2002 AMC 10A Problem 12

At $40$ mph he is $3$ minutes late, and at $60$ mph he is $3$ minutes early. So the two trips differ by $6$ minutes. That difference tells us how far he drives.

1. Try one mile. At $40$ mph, one mile takes $\frac{60}{40} = 1.5$ minutes. At $60$ mph, one mile takes $1$ minute. The difference is $0.5$ minutes per mile.

2. Find the distance. We need a difference of $6$ minutes, and each mile adds $0.5$ minutes, so the distance is
$$6 \div 0.5 = 12 \text{ miles}.$$

1. Find the on-time trip length. At $40$ mph, $12$ miles take $12 \times 1.5 = 18$ minutes, and that is $3$ minutes late. So arriving on time means taking $15$ minutes. (Check: at $60$ mph the trip takes $12$ minutes, which is $3$ minutes early.)

4. Find the speed. $15$ minutes is $\frac{1}{4}$ of an hour, so
$$\text{speed} = 12 \div \frac14 = 48 \text{ miles per hour}.$$

The answer is $\boxed{\textbf{(B)}\ 48}$.

Two trips over the same road at different speeds, with a known difference in time, always pin down the distance. After that, the needed speed is just distance divided by time. Averaging the speeds is wrong because he spends more time at the slower speed; average the times instead.

Use an equation in hours. Let $d$ be the distance. The trip times are $\frac{d}{40}$ and $\frac{d}{60}$ hours, and $6$ minutes is $\frac{1}{10}$ hour:
$$\frac{d}{40} - \frac{d}{60} = \frac{1}{10} \quad\Longrightarrow\quad \frac{3d - 2d}{120} = \frac{1}{10} \quad\Longrightarrow\quad d = 12.$$
Then continue as in steps 3 and 4.