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2010 AMC 10B Problem 14

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2010 AMC 10B Problem 14AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 2 of 5 · about 2 min · position P11-15

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

A)
$\dfrac{49}{101}$
B)
$\dfrac{50}{101}$
C)
$\dfrac{1}{2}$
D)
$\dfrac{51}{101}$
E)
$\dfrac{50}{99}$

Answer

Key insight

There are 100 numbers whose sum is 4950 + x, so the average condition reads 4950 + x = 10000x, giving x = 4950/9999.

Solution

The list has $99 + 1 = 100$ numbers. The sum of $1$ through $99$ is $\frac{99 \cdot 100}{2} = 4950$ , so the total is $4950 + x$ .

Average equals $100x$ :
$\frac{4950 + x}{100} = 100x \quad\Longrightarrow\quad 4950 + x = 10000x \quad\Longrightarrow\quad 9999x = 4950.$
Both sides are multiples of $99$ : $4950 = 99 \cdot 50$ and $9999 = 99 \cdot 101$ . Hence
$x = \frac{50}{101}.$

The answer is $\boxed{\textbf{(B)}\ \dfrac{50}{101}}$ .

Why this works

"Average of a list equals something" is just "sum equals count times something." The arithmetic-series formula gives the sum instantly, leaving a one-variable linear equation. Recognizing $9999 = 99 \cdot 101$ makes the reduction immediate.

The trap

Dividing by 99 instead of 100 (there are 100 numbers including x), or failing to reduce 4950/9999 by 99.

Common mistakes

Dividing by 99 instead of 100 (there are 100 numbers including x), or failing to reduce 4950/9999 by 99.
Using $\frac{99\cdot100}{2} = 4950$ correctly but then writing the equation as $4950 + x = 100x$ , which forgets the division by the count.

Techniques

Set up the equation/formula and compute; no special trick needed