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

\frac{1}{2}

mile at only 2 mph. At what speed (in mph) must he run the remaining

\frac{1}{2}

mile to still arrive on time?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

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

T_total =

\frac{1 mi}{3 mph}

\frac{1}{3}

hr = 20 min

2STEP 2

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

T_first =

\frac{\frac{1}{2} mi}{2 mph}

\frac{1}{4}

hr = 15 min

3STEP 3

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

T_remain =

\frac{1}{3}

\frac{1}{4}

\frac{4}{12}

\frac{3}{12}

\frac{1}{12}

hr = 5 min

4STEP 4

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

speed =

\frac{\frac{1}{2} mi}{\frac{1}{12} hr}

\frac{1}{2}

× 12 = 6 mph → (B)

Answer

Sanity check by averaging speeds. George spent equal distance (

\frac{1}{2}

mile) at 2 mph and at 6 mph, so the average speed is the harmonic mean

\frac{2 · 2 · 6}{2 + 6}

\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

\frac{\frac{1}{2}}{4}

\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!

More like this