2022 AMC 10A Problem 2

Because Mike's speed never changes, the fraction of the time that has passed is the same as the fraction of the laps he has done.

1. What part of the ride is $27$ minutes? It is $\frac{27}{57}$ of it. Both numbers divide by $3$, so this is $\frac{9}{19}$.
2. He has done $\frac{9}{19}$ of the $15$ laps: $15 \times \frac{9}{19} = \frac{135}{19}$.
3. Since $19 \times 7 = 133$, we get $\frac{135}{19} = 7\frac{2}{19}$, just a little over $7$.

The closest choice is $7$.

The answer is $\boxed{\textbf{(B)}\ 7}$.

"Constant speed" means twice the time gives twice the distance. So a question about part of the trip is a fraction question: find what fraction of the time has passed, and take that fraction of the distance. Simplify the fraction first so the arithmetic stays small, and remember that only an approximate answer is needed.

Find the time for one lap: $57 \div 15 = 3.8$ minutes. In $27$ minutes he rides $27 \div 3.8 \approx 7.1$ laps.

Or just estimate: $27$ is a bit less than half of $57$, so he did a bit less than half of $15$ laps, which is $7.5$. Only $7$ is close to that.

At constant speed, the number of laps is proportional to the elapsed time. The first $27$ minutes are $\frac{27}{57} = \frac{9}{19}$ of the whole ride, so Mike completed
$$15 \cdot \frac{9}{19} = \frac{135}{19} = 7\tfrac{2}{19} \approx 7.1$$
laps. The closest choice is $7$.

The answer is $\boxed{\textbf{(B)}\ 7}$.