AMC 8 · 2018 · #6

Easy mode Grade 4

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Problem

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?

$\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

100

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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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