AMC 8 · 2025 · #5
Easy mode Grade 3Problem
Imagine a neighborhood laid out like a big grid of city blocks. Betty drives a delivery truck. She has to start at the factory, marked , drop off packages at three other spots, and then come back to .
The three delivery spots are labeled , , and . Betty must visit them in this order: first , then , then . After visiting , she drives back to . The map below shows where each spot is.
Betty can only drive along the streets of the grid. She wants to make her whole trip as short as possible. We measure her trip in blocks — each side of a small grid square counts as one block.
What is the smallest number of blocks she can drive to finish the whole route?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Betty's delivery truck must start at the factory $F$, visit locations $A$, $B$, and $C$ in that order, and return to $F$. The neighborhood is laid out on a grid of city blocks, so the truck can only drive along the horizontal and vertical streets — it cannot cut diagonally. We want the smallest possible total number of blocks driven.
Givens: Street map is a rectangular grid (streets only run horizontally or vertically).; Labeled points on the grid: $F = (6, 5)$, $A = (7, 3)$, $B = (0, 0)$, $C = (2, 4)$.; Route order is fixed: $F \to A \to B \to C \to F$.; Each grid edge is $1$ block.; Answer choices: (A) $20$, (B) $22$, (C) $24$, (D) $26$, (E) $28$.
Unknowns: The shortest total number of blocks Betty drives on the full $F \to A \to B \to C \to F$ loop.
Understand
Restated: Betty's delivery truck must start at the factory $F$, visit locations $A$, $B$, and $C$ in that order, and return to $F$. The neighborhood is laid out on a grid of city blocks, so the truck can only drive along the horizontal and vertical streets — it cannot cut diagonally. We want the smallest possible total number of blocks driven.
Givens: Street map is a rectangular grid (streets only run horizontally or vertically).; Labeled points on the grid: $F = (6, 5)$, $A = (7, 3)$, $B = (0, 0)$, $C = (2, 4)$.; Route order is fixed: $F \to A \to B \to C \to F$.; Each grid edge is $1$ block.; Answer choices: (A) $20$, (B) $22$, (C) $24$, (D) $26$, (E) $28$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The map is already a diagram, so Tool #1 (Draw a Diagram) just asks us to *use* it — read the position of each point by counting blocks right and up from the corner. Because the streets form a grid, the shortest path between any two labeled points is simply (horizontal blocks between them) + (vertical blocks between them); there is no benefit to zig-zagging. Tool #7 (Identify Subproblems) then splits the loop into four independent legs $F \to A$, $A \to B$, $B \to C$, $C \to F$. Solve each tiny leg by counting, then add the four numbers.
Execute — Answer: C
K.G.A.1 Step 1 - Read each labeled point off the grid by counting blocks right from the left edge and blocks up from the bottom.
- This gives $F = (6, 5)$, $A = (7, 3)$, $B = (0, 0)$, $C = (2, 4)$.
💡 Describing where something is on the grid using right/up counts is Kindergarten position language.
3.MD.D.8 Step 2 - Because the truck must follow streets, the shortest distance between two corners is (horizontal blocks apart) $+$ (vertical blocks apart).
- Split the trip into four legs and find each leg separately — this is the Tool #7 (Subproblems) move.
💡 Adding side-distances to get a path length on a grid is the same idea as finding perimeter pieces in Grade 3.
2.OA.A.1 Step 3 - Leg $F \to A$: from column $6$ to column $7$ is $1$ block right; from row $5$ down to row $3$ is $2$ blocks.
- Total $= 1 + 2 = 3$ blocks.
💡 Just counting blocks across and down and adding them is a Grade 2 addition word problem.
2.OA.A.1 Step 4 - Leg $A \to B$: from column $7$ all the way left to column $0$ is $7$ blocks; from row $3$ down to row $0$ is $3$ blocks.
- Total $= 7 + 3 = 10$ blocks.
💡 Counting blocks across and down, then adding, is still Grade 2 arithmetic.
2.OA.A.1 Step 5 - Leg $B \to C$: from column $0$ to column $2$ is $2$ blocks; from row $0$ up to row $4$ is $4$ blocks.
- Total $= 2 + 4 = 6$ blocks.
💡 Same idea: count across, count up, add.
2.OA.A.1 Step 6 - Leg $C \to F$: from column $2$ to column $6$ is $4$ blocks; from row $4$ up to row $5$ is $1$ block.
- Total $= 4 + 1 = 5$ blocks.
💡 One more across-and-up count, then add.
3.NBT.A.2 Step 7 - Add the four leg lengths to get the full loop.
- $3 + 10 + 6 + 5 = 24$ blocks, which matches choice (C).
💡 Summing four small numbers fluently is Grade 3 add-within-$1000$.
K.G.A.1 Read each labeled point off the grid by counting blocks right from the left edge 3.MD.D.8 Because the truck must follow streets, the shortest distance between two corners 2.OA.A.1 Leg $F \to A$: from column $6$ to column $7$ is $1$ block right; from row $5$ do 2.OA.A.1 Leg $A \to B$: from column $7$ all the way left to column $0$ is $7$ blocks; fro 2.OA.A.1 Leg $B \to C$: from column $0$ to column $2$ is $2$ blocks; from row $0$ up to r 2.OA.A.1 Leg $C \to F$: from column $2$ to column $6$ is $4$ blocks; from row $4$ up to r 3.NBT.A.2 Add the four leg lengths to get the full loop. $3 + 10 + 6 + 5 = 24$ blocks, whi Review
Reasonableness: Every leg's length is at least the straight-line block gap between its endpoints, so $24$ blocks is a true lower bound for this $F \to A \to B \to C \to F$ order. The grid is $8 \times 6$, so a route that touches the far corners ($B$ at $(0,0)$, $A$ at $(7,3)$) must include the long $A \to B$ trip of $10$ blocks; $24$ total is consistent with that. The answer sits in the middle of the choices $20$–$28$, which is the typical AMC 8 spread when the trap answers come from miscounting one leg by $\pm 1$ or $\pm 2$.
Alternative: Tool #3 (Eliminate Possibilities) plus a parity check: any closed loop on a grid that returns to its start uses an even number of horizontal moves and an even number of vertical moves, so the total must be even. That alone keeps only $20$, $22$, $24$, $26$, $28$. Then a quick sanity bound — the single leg $A \to B$ already needs $10$ blocks, and the other three legs together need at least $3 + 6 + 5 = 14$, giving $\geq 24$ — pins the answer to (C).
CCSS standards used (min grade 3)
K.G.A.1Describe positions of objects using above, below, beside, in front of (Reading each labeled point's position on the grid by saying how many blocks right and how many blocks up it is from the corner.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Finding each leg's block-length by adding the horizontal block-count and the vertical block-count (e.g. $1 + 2 = 3$, $7 + 3 = 10$).)3.MD.D.8Solve real-world problems involving perimeters of polygons (Treating the closed delivery route $F \to A \to B \to C \to F$ as a polygonal path whose total length is the sum of its sides.)3.NBT.A.2Fluently add and subtract within 1000 (Adding the four leg lengths $3 + 10 + 6 + 5 = 24$ to get the total number of blocks.)
⭐ This AMC 8 problem only needs Grade 3 addition and the idea of "path length around a shape" you already know!
⭐ This AMC 8 problem only needs Grade 3 addition and the idea of "path length around a shape" you already know!
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