AMC 10 · 2022 · #2

Grade 6 rate-ratio
rateratio-proportionestimation dimensional-analysisidentify-subproblems ↑ Prerequisites: rateratio-proportion
📏 Short solution 💡 2 insights
Problem
Mike rode 15 laps in 57 minutes at a steady pace. About how many laps did he finish in the first 27 minutes?

Pick an answer.

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

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Analyze the Units

Constant pace = constant rate, and "rate" is what tool #8 (Analyze the Units) is built for. Track laps/minute and the units carry us straight to the answer. Once we hit 13519\frac{135}{19}, tool #6 (Guess and Check) pins the integer down with one multiplication (19 × 7 = 133), and tool #3 (Eliminate Possibilities) confirms by comparing to the five spaced-out choices. Algebra (#13) is unnecessary — the units do the work.

1STEP 1

Since 27 is just under half of 57, the laps land just under half of 15 — about 7.5.

2757\frac{27}{57}12\frac{1}{2} → laps ≈ 152\frac{15}{2} ≈ 7.5
2STEP 2

Pace is 15 laps per 57 min; multiply by 27 min and the minutes cancel, leaving 15×2757\frac{15 × 27}{57} laps.

15laps57min\frac{15 laps}{57 min} × 27 min = 15×2757\frac{15 × 27}{57} laps
3STEP 3

Both 15 and 57 share a factor of 3, so the rate simplifies and the product shrinks to 13519\frac{135}{19}.

15×2757\frac{15 × 27}{57} = 5×2719\frac{5 × 27}{19} = 13519\frac{135}{19}
4STEP 4

Guess and check: 19 × 7 = 133 leaves remainder 2, so 13519\frac{135}{19} = 7 219\frac{2}{19} ≈ 7.1 — only 7 fits the choices.

19 × 7 = 133, 135 - 133 = 2 → 13519\frac{135}{19} ≈ 7.1 → (B)
Answer
7
Pace check: 15 laps in 57 minutes is roughly one lap per 4 minutes. In 27 minutes you would expect 274\frac{27}{4} ≈ 6.75 laps — practically 7. The estimate, the exact computation 7 219\frac{2}{19}, and the choice (B) 7 all agree.
💡Key takeaway

This AMC 10 problem only needs Grade 6 rate sense — 27 minutes is just under half of 57, so the laps are just under half of 15, which lands on 7.