AMC 10 · 2002 · #11

Grade 8 rate-ratio
Archetype Meeting Time with Piecewise Rates └ Sub-archetype Piecewise-Leg Time Balance
rateunit-conversionlinear-equations-one-var convert-to-algebra ↑ Prerequisites: ratelinear-equations-one-var
📏 Medium solution 💡 2 insights
Problem
Mr. Bird drives the same distance to work each day. At 40 mph he arrives 3 minutes late; at 60 mph he arrives 3 minutes early. Find the constant speed that gets him there exactly on time.

Pick an answer.

(A)
45
(B)
48
(C)
50
(D)
55
(E)
58
How to solve
Strategy Introduce a Variable

The trap is trying to average 40 and 60 to get 50 — but speed and time do not average that simply. The one thing that stays fixed across both trips is the distance. Tool #4 (Introduce a Variable) names the on-time travel time t (in hours); then the late trip takes t+1/20 hour and the early trip takes t-1/20 hour. Tool #8 (Analyze the Units) forces the 3 minutes into hours so it fits with mph. Because both trips share one distance, writing that distance two ways and setting them equal (tool #13, Convert to Algebra) gives one equation in t. Solve for t, recover the distance, then divide distance by the on-time time to get the required speed.

1STEP 1

Name the on-time time, write two distances

Let t be the on-time hours; 3 minutes is 1/20 hour, so the trips are 40(t+1/20) and 60(t-1/20).

3 min=1/20 hr, d=40(t+1/20), d=60(t-1/20)
2STEP 2

Set the two distances equal

Both equal the same distance, so 40(t+1/20) = 60(t-1/20), giving 40t+2 = 60t-3.

40(t+1/20)=60(t-1/20) → 40t+2=60t-3
3STEP 3

Solve for the on-time time

Collecting terms gives 5 = 20t, so t = 1/4 hour, or 15 minutes.

40t+2=60t-3 → 5=20t → t=1/4 hour
4STEP 4

Find the distance, then the on-time speed

The distance is 40(1/4+1/20) = 12 miles, so the on-time speed is 12 ÷ 1/4 = 48 mph, choice (B).

d=40(1/4+1/20)=12 mi, speed=12/1/4=48 mph → (B)
Answer
48
The answer 48 sits between 40 and 60, which it must — the on-time speed has to be faster than the too-slow trip and slower than the too-fast one. It is a bit below the midpoint 50, which makes sense: the same time cushion buys less speed at the high end, so the on-time speed leans toward the slower number. Choice (C) 50 is the trap for anyone who just averages 40 and 60; that average ignores that equal time offsets do not split the speed evenly. Plugging back, 12 miles at 48 mph takes 12/48=1/4 hour =15 minutes — exactly the on-time time, so it checks out.
💡Key takeaway

When the same distance is driven at different speeds, pin down the thing that stays fixed — the distance — and let it tie the two trips into one equation.

  • Name the on-time time, write two distances
  • Set the two distances equal
  • Solve for the on-time time
  • Find the distance, then the on-time speed

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