2010 AMC 10B Problem 14

An average is the sum divided by how many numbers there are, so find the sum and the count.

1. Count. The numbers $1$ through $99$ are $99$ numbers, and $x$ makes $100$.
2. Sum of $1$ through $99$. Pair the first with the last: $1 + 99 = 100$, $2 + 98 = 100$, and so on. There are $49$ such pairs plus the middle number $50$, so the sum is $49 \cdot 100 + 50 = 4950$.
3. The whole list adds to $4950 + x$.
4. The average is $100x$, so
$$\frac{4950 + x}{100} = 100x.$$
Multiply both sides by $100$: $4950 + x = 10000x$, so $9999x = 4950$.
5. Simplify $\frac{4950}{9999}$. Both numbers are multiples of $99$: $4950 = 99 \cdot 50$ and $9999 = 99 \cdot 101$. So
$$x = \frac{4950}{9999} = \frac{50}{101}.$$

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

"The average is something" means "the sum equals the count times that something", which turns the problem into a simple equation. Pairing numbers from the two ends is the quick way to add $1$ through $99$. It is also worth knowing that $9999 = 99 \cdot 101$, which makes the fraction easy to reduce.

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