2022 AMC 10A Problem 3

1. The first number is described using the third, and so is the second. So the third number is the one to name. Call it $c$.
2. The first number is $6$ times the third: it is $6c$.
3. The third is $40$ less than the second, so the second is $40$ more than the third: it is $c + 40$.
4. The three numbers add to $96$:
$$6c + (c + 40) + c = 96.$$
That is $8c + 40 = 96$, so $8c = 56$ and $c = 7$.
5. The numbers are $42$, $47$ and $7$. Check: $42 + 47 + 7 = 96$.
6. The difference between the first and second numbers is $47 - 42 = 5$.

The answer is $\boxed{\textbf{(E)}\ 5}$.

When several numbers are all described in terms of one of them, make that one your variable; then the whole problem is a single equation. Be careful with "less than": "$A$ is $40$ less than $B$" means $B$ is the bigger number and $A = B - 40$.

Both other numbers are described relative to the third, so let the third number be $c$. Then the first number is $6c$, and since the third is $40$ less than the second, the second is $c + 40$.

The sum condition gives
$$6c + (c + 40) + c = 96 \quad\Longrightarrow\quad 8c = 56 \quad\Longrightarrow\quad c = 7.$$
So the first number is $42$ and the second is $47$, and $|42 - 47| = 5$.

The answer is $\boxed{\textbf{(E)}\ 5}$.