AMC 10 · 2022 · #2
Grade 6 rate-ratioProblem
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?
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Mike rode $15$ laps in $57$ minutes at a steady pace. About how many laps did he finish in the first $27$ minutes?
Givens: Total laps: $15$; Total time: $57$ minutes; Pace is constant (laps per minute does not change); Answer choices: (A) $5$, (B) $7$, (C) $9$, (D) $11$, (E) $13$
Unknowns: The approximate number of laps Mike completed in $27$ minutes
Understand
Restated: Mike rode $15$ laps in $57$ minutes at a steady pace. About how many laps did he finish in the first $27$ minutes?
Givens: Total laps: $15$; Total time: $57$ minutes; Pace is constant (laps per minute does not change); Answer choices: (A) $5$, (B) $7$, (C) $9$, (D) $11$, (E) $13$
Plan
Primary tool: #8 Analyze the Units
Secondary: #6 Guess and Check, #3 Eliminate Possibilities
Constant pace = constant rate, and "rate" is what tool #8 (Analyze the Units) is built for. Track $\dfrac{\text{laps}}{\text{minute}}$ and the units carry us straight to the answer. Once we hit $\dfrac{135}{19}$, tool #6 (Guess and Check) pins the integer down with one multiplication ($19 \times 7 = 133$), and tool #3 (Eliminate Possibilities) confirms by comparing to the five spaced-out choices. Algebra (#13) is unnecessary — the units do the work.
Execute — Answer: B
6.RP.A.3 Step 1 - Notice that $27$ is almost exactly half of $57$ — in fact $27/57 \approx 0.474$, just under one-half.
- So the answer should be just under half of $15$ laps, i.e.
- just under $7.5$.
- That already isolates the answer to roughly $7$.
💡 If half the time gives half the laps, then $27$ minutes (just under half of $57$) gives just under half of $15$ laps — Grade 6 rate reasoning at a glance.
6.RP.A.2 Step 2 - Make it exact.
- Mike's pace is $\dfrac{15 \text{ laps}}{57 \text{ min}}$.
- In $27$ minutes the laps are pace $\times$ time.
- The minutes cancel and we are left with laps.
💡 Treating laps/min as a unit rate and multiplying by minutes — the standard Grade 6 unit-rate move.
4.OA.B.4 Step 3 - Simplify the fraction.
- Both $15$ and $57$ are divisible by $3$ ($15 = 3 \cdot 5$, $57 = 3 \cdot 19$), so the rate simplifies and the multiplication shrinks.
💡 Spotting the shared factor of $3$ keeps the numbers small — Grade 4 factor-pair / multiples awareness.
5.NBT.B.6 Step 4 - Now identify the closest integer.
- Use guess and check on $19 \times ?$: $19 \times 7 = 133$, leaving a tiny remainder of $2$.
- So $\dfrac{135}{19} = 7\dfrac{2}{19} \approx 7.1$.
- Among the choices ($5, 7, 9, 11, 13$) only $7$ is close.
💡 Guessing $7$ and multiplying is faster than long division — Grade 5 "divide with strategies" handles the rest.
6.RP.A.3 Notice that $27$ is almost exactly half of $57$ — in fact $27/57 \approx 0.474$, 6.RP.A.2 Make it exact. Mike's pace is $\dfrac{15 \text{ laps}}{57 \text{ min}}$. In $27$ 4.OA.B.4 Simplify the fraction. Both $15$ and $57$ are divisible by $3$ ($15 = 3 \cdot 5$ 5.NBT.B.6 Now identify the closest integer. Use guess and check on $19 \times ?$: $19 \tim Review
Reasonableness: Pace check: $15$ laps in $57$ minutes is roughly one lap per $4$ minutes. In $27$ minutes you would expect $27/4 \approx 6.75$ laps — practically $7$. The estimate, the exact computation $7\dfrac{2}{19}$, and the choice (B) $7$ all agree.
Alternative: Tool #3 (Eliminate Possibilities). Use the half-time anchor: half of $57$ is $28.5$ min, which would give $7.5$ laps. Since $27 < 28.5$, the answer is just under $7.5$ — too small for (C) $9$ and way too small for (D), (E); too large for (A) $5$. Only (B) $7$ survives.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs for a whole number; recognize multiples (Spotting $3$ as a common factor of $15$ and $57$ so $\dfrac{15}{57}$ simplifies to $\dfrac{5}{19}$.)5.NBT.B.6Find whole-number quotients with up to four-digit dividends and two-digit divisors (Dividing $135$ by $19$ via $19 \times 7 = 133$ to read off the quotient $7$ with remainder $2$.)6.RP.A.2Understand the concept of a unit rate and use rate language (Writing the constant pace as a unit rate $\dfrac{15 \text{ laps}}{57 \text{ min}}$ and multiplying by $27$ min to get laps.)6.RP.A.3Use ratio and rate reasoning to solve real-world problems (Reasoning that $27$ min is almost half of $57$ min, so the laps are almost half of $15$, anchoring the estimate near $7$.)
⭐ 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 $7$.
⭐ 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 $7$.
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