AMC 10 · 2002 · #12
Grade 8 rate-ratioMr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The distance from home to work is the same in every case; At $40$ mph he is $3$ minutes late; At $60$ mph he is $3$ minutes early; Answer choices: (A) $45$, (B) $48$, (C) $50$, (D) $55$, (E) $58$ (miles per hour)
Unknowns: The speed in miles per hour that makes the arrival time exactly on time
Understand
Restated: 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.
Givens: The distance from home to work is the same in every case; At $40$ mph he is $3$ minutes late; At $60$ mph he is $3$ minutes early; Answer choices: (A) $45$, (B) $48$, (C) $50$, (D) $55$, (E) $58$ (miles per hour)
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #13 Convert to Algebra
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 $t+\tfrac{1}{20}$ hour and the early trip takes $t-\tfrac{1}{20}$ 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.
Execute — Answer: B
6.RP.A.3 Step 1 Name the on-time time, write two distances
- Let $t$ be the number of hours the trip should take to arrive exactly on time.
- First put the $3$ minutes in the same units as mph: $3$ minutes $=\tfrac{3}{60}=\tfrac{1}{20}$ hour.
- Driving $40$ mph makes him $3$ minutes late, so that slow trip takes $t+\tfrac{1}{20}$ hour and covers $40\left(t+\tfrac{1}{20}\right)$ miles.
- Driving $60$ mph makes him $3$ minutes early, so that fast trip takes $t-\tfrac{1}{20}$ hour and covers $60\left(t-\tfrac{1}{20}\right)$ miles.
💡 The road to work never gets longer or shorter — only the clock changes — so both trips must cover one and the same distance.
8.EE.C.7 Step 2 Set the two distances equal
- Since both expressions equal the same distance $d$, they equal each other: $40\left(t+\tfrac{1}{20}\right)=60\left(t-\tfrac{1}{20}\right)$.
- Multiply out each side.
- On the left, $40t+40\cdot\tfrac{1}{20}=40t+2$.
- On the right, $60t-60\cdot\tfrac{1}{20}=60t-3$.
- So $40t+2=60t-3$.
💡 Two names for the same distance can be set side by side, turning a word problem into one clean equation.
8.EE.C.7 Step 3 Solve for the on-time time
- Collect the $t$ terms on one side and the numbers on the other.
- From $40t+2=60t-3$, subtract $40t$ from both sides to get $2=20t-3$, then add $3$: $5=20t$.
- Divide by $20$: $t=\tfrac{5}{20}=\tfrac{1}{4}$ hour, which is $15$ minutes.
💡 Sliding the variable to one side and the plain numbers to the other unwraps the unknown time in one move.
6.RP.A.3 Step 4 Find the distance, then the on-time speed
- Put $t=\tfrac{1}{4}$ back into either distance formula: $d=40\left(\tfrac{1}{4}+\tfrac{1}{20}\right)=40\cdot\tfrac{6}{20}=12$ miles.
- (Check with the other: $60\left(\tfrac{1}{4}-\tfrac{1}{20}\right)=60\cdot\tfrac{4}{20}=12$ miles — same.) To arrive exactly on time he must cover $12$ miles in $\tfrac{1}{4}$ hour, so his speed is distance $\div$ time $=12\div\tfrac{1}{4}=48$ mph.
- That is choice (B).
💡 Once you know how far and how long the on-time trip is, the speed is just the miles spread evenly over the hours.
6.RP.A.3 Let $t$ be the number of hours the trip should take to arrive exactly on time. F 8.EE.C.7 Since both expressions equal the same distance $d$, they equal each other: $40\l 8.EE.C.7 Collect the $t$ terms on one side and the numbers on the other. From $40t+2=60t- 6.RP.A.3 Put $t=\tfrac{1}{4}$ back into either distance formula: $d=40\left(\tfrac{1}{4}+ Review
Reasonableness: 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 $\tfrac{12}{48}=\tfrac14$ hour $=15$ minutes — exactly the on-time time, so it checks out.
Alternative: Because he is off by the same $3$ minutes in each direction, the on-time speed is the harmonic mean of the two speeds: $\dfrac{2\cdot 40\cdot 60}{40+60}=\dfrac{4800}{100}=48$ mph, again choice (B). This shortcut works only when the early and late offsets are equal, but it lands the answer without ever finding the distance or time.
CCSS standards used (min grade 8)
6.RP.A.3Solve real-world problems using ratio and rate reasoning (Using distance $=$ speed $\times$ time, converting $3$ minutes to $\tfrac{1}{20}$ hour, and dividing $12$ miles by $\tfrac14$ hour to get the on-time speed.)8.EE.C.7Solve linear equations in one variable (Setting the two distance expressions equal and solving $40t+2=60t-3$ (variable on both sides) for $t=\tfrac14$ hour.)
⭐ 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.
⭐ 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.
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