2010 AMC 10B Problem 10

The times in the problem are in minutes but the speeds are per hour, so first put the speeds in miles per minute.

1. $30$ miles per hour is $30$ miles in $60$ minutes, which is $\frac12$ mile per minute.
2. $20$ miles per hour is $20$ miles in $60$ minutes, which is $\frac13$ mile per minute.
3. Let $t$ be the number of minutes in the rain (the thing we want). The rest of the $40$ minutes, $40 - t$, were in the sun.
4. Distance is speed times time, and the two distances add to $16$ miles:
$$\frac{40 - t}{2} + \frac{t}{3} = 16.$$
5. Multiply both sides by $6$ to clear the fractions:
$$3(40 - t) + 2t = 96 \quad\Longrightarrow\quad 120 - t = 96 \quad\Longrightarrow\quad t = 24.$$
6. Check: $16$ minutes of sun at $\frac12$ mile per minute is $8$ miles, and $24$ minutes of rain at $\frac13$ mile per minute is $8$ miles. Total $16$ miles.

The answer is $\boxed{\textbf{(C)}\ 24}$.

A trip at two speeds becomes one equation once you call one of the times $t$ and write the other time as "total minus $t$". Converting the speeds to the same time unit as the data (minutes) keeps the numbers small and avoids mistakes.

Start with a guess and fix it. If all $40$ minutes were sunny she would go $40 \cdot \frac12 = 20$ miles, which is $4$ miles too many. Each minute moved from sun to rain loses $\frac12 - \frac13 = \frac16$ of a mile. To lose $4$ miles she needs $4 \div \frac16 = 24$ rainy minutes.