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

Revision as of 21:14, 2 February 2012

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}} (Error compiling LaTeX. Unknown error_msg)

2010 AMC 10B (Problems • Answer Key • Resources)
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

fuck all

@@ Line 16: / Line 16: @@
 ==See Also==
 {{AMC10 box|year=2010|ab=B|num-b=13|num-a=15}}
+fuck all

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

Revision as of 21:14, 2 February 2012

Problem

Solution

See Also