AMC 8 · 2013 · #11

Grade 6 rate-ratioarithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

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

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Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The question is a straight time = distance / 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 ÷ mph gives hours, so the final answer needs an hours-to-minutes conversion.

1STEP 1

Find each day's actual time with time = distance / speed, using the same 2 miles every day.

t_Mon =

\frac{2}{5}

hr, t_Wed =

\frac{2}{3}

hr, t_Fri =

\frac{2}{4}

\frac{1}{2}

2STEP 2

Add the three times over the common denominator 30 to get $\frac{47}{30}$ hr.

\frac{2}{5}

\frac{2}{3}

\frac{1}{2}

\frac{12}{30}

\frac{20}{30}

\frac{15}{30}

\frac{47}{30}

3STEP 3

In the all-4 mph case each day's 2 miles takes $\frac{1}{2}$ hr, so three days total $\frac{3}{2}$ hr.

3 ×

\frac{1}{2}

\frac{3}{2}

4STEP 4

Subtract the totals over denominator 30: $\frac{47}{30}$ - $\frac{45}{30}$ = $\frac{1}{15}$ hr.

\frac{47}{30}

\frac{45}{30}

\frac{2}{30}

\frac{1}{15}

5STEP 5

Convert the saved hours to minutes by multiplying by 60 so it matches the choice units.

\frac{1}{15}

hr × 60 min/hr =

\frac{60}{15}

= 4 min → (D)

Answer

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).

💡Key takeaway

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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