2021 AMC 10B Problem 1

1. First find out how big $3\pi$ is. Since $\pi \approx 3.14$, we get $3\pi \approx 9.42$.
2. $|x|$ is the distance from $x$ to $0$ on the number line. So $|x| < 9.42$ means $x$ is less than $9.42$ steps away from $0$, on either side.
3. Check the edge: $|9| = 9$ is less than $9.42$, so $9$ works. But $|10| = 10$ is too big. The same holds for $-9$ and $-10$.
4. So the whole numbers that work are
$$-9, -8, \ldots, -1, \quad 0, \quad 1, 2, \ldots, 9.$$
5. Count them: $9$ negative numbers, $9$ positive numbers, and $0$. That is $9 + 9 + 1 = 19$.

The answer is $\boxed{\textbf{(D)}\ 19}$.

"$|x|$ is less than some number" always gives a stretch of the number line that is the same on both sides of $0$. So count the positive numbers, double that, and add one for $0$. The only real work is deciding which two whole numbers $3\pi$ sits between, because that tells you where the list stops.

Since $\pi \approx 3.14$, we have $3\pi \approx 9.42$. The inequality $|x| < 9.42$ means $-9.42 < x < 9.42$.

The integers in this range are $-9, -8, \dots, -1, 0, 1, \dots, 8, 9$: nine negative integers, nine positive integers, and zero. That is $9 + 9 + 1 = 19$ integers.

The answer is $\boxed{\textbf{(D)}\ 19}$.