AMC 8 · 2013 · #11

Easy mode Grade 6

Problem

Ted's grandfather used his treadmill on $3$ days this week. Each day he traveled exactly $2$ miles.

Here are the speeds he actually used:

Monday: he jogged at $5$ miles per hour
Wednesday: he walked at $3$ miles per hour
Friday: he walked at $4$ miles per hour

Now imagine instead that he had walked at $4$ miles per hour on all three days. That would have changed the total time he spent on the treadmill.

How many fewer minutes would he have spent on the treadmill in total?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2013 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.

View mode:

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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