AMC 8 · 2018 · #6

Grade 4 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

rateratio-proportionunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

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Problem

On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

100

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.

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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

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

#8 Analyze the Units 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.)

$$\text{time per highway mile} = \tfrac{1}{3} \times \text{time per coastal mile}$$

💡 "$3$ times as fast" is a Grade 4 multiplicative comparison — and the speed/time inverse just flips the multiplier.

#9 Solve an Easier Related Problem 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.

$$\tfrac{1}{3} \times 30 \text{ min} = 10 \text{ min for } 10 \text{ highway miles}$$

💡 Multiplying a whole number by the fraction $\tfrac{1}{3}$ is exactly the Grade 4 "fraction times a whole number" standard.

#8 Analyze the Units 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.

$$5 \times 10 \text{ min} = 50 \text{ min on the highway}$$

💡 Scaling time by the same factor as distance (at constant speed) is the Grade 4 distance-time word-problem move.

#7 Identify Subproblems 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.

$$50 + 30 = 80 \text{ min} \;\Rightarrow\; \textbf{(C)}$$

💡 Combining two sub-trip times into one total trip time is the addition piece of a multi-step distance-time word problem.

[1] #8 4.OA.A.1 Translate "$3$ times as fast" into per-mile time. If the highway speed is $3$ ti

[2] #9 4.NF.B.4 Find the time for $10$ miles on the highway by the easier-problem move. The coas

[3] #8 4.MD.A.2 Scale up from $10$ highway miles to the real $50$ highway miles. Since $50$ mile

[4] #7 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.1 Interpret 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.4 Apply 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.2 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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