AMC 10 · 2002 · #12

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)

(B)

(C)

(D)

(E)

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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 travel time in hours. Since $3$ min $=\frac{1}{20}$ hr, the late trip takes $t+\frac{1}{20}$ and the early trip $t-\frac{1}{20}$ .

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

2STEP 2

Set the two distances equal

Both expressions name the same distance, so $40\left(t+\frac{1}{20}\right)=60\left(t-\frac{1}{20}\right)$ , which expands to $40t+2=60t-3$ .

40(t+1/20)=60(t-1/20) → 40t+2=60t-3

3STEP 3

Solve for the on-time time

Move the $t$ terms one way and the numbers the other: $5=20t$ , so $t=\frac{1}{4}$ hour — that is $15$ minutes.

40t+2=60t-3 → 5=20t → t=1/4 hour

4STEP 4

Find the distance, then the on-time speed

Then $d=40\left(\frac{1}{4}+\frac{1}{20}\right)=12$ miles, and dividing by $\frac{1}{4}$ hour gives $48$ mph — choice (B).

d=40(1/4+1/20)=12 mi, speed=12/1/4=48 mph → (B)

Answer

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

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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