2010 AMC 12B Problem 10

Problem 10 of 25EasierAlgebra

The average of the numbers

1,

2,

3,

\ldots,

98,

99,

and

x

100x.

What is

x?

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Solution

The numbers

1

through

99

sum to

\dfrac{99\cdot100}{2}=4950.

The average condition is

\frac{4950+x}{100}=100x,

4950+x=10000x

and

9999x=4950.

Thus

x=\dfrac{4950}{9999}=\dfrac{50}{101}.

Thus, the correct answer is B.

Concepts: mean · summation · linear equation

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.