AMC 8 · 2016 · #4

Easy mode Grade 6

Problem

When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. So back then, each mile took him a certain number of minutes.

Now, as an old man, he walks $10$ miles in $4$ hours. Each mile now also takes some number of minutes.

How many more minutes does one mile take him now than it did when he was a boy?

$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: As a boy, Cheenu ran $15$ miles in $3$ hours $30$ minutes. As an old man, he walks $10$ miles in $4$ hours. How many more minutes does it now take him to cover one mile than it did when he was a boy?

Givens: Boy: $15$ miles in $3$ h $30$ min; Old man: $10$ miles in $4$ h; Answer choices: (A) $6$, (B) $10$, (C) $15$, (D) $18$, (E) $30$ (minutes)

Unknowns: The difference, in minutes per mile, between his old-man pace and his boyhood pace

Understand

Givens: Boy: $15$ miles in $3$ h $30$ min; Old man: $10$ miles in $4$ h; Answer choices: (A) $6$, (B) $10$, (C) $15$, (D) $18$, (E) $30$ (minutes)

Plan

Primary tool: #8 Analyze the Units

Secondary: #7 Identify Subproblems

The question asks for a difference in minutes per mile, but the data are mixed in hours and minutes and use totals (not per-mile rates). Tool #8 (Analyze the Units) tells us to first convert each total time to minutes, then divide by miles so the unit "minutes per mile" pops out — only then can we subtract. Tool #7 (Identify Subproblems) splits the work into three clean pieces: (1) boy's pace, (2) old man's pace, (3) the difference.

Execute — Answer: B

#8 Analyze the Units 5.MD.A.1 Step 1

Find the boy's pace in minutes per mile.
First convert $3$ hours $30$ minutes to minutes, then divide by $15$ miles so the answer comes out in minutes per mile.

$$3 \text{ h} \times 60 \tfrac{\text{min}}{\text{h}} + 30 \text{ min} = 210 \text{ min}; \quad \dfrac{210 \text{ min}}{15 \text{ mi}} = 14 \tfrac{\text{min}}{\text{mi}}$$

💡 Converting hours to minutes within the same time system is exactly the Grade 5 "convert measurement units" move.

#7 Identify Subproblems 6.RP.A.3 Step 2

Find the old man's pace in minutes per mile the same way.
Convert $4$ hours to minutes, then divide by $10$ miles.

$$4 \text{ h} \times 60 \tfrac{\text{min}}{\text{h}} = 240 \text{ min}; \quad \dfrac{240 \text{ min}}{10 \text{ mi}} = 24 \tfrac{\text{min}}{\text{mi}}$$

💡 Computing a unit rate (minutes per mile) from a total time and a total distance is Grade 6 rate reasoning.

#8 Analyze the Units 4.OA.A.3 Step 3

Subtract the two paces.
Both are now in the same unit (minutes per mile), so the difference is the number of extra minutes needed per mile.

$$24 \tfrac{\text{min}}{\text{mi}} - 14 \tfrac{\text{min}}{\text{mi}} = 10 \tfrac{\text{min}}{\text{mi}} \;\Rightarrow\; \textbf{(B)}$$

💡 "How many more" with matching units is a Grade 4 multi-step word-problem subtraction.

[1] #8 5.MD.A.1 Find the boy's pace in minutes per mile. First convert $3$ hours $30$ minutes to

[2] #7 6.RP.A.3 Find the old man's pace in minutes per mile the same way. Convert $4$ hours to m

[3] #8 4.OA.A.3 Subtract the two paces. Both are now in the same unit (minutes per mile), so the

Review

Reasonableness: Boy's pace $= 14$ min/mi corresponds to a speed of $\tfrac{60}{14} \approx 4.3$ mph — a slow run, which fits "run." Old man's pace $= 24$ min/mi corresponds to $\tfrac{60}{24} = 2.5$ mph — a slow walk, which fits "walk." The $10$-minute gap is the right order of magnitude (single digits to mid-teens), matching choice (B).

Alternative: Tool #5 (Find a Pattern / equivalent ratios). Boy: $15$ miles in $210$ min $\Rightarrow$ $1$ mile in $\tfrac{210}{15} = 14$ min by scaling the ratio down by $15$. Old man: $10$ miles in $240$ min $\Rightarrow$ $1$ mile in $\tfrac{240}{10} = 24$ min by scaling down by $10$. The same answer, $24 - 14 = 10$ min, falls out from equivalent-ratio reasoning instead of explicit division.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multistep word problems using the four operations (Subtracting the two per-mile paces ($24 - 14 = 10$ min/mi) to answer the "how many minutes longer" question.)
5.MD.A.1 Convert among different-sized standard measurement units within a given system (Converting $3$ h $30$ min to $210$ min and $4$ h to $240$ min so both times share the unit "minutes.")
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Computing the unit rates $\tfrac{210}{15} = 14$ and $\tfrac{240}{10} = 24$ minutes per mile from total time and total distance.)

⭐ This AMC 8 problem only needs the Grade 6 idea that "unit rate = total $\div$ count" — divide minutes by miles, then subtract.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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