AMC 8 · 2023 · #13

Easy mode Grade 4

Problem

Picture a long straight bicycle race course. There's a start line at one end and a finish line at the other.

Between the start and finish, $7$ water stations are placed in a row, all with equal gaps between them. The first water station is the same distance from the start as the last one is from the finish.

In the same race course, $2$ repair stations are also placed between the start and finish, with equal gaps between them. (The first repair station is the same distance from the start as the second is from the finish.)

We're told one extra fact: the $3$ rd water station is exactly $2$ miles past the $1$ st repair station.

How long is the whole race, in miles?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: A race route runs straight from Start to Finish. Seven water stations and two repair stations are each evenly spaced between Start and Finish. The 3rd water station is 2 miles past the 1st repair station. Find the total race length $d$ in miles.

Givens: 7 water stations are evenly spaced between Start and Finish, splitting the route into 8 equal pieces; 2 repair stations are evenly spaced between Start and Finish, splitting the route into 3 equal pieces; Distance from 1st repair station to 3rd water station = 2 miles (water is the farther one from Start); Answer choices: 8, 16, 24, 48, 96

Unknowns: The total length $d$ of the race, in miles

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

We have only five answer choices, and for each candidate length $d$ it is easy to find where the 1st repair station ($d \div 3$ from Start) and the 3rd water station ($3 \times d \div 8$ from Start) sit. So we can simply try each choice and keep the one where the gap is exactly 2 miles. A quick diagram first makes the 'divides into 8 (or 3) equal pieces' idea concrete, and the multiple-choice format lets us eliminate the rest instead of solving an algebra equation.

Execute — Answer: D

#1 Draw a Diagram 3.NF.A.1 Step 1

Draw a straight line from Start (S) to Finish (F).
Place 7 water-station tick marks that cut the line into 8 equal pieces, and on a second copy place 2 repair-station tick marks that cut the line into 3 equal pieces.
This makes it clear that each water gap is $\tfrac{1}{8}$ of the route and each repair gap is $\tfrac{1}{3}$ of the route.

$$\text{water gap} = \tfrac{d}{8},\quad \text{repair gap} = \tfrac{d}{3}$$

💡 Putting stations between Start and Finish always creates one more piece than the number of stations.

#1 Draw a Diagram 4.NF.B.4 Step 2

From the diagram, the 1st repair station is $1 \times \tfrac{d}{3} = \tfrac{d}{3}$ miles from Start, and the 3rd water station is $3 \times \tfrac{d}{8} = \tfrac{3d}{8}$ miles from Start.
Both are just 'so many gaps' from Start.

$$R_1 = \tfrac{d}{3},\qquad W_3 = \tfrac{3d}{8}$$

💡 Multiplying a unit gap by a whole number gives the position of the k-th station.

#6 Guess and Check 4.NBT.B.6 Step 3

Try the answer choices one by one.
For each candidate $d$, compute $R_1 = d \div 3$ and $W_3 = 3 \times (d \div 8)$, then see if $W_3 - R_1 = 2$.
Choices 8, 16, and 24 are too small to give a gap that big.

$$\begin{aligned} d=24:&\ R_1 = 8,\ W_3 = 9,\ W_3 - R_1 = 1 \\ d=48:&\ R_1 = 16,\ W_3 = 18,\ W_3 - R_1 = 2 \ \checkmark \\ d=96:&\ R_1 = 32,\ W_3 = 36,\ W_3 - R_1 = 4 \end{aligned}$$

💡 Plugging the choices in turns the problem into a few easy divisions and a comparison.

#3 Eliminate Possibilities 4.MD.A.2 Step 4

Only $d = 48$ produces $W_3 - R_1 = 2$ miles, matching the problem's condition.
The other four choices are eliminated.

$$d = 48 = \textbf{(D)}$$

💡 On a multiple-choice problem, finding exactly one choice that fits every condition pins down the answer.

[1] #1 3.NF.A.1 Draw a straight line from Start (S) to Finish (F). Place 7 water-station tick ma

[2] #1 4.NF.B.4 From the diagram, the 1st repair station is $1 \times \tfrac{d}{3} = \tfrac{d}{3

[3] #6 4.NBT.B.6 Try the answer choices one by one. For each candidate $d$, compute $R_1 = d \div

[4] #3 4.MD.A.2 Only $d = 48$ produces $W_3 - R_1 = 2$ miles, matching the problem's condition.

Review

Reasonableness: Check the winner against the original setup. With $d = 48$, water stations sit at $6, 12, 18, 24, 30, 36, 42$ miles (each $\tfrac{1}{8}$ of 48), and repair stations sit at $16$ and $32$ miles (each $\tfrac{1}{3}$ of 48). The 3rd water station is at 18 and the 1st repair station is at 16, so the water station is exactly 2 miles past the repair station — matches the problem. The race length 48 miles is also a believable distance for a bike race.

Alternative: Tool 13 (Convert to Algebra) gives a one-shot solution: setting $\tfrac{3d}{8} = \tfrac{d}{3} + 2$ leads to $\tfrac{d}{24} = 2$, so $d = 48$. That route is faster for someone comfortable with unlike-denominator fractions, but Guess-and-Check with the answer choices keeps the arithmetic at grade 4 and is just as reliable.

CCSS standards used (min grade 4)

3.NF.A.1 Understand a fraction as quantity formed by parts of a whole (Interpreted the route as being cut into 8 equal water-gaps and 3 equal repair-gaps, so each gap is a unit fraction ($\tfrac{1}{8}$ or $\tfrac{1}{3}$) of the total length.)
4.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a whole number (Expressed the position of the k-th station as $k$ times a unit gap, giving $W_3 = \tfrac{3d}{8}$ and $R_1 = \tfrac{d}{3}$.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Computed $d \div 8$ and $d \div 3$ for each candidate $d \in \{8, 16, 24, 48, 96\}$ during the Guess-and-Check pass.)
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money (Compared the two computed distances along the race route and identified the candidate where the gap equals 2 miles.)

⭐ This AMC 8 problem only needs Grade 4 division and 'fraction of a whole' you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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