2002 AMC 10A Problem 12

Let the distance to work be $d$ miles. At $40$ mph the trip takes $\frac{d}{40}$ hours; at $60$ mph it takes $\frac{d}{60}$ hours. The first is $3$ minutes late and the second $3$ minutes early, so they differ by $6$ minutes $= \frac{1}{10}$ hour:
$$\frac{d}{40} - \frac{d}{60} = \frac{1}{10} \quad\Longrightarrow\quad \frac{3d - 2d}{120} = \frac{1}{10} \quad\Longrightarrow\quad d = 12.$$

The on-time duration is the $40$ mph time minus $3$ minutes: $\frac{12}{40}$ hours $= 18$ minutes, less $3$ minutes, is $15$ minutes $= \frac14$ hour. The required speed is
$$\frac{12 \text{ miles}}{1/4 \text{ hour}} = 48 \text{ mph}.$$

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

Two trips over the same distance at known speeds, with a known time difference, always determine the distance through $d/v_1 - d/v_2 = \Delta t$. Once $d$ and the target time are known, the needed speed is just distance over time. Keep units consistent by converting minutes to hours at the start.

Because the late and early margins are equal, the on-time speed is the harmonic mean of $40$ and $60$: $\frac{2}{1/40 + 1/60} = \frac{2 \cdot 120}{5} = 48$. This works since the required time is the average of the two trip times, and time is distance over speed.

• Averaging 40 and 60 to get 50, which ignores that the slower trip takes longer and so is weighted more heavily.
• Using $3$ minutes instead of $6$ for the difference between the trips, or leaving the $3$ minutes as $3$ hours in the equation.