AMC 8 · 2013 · #11
Grade 6 rate-ratioarithmeticProblem
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.
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.
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
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)
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
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.
💡 Dividing distance by speed is the basic rate move; doing it once per day is the Tool #7 split.
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.
💡 Combining unlike fractions with a common denominator is the Grade 5 fraction-addition standard.
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.
💡 Same $\text{time} = \text{distance}/\text{speed}$ rule, just held constant across all three days.
5.NF.A.1 Step 4 - Subtract to get the time saved.
- Use denominator $30$ again: $\tfrac{3}{2} = \tfrac{45}{30}$.
💡 Subtracting fractions with a common denominator is the same Grade 5 move as adding them.
5.MD.A.1 Step 5 Convert the saved time from hours to minutes so the answer matches the units in the choices.
💡 Multiplying hours by $60$ min/hr cancels "hours" and leaves "minutes" — straight unit conversion.
6.RP.A.3 Find each day's actual time using $\text{time} = \text{distance} / \text{speed}$ 5.NF.A.1 Add the three actual times. The least common denominator of $5$, $3$, and $2$ is 6.RP.A.3 Find the hypothetical total time. At $4$ mph, each day's $2$ miles takes $\tfrac 5.NF.A.1 Subtract to get the time saved. Use denominator $30$ again: $\tfrac{3}{2} = \tfr 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.1Add 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.1Convert 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.3Use 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.
⭐ 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.
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