AMC 10 · 2022 · #2

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition

rateratio-proportionestimation dimensional-analysisidentify-subproblems ↑ Prerequisites: rateratio-proportion

📏 Short solution 💡 2 insights

Problem

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?

$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

#8 Analyze the Units 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 $7$.

$$\dfrac{27}{57} \approx \dfrac{1}{2} \;\Rightarrow\; \text{laps} \approx \dfrac{15}{2} \approx 7.5$$

💡 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.

#8 Analyze the Units 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.

$$\dfrac{15 \text{ laps}}{57 \text{ min}} \times 27 \text{ min} = \dfrac{15 \times 27}{57} \text{ laps}$$

💡 Treating laps/min as a unit rate and multiplying by minutes — the standard Grade 6 unit-rate move.

#8 Analyze the Units 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.

$$\dfrac{15 \times 27}{57} = \dfrac{5 \times 27}{19} = \dfrac{135}{19}$$

💡 Spotting the shared factor of $3$ keeps the numbers small — Grade 4 factor-pair / multiples awareness.

#6 Guess and Check 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.

$$19 \times 7 = 133, \quad 135 - 133 = 2 \;\Rightarrow\; \dfrac{135}{19} \approx 7.1 \;\Rightarrow\; \textbf{(B)}$$

💡 Guessing $7$ and multiplying is faster than long division — Grade 5 "divide with strategies" handles the rest.

[1] #8 6.RP.A.3 Notice that $27$ is almost exactly half of $57$ — in fact $27/57 \approx 0.474$,

[2] #8 6.RP.A.2 Make it exact. Mike's pace is $\dfrac{15 \text{ laps}}{57 \text{ min}}$. In $27$

[3] #8 4.OA.B.4 Simplify the fraction. Both $15$ and $57$ are divisible by $3$ ($15 = 3 \cdot 5$

[4] #6 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.4 Find 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.6 Find 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.2 Understand 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.3 Use 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 $7$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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