Competition · AMC preparation · step 4 of 4

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)
1
(B)
2
(C)
3
(D)
4
(E)
5

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 riding time

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

t_Mon = 2/5 hr, t_Wed = 2/3 hr, t_Fri = 2/4 = 1/2 hr
2STEP 2

Add the three times

Add the three times over the common denominator 30 to get 4730\frac{47}{30} hr.

2/5 + 2/3 + 1/2 = 12/30 + 20/30 + 15/30 = 47/30 hr
3STEP 3

Find the ideal total time

In the all-4 mph case each day's 2 miles takes 12\frac{1}{2} hr, so three days total 32\frac{3}{2} hr.

3 × 1/2 = 3/2 hr
4STEP 4

Subtract to find the savings

Subtract the totals over denominator 30: 4730\frac{47}{30} - 4530\frac{45}{30} = 115\frac{1}{15} hr.

47/30 - 45/30 = 2/30 = 1/15 hr
5STEP 5

Convert hours to minutes

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

1/15 hr × 60 min/hr = 60/15 = 4 min → (D)
Answer
4
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.

  • Find each day's riding time
  • Add the three times
  • Find the ideal total time
  • Subtract to find the savings
  • Convert hours to minutes

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