Competition · AMC preparation · step 4 of 4

AMC 8 · 2014 · #17

Grade 6 rate-ratio
Archetype Meeting Time with Piecewise Rates └ Sub-archetype Piecewise-Leg Time Balance
ratefraction-arithmeticunit-conversion identify-subproblemsdimensional-analysis ↑ Prerequisites: ratefraction-arithmetic
📏 Medium solution 💡 3 insights
Problem
George walks 1 mile to school each day at a steady 3 mph, arriving right when school begins. Today he walked the first 12\frac{1}{2} mile at only 2 mph. At what speed (in mph) must he run the remaining 12\frac{1}{2} mile to still arrive on time?

Pick an answer.

(A)
4
(B)
6
(C)
8
(D)
10
(E)
12

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

This is a rate problem built on time = distance / speed. Tool #7 (Identify Subproblems) is the key move: split the trip into the usual full trip (to get the time budget), the slow first half (to get the time already used), and the fast second half (the unknown). Once each piece is solved, the answer follows by simple subtraction and one more speed = distance / time step. Tool #8 (Analyze the Units) keeps everything in miles and hours so the final number is automatically in mph.

1STEP 1

Find the total time allowed

The usual 1-mile trip at 3 mph sets the whole time budget: 13\frac{1}{3} hour (20 minutes).

T_total = (1 mi)/(3 mph) = 1/3 hr = 20 min
2STEP 2

Find the time already used

The slow first half — 12\frac{1}{2} mile at 2 mph — already eats 14\frac{1}{4} hour (15 minutes).

T_first = (1/2 mi)/(2 mph) = 1/4 hr = 15 min
3STEP 3

Subtract for the time left

Subtract: 13\frac{1}{3} - 14\frac{1}{4} = 412\frac{4}{12} - 312\frac{3}{12}, so only 112\frac{1}{12} hour (5 minutes) remains for the second half.

T_remain = 1/3 - 1/4 = 4/12 - 3/12 = 1/12 hr = 5 min
4STEP 4

Divide to get the speed

Speed = distance / time = 12mile112hour\frac{\frac{1}{2} mile}{\frac{1}{12} hour} = 12\frac{1}{2} × 12 = 6 mph, choice (B).

speed = (1/2 mi)/(1/12 hr) = 1/2 × 12 = 6 mph → (B)
Answer
6
Sanity check by averaging speeds. George spent equal distance (12\frac{1}{2} mile) at 2 mph and at 6 mph, so the average speed is the harmonic mean 2⋅2⋅62+6\frac{2 · 2 · 6}{2 + 6} = 248\frac{24}{8} = 3 mph — exactly his usual pace. That confirms today's total time matches the normal time, so (B) 6 mph is correct. (A) 4 would be too slow — only 124\frac{\frac{1}{2}}{4} = 18\frac{1}{8} hr = 7.5 min, overshooting the 5-min budget.
💡Key takeaway

This AMC 8 problem only needs Grade 6 rate reasoning — distance, time, and speed — plus a single Grade 5 fraction subtraction!

  • Find the total time allowed
  • Find the time already used
  • Subtract for the time left
  • Divide to get the speed

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