AMC 8 · 2015 · #3

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

rateunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

Jack and Jill both leave home at the same instant for a pool that is 1 mile away. Jill bikes at a constant 10 mph; Jack walks at a constant 4 mph. How many minutes earlier does Jill arrive?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Analyze the Units

This is a rate problem with time = $\frac{distance}{speed}$ . The catch is that distance is in miles and speed is in mph, so dividing gives hours — but the answer must be in minutes. Tool #8 (Analyze the Units) keeps the conversion honest: $\frac{mile}{\frac{mile}{hour}}$ = hour, then hour × 60 = minutes. Tool #7 (Identify Subproblems) splits the work into three clean pieces — Jill's time, Jack's time, and the difference — so each piece is a single short calculation.

1STEP 1

Jill's time: divide 1 mile by 10 mph to get $\frac{1}{10}$ hour, then multiply by 60 to reach 6 minutes.

t_Jill =

\frac{1 mi}{10 mph}

\frac{1}{10}

hr =

\frac{1}{10}

× 60 = 6 min

2STEP 2

Jack's time, same method: divide 1 mile by 4 mph to get $\frac{1}{4}$ hour, then multiply by 60 to reach 15 minutes.

t_Jack =

\frac{1 mi}{4 mph}

\frac{1}{4}

hr =

\frac{1}{4}

× 60 = 15 min

3STEP 3

Both start together, so the head start equals the time gap: 15 - 6 = 9 minutes → (D).

t_Jack - t_Jill = 15 - 6 = 9 min → (D)

Answer

Jill is

\frac{10}{4}

= 2.5 times faster than Jack, so her travel time should be 2.5 times shorter:

\frac{15}{2.5}

= 6 min, matching the calculation. The gap of 9 minutes is sensible for a 1-mile trip at very different speeds — Jill finishes in 6 min while Jack still has more than half the mile to go.

💡Key takeaway

This AMC 8 problem only needs the Grade 6 rate rule "time = distance ÷ speed" plus a minute-to-hour conversion you learned in Grade 5.

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