{99\times50}{99}$ , which equals 50. Since the average is $100x$ , we set set $100x$ equal to $50$ and solve for $x=\frac{1}{2}$ . Thus, the answer is $\boxed{\text{C}}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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

This article is a stub. Help us out by expanding it.

@@ Line 5: / Line 5: @@
 == Solution ==
-We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\times50=4,950</math>. With 99 terms, the average is <math>\frac{99\times50}{99}</math>, which equals 50. Since the average is <math>100x</math>, we set set <math>100x</math> equal to <math>50</math> and solve for <math>x</math> = <math>\frac{1}{2}</math>. Thus, the answer is <math>\boxed{\text{C}}</math>.
+We first sum the first <math>99</math> numbers: <math>\frac{99(100)}{2}=99\times50=4,950</math>. With 99 terms, the average is <math>\frac{99\times50}{99}</math>, which equals 50. Since the average is <math>100x</math>, we set set <math>100x</math> equal to <math>50</math> and solve for <math>x=\frac{1}{2}</math>. Thus, the answer is <math>\boxed{\text{C}}</math>.
 == See also ==
 {{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}
 {{stub}}

Page

Toolbox

Search

Difference between revisions of "2010 AMC 12B Problems/Problem 10"

Revision as of 14:13, 3 February 2011

Problem 10

Solution

See also