2021 AMC Fall 10A Problem 6

1. Count the gaps. Try a small case first: from pole $1$ to pole $3$ there are $2$ gaps (1 to 2, and 2 to 3). In the same way, from pole $1$ to pole $41$ there are $41 - 1 = 40$ gaps.
2. Length of one gap. The $40$ equal gaps add up to one mile, $5280$ feet. So one gap is
$$\frac{5280}{40} = 132 \text{ feet}.$$
3. Elmer's stride. Elmer takes $44$ strides to cover one gap, so one stride is $\frac{132}{44} = 3$ feet.
4. Oscar's leap. Oscar takes $12$ leaps to cover one gap, so one leap is $\frac{132}{12} = 11$ feet.
5. Difference. $11 - 3 = 8$ feet.

The answer is $\boxed{\textbf{(B)}\ 8}$.

Both animals are measured against the same length, the space between two poles. Find that one length first, then split it into strides and into leaps. The only tricky part is the "fence post" count: $41$ poles in a row have $40$ spaces between them. A good check: $5280 \div 41$ is not a whole number, a sign that $41$ is the wrong divisor.