2018 AMC 10B Problem 2

The speeds are in miles per hour, so first turn $30$ minutes into hours: $30$ minutes is $\tfrac12$ hour.

1. First half hour at $60$ mph: $60 \times \tfrac12 = 30$ miles.
2. Second half hour at $65$ mph: $65 \times \tfrac12 = 32.5$ miles.
3. The whole trip was $96$ miles, so the last half hour covered
$$96 - 30 - 32.5 = 33.5 \text{ miles}.$$
4. Going $33.5$ miles in half an hour means going twice that in a full hour: $33.5 \times 2 = 67$ mph.

The answer is $\boxed{\textbf{(D)}\ 67}$.

A trip in pieces is easiest handled one piece at a time: distance equals speed times time for each piece, and the pieces add up to the whole trip. Always put the time in hours before multiplying by a speed in miles per hour.

The three stretches are all the same length of time, so the average speed for the whole trip is the plain average of the three speeds. The whole trip averages $96$ miles in $1.5$ hours, which is $96 \div 1.5 = 64$ mph. So the three speeds add up to $3 \times 64 = 192$, and the last speed is $192 - 60 - 65 = 67$.

Each $30$-minute stretch is half an hour.

• First half hour at $60$ mph: $60 \cdot \tfrac12 = 30$ miles.
• Second half hour at $65$ mph: $65 \cdot \tfrac12 = 32.5$ miles.

That leaves $96 - 30 - 32.5 = 33.5$ miles for the final half hour, so the speed then was
$$\frac{33.5}{1/2} = 67 \text{ mph}.$$

The answer is $\boxed{\textbf{(D)}\ 67}$.