AMC 8 · 2016 · #4

Grade 6 rate-ratio
rateunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticunit-conversion
📏 Short solution 💡 3 insights
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Problem
As a boy, Cheenu ran 15 miles in 3 hours 30 minutes. As an old man, he walks 10 miles in 4 hours. How many more minutes does it now take him to cover one mile than it did when he was a boy?

Pick an answer.

(A)
6
(B)
10
(C)
15
(D)
18
(E)
30

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Analyze the Units

The question asks for a difference in minutes per mile, but the data are mixed in hours and minutes and use totals (not per-mile rates). Tool #8 (Analyze the Units) tells us to first convert each total time to minutes, then divide by miles so the unit "minutes per mile" pops out — only then can we subtract. Tool #7 (Identify Subproblems) splits the work into three clean pieces: (1) boy's pace, (2) old man's pace, (3) the difference.

1STEP 1

Convert the boy's 3 h 30 min to 210 minutes, then divide by 15 miles to get 14 minutes per mile.

3 h × 60 min/h + 30 min = 210 min; (210 min)/(15 mi) = 14 min/mi
2STEP 2

Do the same for the old man: 4 hours is 240 minutes, divided by 10 miles gives 24 minutes per mile.

4 h × 60 min/h = 240 min; (240 min)/(10 mi) = 24 min/mi
3STEP 3

Both paces share the unit minutes per mile, so subtract: 24 - 14 = 10 minutes per mile more.

24 min/mi - 14 min/mi = 10 min/mi → (B)
Answer
10
Boy's pace = 14 min/mi corresponds to a speed of 6014\frac{60}{14} ≈ 4.3 mph — a slow run, which fits "run." Old man's pace = 24 min/mi corresponds to 6024\frac{60}{24} = 2.5 mph — a slow walk, which fits "walk." The 10-minute gap is the right order of magnitude (single digits to mid-teens), matching choice (B).
💡Key takeaway

This AMC 8 problem only needs the Grade 6 idea that "unit rate = total ÷ count" — divide minutes by miles, then subtract.