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2022 AMC 10A Problem 2

2022 AMC 10A Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?

A)
$5$
B)
$7$
C)
$9$
D)
$11$
E)
$13$

Answer

Key insight

Laps are proportional to time at constant speed: 15 laps times 27/57 is 135/19, slightly more than 7.

Solution

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 $\boxed{\textbf{(B)}\ 7}$ .

Why this works

"Constant speed" turns a distance question into a proportion: the fraction of the time elapsed equals the fraction of the distance covered. Reduce the fraction first ( $\frac{27}{57} = \frac{9}{19}$ ) so the arithmetic stays small, and remember the answer only needs to be approximate.

Alternative approach

Rate per lap: $57/15 = 3.8$ minutes per lap. In $27$ minutes that is $27/3.8 \approx 7.1$ laps. Either way the estimate lands just above $7$ , well short of $9$ .

The trap

Approximating 27/57 as one half to get 7.5 laps, then hesitating between 7 and 9 without noticing 27/57 is less than one half.

Common mistakes

Approximating 27/57 as one half to get 7.5 laps, then hesitating between 7 and 9 without noticing 27/57 is less than one half.
Inverting the proportion (computing $57 \cdot 15 / 27$ or similar) and getting a number of laps larger than $15$ .

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed