AMC 8 · 2018 · #6
Easy mode Grade 4Problem
Anh is driving to the beach. The trip has two parts.
Part one is on the highway. He drives 50 miles on the highway.
Part two is on a small coastal road. He drives 10 miles on the coastal road, and it takes him 30 minutes.
On the highway, Anh drives 3 times as fast as he does on the coastal road.
How many minutes did the entire trip take?
Pick an answer.
AMC 8 2018 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: Anh drives $50$ miles on the highway and $10$ miles on a coastal road. His highway speed is exactly $3$ times his coastal speed. The $10$-mile coastal stretch took $30$ minutes. How many minutes did the entire $60$-mile trip take?
Givens: Highway distance $= 50$ miles; Coastal distance $= 10$ miles; Highway speed $= 3 \times$ coastal speed; Coastal driving time $= 30$ minutes; Answer choices: (A) $50$, (B) $70$, (C) $80$, (D) $90$, (E) $100$ minutes
Unknowns: The total trip time in minutes (highway time plus coastal time)
Understand
Restated: Anh drives $50$ miles on the highway and $10$ miles on a coastal road. His highway speed is exactly $3$ times his coastal speed. The $10$-mile coastal stretch took $30$ minutes. How many minutes did the entire $60$-mile trip take?
Givens: Highway distance $= 50$ miles; Coastal distance $= 10$ miles; Highway speed $= 3 \times$ coastal speed; Coastal driving time $= 30$ minutes; Answer choices: (A) $50$, (B) $70$, (C) $80$, (D) $90$, (E) $100$ minutes
Plan
Primary tool: #8 Analyze the Units
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
This is a classic rate problem (distance, speed, time). Tool #8 (Analyze the Units) keeps the bookkeeping honest: $\tfrac{\text{miles}}{\text{minutes-per-mile}} = \text{minutes}$, so working with the rate "minutes per mile" lets us avoid converting to mph at all. Tool #9 (Easier Related Problem) is the elegance move: instead of computing the highway speed, first ask the smaller question "how long would $10$ highway miles take?" — that gives a per-mile rate we can scale up by $5$ to reach the actual $50$ miles. Tool #7 (Identify Subproblems) splits the trip into the highway leg and the coastal leg, which are summed at the very end. This avoids reaching for Tool #13 (Algebra) when scaling and addition are enough.
Execute — Answer: C
4.OA.A.1 Step 1 - Translate "$3$ times as fast" into per-mile time.
- If the highway speed is $3$ times the coastal speed, then each highway mile takes $\tfrac{1}{3}$ as much time as each coastal mile.
- (Faster speed means less time per mile.)
💡 "$3$ times as fast" is a Grade 4 multiplicative comparison — and the speed/time inverse just flips the multiplier.
4.NF.B.4 Step 2 - Find the time for $10$ miles on the highway by the easier-problem move.
- The coastal road took $30$ minutes for $10$ miles, so the same $10$ miles on the highway take $\tfrac{1}{3}$ of that time.
💡 Multiplying a whole number by the fraction $\tfrac{1}{3}$ is exactly the Grade 4 "fraction times a whole number" standard.
4.MD.A.2 Step 3 - Scale up from $10$ highway miles to the real $50$ highway miles.
- Since $50$ miles is $5$ times $10$ miles at the same speed, the time is $5$ times as long.
💡 Scaling time by the same factor as distance (at constant speed) is the Grade 4 distance-time word-problem move.
4.MD.A.2 Step 4 - Add the two legs to get the total trip time.
- The highway took $50$ minutes and the coastal road took $30$ minutes.
💡 Combining two sub-trip times into one total trip time is the addition piece of a multi-step distance-time word problem.
4.OA.A.1 Translate "$3$ times as fast" into per-mile time. If the highway speed is $3$ ti 4.NF.B.4 Find the time for $10$ miles on the highway by the easier-problem move. The coas 4.MD.A.2 Scale up from $10$ highway miles to the real $50$ highway miles. Since $50$ mile 4.MD.A.2 Add the two legs to get the total trip time. The highway took $50$ minutes and t Review
Reasonableness: Sanity-check the speeds. Coastal: $10$ miles in $30$ min $= 20$ mph (a slow coastal road). Highway: $3 \times 20 = 60$ mph (a normal highway speed). Both numbers are realistic. The highway leg, $50$ miles at $60$ mph, takes $\tfrac{50}{60} = \tfrac{5}{6}$ hour $= 50$ minutes, matching our scaled answer. Total $= 50 + 30 = 80$ minutes, which is choice (C) and lies in the middle of the answer choices — not suspiciously extreme.
Alternative: Tool #13 (Convert to Algebra) with Tool #8 (Units) gives the textbook version. Coastal speed $v = \tfrac{10 \text{ mi}}{0.5 \text{ hr}} = 20$ mph, so highway speed $= 3v = 60$ mph. Highway time $= \tfrac{50}{60}$ hr $= 50$ min. Total $= 80$ min. The proportional path above avoids the mph detour entirely, which is why it's preferable for younger students.
CCSS standards used (min grade 4)
4.OA.A.1Interpret a multiplication equation as a comparison (Translating "highway speed is $3$ times the coastal speed" into the inverse statement that each highway mile takes $\tfrac{1}{3}$ the time of a coastal mile.)4.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a whole number (Computing $\tfrac{1}{3} \times 30 = 10$ minutes for $10$ miles on the highway.)4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Scaling the $10$-highway-mile time up by $5$ to get the $50$-mile time, and adding the highway and coastal times for the total trip.)
⭐ This AMC 8 problem only needs Grade 4 multiplicative comparison ("$3$ times as fast means $\tfrac{1}{3}$ the time") that you already know!
⭐ This AMC 8 problem only needs Grade 4 multiplicative comparison ("$3$ times as fast means $\tfrac{1}{3}$ the time") that you already know!
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