AMC 10 · 2014 · #15
Grade 7 rate-ratioDavid drives from his home to the airport to catch a flight. He drives 35 miles in the first hour, but realizes that he will be 1 hour late if he continues at this speed. He increases his speed by 15 miles per hour for the rest of the way to the airport and arrives 30 minutes early. How many miles is the airport from his home?
Pick an answer.
AMC 10 2014 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: David drives $35$ miles in his first hour. At that speed he would arrive $1$ hour late, so he speeds up by $15$ mph for the rest of the trip and instead arrives $30$ minutes early. Find the total distance from his home to the airport.
Givens: In the first hour he drives $35$ miles, so his starting speed is $35$ mph; Staying at $35$ mph would make him arrive $1$ hour late; He raises his speed by $15$ mph, to $50$ mph, for the rest of the way; At $50$ mph for the rest he arrives $30$ minutes early; Answer choices: (A) $140$, (B) $175$, (C) $210$, (D) $245$, (E) $280$
Unknowns: The total distance in miles from David's home to the airport
Understand
Restated: David drives $35$ miles in his first hour. At that speed he would arrive $1$ hour late, so he speeds up by $15$ mph for the rest of the trip and instead arrives $30$ minutes early. Find the total distance from his home to the airport.
Givens: In the first hour he drives $35$ miles, so his starting speed is $35$ mph; Staying at $35$ mph would make him arrive $1$ hour late; He raises his speed by $15$ mph, to $50$ mph, for the rest of the way; At $50$ mph for the rest he arrives $30$ minutes early; Answer choices: (A) $140$, (B) $175$, (C) $210$, (D) $245$, (E) $280$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #13 Convert to Algebra
The whole problem turns on the distance still left after the first hour, so Tool #4 (Introduce a Variable) names that leftover distance $d$ and lets me write its travel time at each speed. Tool #8 (Analyze the Units) keeps the rates honest: miles divided by miles-per-hour gives hours, so $d/35$ and $d/50$ are times I can compare. Tool #13 (Convert to Algebra) turns the phrase "from $1$ hour late to $30$ minutes early" into a single equation about saved time.
Execute — Answer: C
6.RP.A.3 Step 1 Read off the two speeds
- In the first hour David covers $35$ miles, so his first speed is $35$ mph.
- He then increases it by $15$ mph, giving $35+15=50$ mph for the rest of the way.
- Those $35$ miles from the first hour are already behind him; only the leftover distance is driven at $50$ mph.
💡 Driving $35$ miles in exactly one hour is what "$35$ mph" means, so the first hour hands us both the speed and the miles.
6.EE.B.6 Step 2 Name the leftover distance
- Let $d$ be the distance still left after the first hour.
- Covering $d$ at the old speed of $35$ mph would take $d/35$ hours, while covering the same $d$ at the new speed of $50$ mph takes $d/50$ hours.
- Both expressions describe the same stretch of road, just at different speeds.
💡 One letter for the unknown leftover distance lets both travel times be written from the single rule time = distance / speed.
7.EE.B.4 Step 3 Turn the timing into an equation
- At $35$ mph he is $1$ hour late; at $50$ mph he is $30$ minutes (half an hour) early.
- So switching speeds saves him $1+\tfrac12=1.5$ hours on the leftover distance.
- The time saved is the slow time minus the fast time, which gives the equation $\dfrac{d}{35}-\dfrac{d}{50}=1.5$.
💡 Going from late to early is nothing more than saved minutes, so the gap between the two travel times equals that saving.
6.EE.B.7 Step 4 Solve for the leftover distance
- Put the fractions over a common denominator of $350$: $\dfrac{d}{35}=\dfrac{10d}{350}$ and $\dfrac{d}{50}=\dfrac{7d}{350}$, so their difference is $\dfrac{3d}{350}$.
- Setting $\dfrac{3d}{350}=1.5=\dfrac{3}{2}$ gives $d=350\cdot\dfrac{1}{2}=175$ miles left after the first hour.
💡 Once both times share a denominator, the equation collapses to a single fraction equal to a number, and one step frees $d$.
6.RP.A.3 Step 5 Add back the first hour
- The $175$ miles is only the part after the first hour.
- Add the $35$ miles already driven in that first hour to get the full distance: $35+175=210$ miles.
- That matches choice (C).
- Choice (B) $175$ is the trap that forgets the first-hour miles.
💡 The variable only tracked the road after the first hour, so the first-hour miles still have to be added on.
6.RP.A.3 In the first hour David covers $35$ miles, so his first speed is $35$ mph. He th 6.EE.B.6 Let $d$ be the distance still left after the first hour. Covering $d$ at the old 7.EE.B.4 At $35$ mph he is $1$ hour late; at $50$ mph he is $30$ minutes (half an hour) e 6.EE.B.7 Put the fractions over a common denominator of $350$: $\dfrac{d}{35}=\dfrac{10d} 6.RP.A.3 The $175$ miles is only the part after the first hour. Add the $35$ miles alread Review
Reasonableness: Check $210$ directly. After the first $35$ miles, $175$ miles remain. At $50$ mph that takes $175/50=3.5$ hours, so the whole trip is $1+3.5=4.5$ hours. At the old $35$ mph the leftover would take $175/35=5$ hours, a total of $6$ hours; being $1$ hour late means the flight time is $5$ hours after leaving. Arriving in $4.5$ hours is exactly $0.5$ hour before $5$ — that is $30$ minutes early, just as stated. Everything lines up, so (C) is correct.
Alternative: Test the answer choices with Tool #6 (Guess and Check). For $210$: leftover $=175$, slow time $=175/35=5$ h, fast time $=175/50=3.5$ h, difference $=1.5$ h, which is exactly the late-to-early swing of $1.5$ hours. Smaller choices like $140$ give a leftover of $105$ and a difference of only $0.9$ h, too small, so $210$ is the one that fits.
CCSS standards used (min grade 7)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Reading $35$ mph and $50$ mph as rates and using distance = rate x time to relate distance and travel time.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting $d$ be the leftover distance and writing the two travel times $d/35$ and $d/50$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Turning "$1$ hour late to $30$ minutes early" into the equation $d/35 - d/50 = 1.5$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $3d/350 = 3/2$ to find $d = 175$.)
⭐ Going from late to early just means saved time, so compare how long the leftover road takes at each speed, solve for that leftover distance, then add back the first-hour miles.
⭐ Going from late to early just means saved time, so compare how long the leftover road takes at each speed, solve for that leftover distance, then add back the first-hour miles.
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