2010 AMC 10B Problem 14

Question

The average of the numbers

1,

2,

3,

\cdots,

98,

99,

and

x

is

100x.

What is

x?

\dfrac{49}{101}

\dfrac{50}{101}

\dfrac{1}{2}

\dfrac{51}{101}

\dfrac{50}{99}

Use the sum

1+2+\cdots+99

Set

\dfrac{99\cdot50+x}{100}=100x

Recall that the sum of the first

n

integers is

\dfrac{n(n + 1)}{2}.

Then, we have that

\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x,

which simplifies to

99 \cdot 50 = (100^2 - 1)x

= 101 \cdot 99x,

by difference of squares. Dividing gives us

x = \dfrac{50}{101}.

Thus, B is the correct answer.

Problem 14 in other years