2003 AMC 10B Problem 5

1. The mower cuts $28$ inches, but $4$ inches of each pass is grass that was already cut. So each pass mows only $28 - 4 = 24$ inches of new grass. That is exactly $2$ feet.
2. In one hour Moe walks $5000$ feet. So in one hour he mows a strip $5000$ feet long and $2$ feet wide:
$$5000 \times 2 = 10000 \text{ square feet per hour}.$$
3. The lawn is $90 \times 150 = 13500$ square feet.
4. Time needed:
$$13500 \div 10000 = 1.35 \text{ hours}.$$

The answer is $\boxed{\textbf{(C)}\ 1.35}$.

Mowing speed in area per hour is walking speed times the useful width of the cut. The overlap only shrinks that useful width. Put everything in the same unit (feet) before you multiply, and the rest is one division.

Picture the strips running along the $90$-foot side. Each strip covers $2$ feet of the $150$-foot side, so Moe needs $150 \div 2 = 75$ strips of $90$ feet. He walks $75 \times 90 = 6750$ feet, which takes $6750 \div 5000 = 1.35$ hours.

Because each pass overlaps the previous one by $4$ inches, every pass adds only $28 - 4 = 24$ inches of new grass, which is exactly $2$ feet.

While walking $5000$ feet in an hour, Moe therefore clears a strip $5000$ feet long and $2$ feet wide: $10000$ square feet per hour.

The lawn has area $90 \cdot 150 = 13500$ square feet, so the time needed is
$$\frac{13500}{10000} = 1.35 \text{ hours}.$$

The answer is $\boxed{\textbf{(C)}\ 1.35}$.