AMC 8 · 2023 · #13
Grade 4 geometry-2dProblem
Along the route of a bicycle race, 7 water stations are evenly spaced between the start and finish lines,
as shown in the figure below. There are also 2 repair stations evenly spaced between the start and
finish lines. The 3rd water station is located 2 miles after the 1st repair station. How long is the race
in miles?
Pick an answer.
AMC 8 2023 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: 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
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
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
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.
💡 Putting stations between Start and Finish always creates one more piece than the number of stations.
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.
💡 Multiplying a unit gap by a whole number gives the position of the k-th station.
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.
💡 Plugging the choices in turns the problem into a few easy divisions and a comparison.
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.
💡 On a multiple-choice problem, finding exactly one choice that fits every condition pins down the answer.
3.NF.A.1 Draw a straight line from Start (S) to Finish (F). Place 7 water-station tick ma 4.NF.B.4 From the diagram, the 1st repair station is $1 \times \tfrac{d}{3} = \tfrac{d}{3 4.NBT.B.6 Try the answer choices one by one. For each candidate $d$, compute $R_1 = d \div 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.1Understand 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.4Apply 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.6Find 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.2Solve 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!
⭐ This AMC 8 problem only needs Grade 4 division and 'fraction of a whole' you already know!
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