AMC 8 · 2013 · #11

Grade 6 rate-ratioarithmetic

Archetype Arithmetic Expression Decomposition

ratefraction-arithmeticunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmeticrate

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Problem

Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Ted's grandfather walked or jogged $2$ miles on each of three days. His speeds were $5$ mph on Monday, $3$ mph on Wednesday, and $4$ mph on Friday. If he had instead walked at $4$ mph on all three days, how many fewer minutes would he have spent on the treadmill?

Givens: Distance per day = $2$ miles, on each of $3$ days; Actual speeds: Monday $5$ mph, Wednesday $3$ mph, Friday $4$ mph; Hypothetical scenario: $4$ mph on all three days, same $2$ miles per day; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$ (minutes)

Unknowns: Difference (in minutes) between the actual total treadmill time and the hypothetical total time at $4$ mph

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #8 Analyze the Units

The question is a straight $\text{time} = \text{distance} / \text{speed}$ setup, but it bundles three separate days plus a hypothetical comparison. Tool #7 (Identify Subproblems) keeps the work clean: compute each day's actual time, sum them, then compare with the hypothetical total. Tool #8 (Analyze the Units) reminds us that miles $\div$ mph gives hours, so the final answer needs an hours-to-minutes conversion.

Execute — Answer: D

#7 Identify Subproblems 6.RP.A.3 Step 1

Find each day's actual time using $\text{time} = \text{distance} / \text{speed}$.
The distance is $2$ miles every day.

$$t_\text{Mon} = \tfrac{2}{5} \text{ hr}, \quad t_\text{Wed} = \tfrac{2}{3} \text{ hr}, \quad t_\text{Fri} = \tfrac{2}{4} = \tfrac{1}{2} \text{ hr}$$

💡 Dividing distance by speed is the basic rate move; doing it once per day is the Tool #7 split.

#7 Identify Subproblems 5.NF.A.1 Step 2

Add the three actual times.
The least common denominator of $5$, $3$, and $2$ is $30$, so rewrite each fraction with denominator $30$ and add.

$$\tfrac{2}{5} + \tfrac{2}{3} + \tfrac{1}{2} = \tfrac{12}{30} + \tfrac{20}{30} + \tfrac{15}{30} = \tfrac{47}{30} \text{ hr}$$

💡 Combining unlike fractions with a common denominator is the Grade 5 fraction-addition standard.

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

Find the hypothetical total time.
At $4$ mph, each day's $2$ miles takes $\tfrac{1}{2}$ hour, and there are three such days.

$$3 \times \tfrac{1}{2} = \tfrac{3}{2} \text{ hr}$$

💡 Same $\text{time} = \text{distance}/\text{speed}$ rule, just held constant across all three days.

#7 Identify Subproblems 5.NF.A.1 Step 4

Subtract to get the time saved.
Use denominator $30$ again: $\tfrac{3}{2} = \tfrac{45}{30}$.

$$\tfrac{47}{30} - \tfrac{45}{30} = \tfrac{2}{30} = \tfrac{1}{15} \text{ hr}$$

💡 Subtracting fractions with a common denominator is the same Grade 5 move as adding them.

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

Convert the saved time from hours to minutes so the answer matches the units in the choices.

$$\tfrac{1}{15} \text{ hr} \times 60 \tfrac{\text{min}}{\text{hr}} = \tfrac{60}{15} = 4 \text{ min} \;\Rightarrow\; \textbf{(D)}$$

💡 Multiplying hours by $60$ min/hr cancels "hours" and leaves "minutes" — straight unit conversion.

[1] #7 6.RP.A.3 Find each day's actual time using $\text{time} = \text{distance} / \text{speed}$

[2] #7 5.NF.A.1 Add the three actual times. The least common denominator of $5$, $3$, and $2$ is

[3] #7 6.RP.A.3 Find the hypothetical total time. At $4$ mph, each day's $2$ miles takes $\tfrac

[4] #7 5.NF.A.1 Subtract to get the time saved. Use denominator $30$ again: $\tfrac{3}{2} = \tfr

[5] #8 5.MD.A.1 Convert the saved time from hours to minutes so the answer matches the units in

Review

Reasonableness: On Monday he was faster than $4$ mph ($5$ vs $4$), so he saved a bit of time; on Wednesday he was slower ($3$ vs $4$), so he lost time; on Friday he was already at $4$ mph, so no change. The Wednesday loss should outweigh the Monday save because dropping from $4$ to $3$ mph stretches $2$ miles from $30$ min to $40$ min (a $10$-min gain), while jumping from $4$ to $5$ mph shrinks $2$ miles from $30$ min to $24$ min (a $6$-min save). Net extra time actually spent: $10 - 6 = 4$ min, matching (D).

Alternative: Tool #2 (Find a Pattern / Compare Cases): compute each day's time directly in minutes — Monday $= \tfrac{2}{5} \times 60 = 24$, Wednesday $= \tfrac{2}{3} \times 60 = 40$, Friday $= \tfrac{1}{2} \times 60 = 30$, total $= 94$ min. Hypothetical $= 3 \times 30 = 90$ min. Difference $= 94 - 90 = 4$ min, again (D). Converting to minutes early avoids the LCD-$30$ fraction work.

CCSS standards used (min grade 6)

5.NF.A.1 Add and subtract fractions with unlike denominators (Adding $\tfrac{2}{5} + \tfrac{2}{3} + \tfrac{1}{2}$ and subtracting $\tfrac{47}{30} - \tfrac{45}{30}$ via the common denominator $30$.)
5.MD.A.1 Convert among different-sized standard measurement units within a given system (Converting the time saved from $\tfrac{1}{15}$ hour into $4$ minutes by multiplying by $60$ min/hr.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Applying $\text{time} = \text{distance} / \text{speed}$ to each day's $2$-mile walk to find the time at each speed.)

⭐ This AMC 8 problem only needs the Grade 6 rate idea time = distance ÷ speed, plus the Grade 5 trick of adding fractions with a common denominator.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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