The average of the numbers 1, 2, 3, …, 98, 99, and x is 100x. What is x?

2010 AMC 12B Problem 10

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10.

The average of the numbers

1,

2,

3,

\ldots,

98,

99,

and

x

100x.

What is

x?

\dfrac{49}{101}

\dfrac{50}{101}

\dfrac{1}{2}

\dfrac{51}{101}

\dfrac{50}{99}

The sum

1+2+\cdots+99

equals

4950

\dfrac{4950+x}{100}=100x

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.