AMC 8 · 2005 · #10

Grade 4 rate-ratio

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

rateratio-proportionfraction-arithmetic identify-subproblemsratio-proportion ↑ Prerequisites: ratefraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

Joe walks the first half of the way to school, then runs the second half. He runs 3 times as fast as he walks, and walking the first half took 6 minutes. How many minutes is the whole trip?

Pick an answer.

(A)

(B)

7.3

(C)

7.7

(D)

(E)

8.3

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Identify Subproblems) splits the trip into two clean pieces: (a) the walking half, whose time is given, and (b) the running half, whose time we need. Each piece is one small calculation, and the total is just their sum. For the running piece, Tool #5 (Look for a Pattern) catches the inverse-proportion pattern: the running half covers the same distance as the walking half, but 3 times as fast — so it takes $\frac{1}{3}$ as much time. Picking Tool #7 plus Tool #5 over Tool #13 (Convert to Algebra) keeps the whole solution at Grade 4-5 arithmetic with no equation-solving.

1STEP 1

The walking half's time is handed to us: Joe walks the first half in 6 minutes.

t_walk = 6 minutes

2STEP 2

Same distance, 3 times the speed, so the running half takes one-third the time: 2 minutes.

t_run =

\frac{1}{3}

× t_walk =

\frac{1}{3}

× 6 = 2 minutes

3STEP 3

Add the two halves: 6 + 2 = 8 minutes, choice (D).

t_total = t_walk + t_run = 6 + 2 = 8 minutes → (D)

Answer

Sanity check with concrete numbers. Suppose the half-distance is 6 blocks. Walking 6 blocks in 6 minutes is 1 block per minute. Running 3 times as fast is 3 blocks per minute, so the same 6 blocks take 6 ÷ 3 = 2 minutes. Total: 6 + 2 = 8 minutes. Magnitude check: the answer must be more than 6 (Joe still has half the trip to go after walking) and less than 12 (he would only need 12 if he walked the whole way), so 8 sits in the right range. The trap answers 7.3, 7.7, and 8.3 look like "6 +

\frac{6}{3}

" miscomputed with decimals, and 7 would be the answer if Joe ran the second half 6 times as fast.

💡Key takeaway

Split the trip into the walking half and the running half. The running half is the same distance covered 3 times as fast, so it takes $\frac{1}{3}$ the time — that one inverse-proportion idea turns this AMC 8 problem into a Grade 4 multiplicative-comparison exercise.

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