AMC 8 · 2024 · #14
Easy mode Grade 2Problem
Imagine six towns named A, M, C, X, Y, and Z. They are connected by one-way roads, shown in the figure below. The number beside each road tells you its length in kilometers.
You want to travel from town A to town Z by following these roads. There are several possible routes, and each has a total length.
What is the shortest possible total distance from A to Z, in kilometers?
Diagram
Six towns connected by one-way roads; the number on each arrow shows that road's length in kilometers.
Pick an answer.
AMC 8 2024 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: Six towns $A, M, C, X, Y, Z$ are connected by one-way roads with given distances (km). Using only these roads (in the direction of the arrows), we want the **shortest total distance** from $A$ to $Z$ in km.
Givens: Nodes (towns): $A, M, C, X, Y, Z$; Directed edges with distances: $A\to M = 8$, $A\to X = 5$, $X\to M = 2$, $X\to Y = 10$, $M\to Y = 6$, $M\to C = 14$, $Y\to C = 5$, $C\to Z = 10$, $Y\to Z = 17$, $M\to Z = 25$; All roads are one-way (you can only travel in the arrow direction); Answer choices: (A) 28, (B) 29, (C) 30, (D) 31, (E) 32
Unknowns: The shortest total distance from $A$ to $Z$ in km
Understand
Restated: Six towns $A, M, C, X, Y, Z$ are connected by one-way roads with given distances (km). Using only these roads (in the direction of the arrows), we want the **shortest total distance** from $A$ to $Z$ in km.
Givens: Nodes (towns): $A, M, C, X, Y, Z$; Directed edges with distances: $A\to M = 8$, $A\to X = 5$, $X\to M = 2$, $X\to Y = 10$, $M\to Y = 6$, $M\to C = 14$, $Y\to C = 5$, $C\to Z = 10$, $Y\to Z = 17$, $M\to Z = 25$; All roads are one-way (you can only travel in the arrow direction); Answer choices: (A) 28, (B) 29, (C) 30, (D) 31, (E) 32
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #3 Eliminate Possibilities
This is a classic **map problem** about positions and connections, so the first move is Tool #1 — get the towns, arrows, and distances laid out in a clean picture. Then break the big question "$A\to Z$" into **smaller subproblems**: the shortest distance from $A$ to each intermediate town (Tool #7). That lets us reuse partial sums instead of re-adding the same numbers. Finally, at the last step we **list all routes** into $Z$ exhaustively (Tool #2) and pick the smallest, then check it against the answer choices with Tool #3.
Execute — Answer: A
K.G.A.1 Step 1 - Start by getting the arrows organized in a picture.
- Only two arrows leave $A$ directly: $A\to X = 5$ and $A\to M = 8$.
- Only three arrows enter $Z$ directly: $M\to Z = 25$, $Y\to Z = 17$, $C\to Z = 10$.
- So every route must finish with one of those three final hops from $M$, $Y$, or $C$ into $Z$.
- That observation gives the whole solution its shape.
💡 Describing how towns and arrows sit relative to each other (which arrow leaves which town) is the basic position-vocabulary skill from Kindergarten geometry.
2.NBT.B.5 Step 2 - Break the big problem into **smaller pieces**: first find the shortest distance from $A$ to the towns nearest $A$, which are $X$ and $M$.
- There's only one way into $X$, so $A\to X = 5$.
- For $M$ there are two ways: directly $A\to M = 8$, or via $X$: $A\to X\to M = 5 + 2 = 7$.
- Since $7 < 8$, the shortest distance from $A$ to $M$ is $7$.
💡 Adding two small two-digit numbers $5+2=7$ and comparing $7$ with $8$ is exactly Grade 2 fluent addition and comparison within 100.
2.OA.A.1 Step 3 - Push the same idea one layer further to find the shortest distance to $Y$ and $C$.
- $Y$ is reached by $X\to Y = 10$ or $M\to Y = 6$: $A\to X\to Y = 5 + 10 = 15$, $A\to \cdots \to M\to Y = 7 + 6 = 13$.
- So $A\to Y$ shortest $= 13$.
- $C$ is reached by $M\to C = 14$ or $Y\to C = 5$: $A\to \cdots \to M\to C = 7 + 14 = 21$, $A\to \cdots \to Y\to C = 13 + 5 = 18$.
- So $A\to C$ shortest $= 18$.
💡 Adding distances in two steps and keeping the smaller running total is a Grade 2 two-step word-problem move.
2.OA.A.1 Step 4 - Now consider the **last hop** into $Z$ — there are exactly three options ($M\to Z$, $Y\to Z$, $C\to Z$), so we list all three: ① via $M$: $7 + 25 = 32$, ② via $Y$: $13 + 17 = 30$, ③ via $C$: $18 + 10 = 28$.
- The smallest of $32, 30, 28$ is $28$, so the shortest distance from $A$ to $Z$ is $28$.
💡 Listing every way to reach $Z$ and adding the two-digit pieces is a Grade 2 two-step addition exercise within 100.
1.NBT.B.3 Step 5 - Match $28$ to the answer choices.
- Of $28, 29, 30, 31, 32$, our value $28$ is exactly choice (A).
- Choices (C) $30$ and (E) $32$ are the longer routes we already computed (via $Y$ and via $M$), and (B) $29$ and (D) $31$ never appear as a sum of any combination of integer edge lengths along an actual path, so they can be ruled out.
💡 Comparing a handful of two-digit numbers and picking the smallest is exactly the Grade 1 two-digit comparison skill.
K.G.A.1 Start by getting the arrows organized in a picture. Only two arrows leave $A$ di 2.NBT.B.5 Break the big problem into **smaller pieces**: first find the shortest distance 2.OA.A.1 Push the same idea one layer further to find the shortest distance to $Y$ and $C 2.OA.A.1 Now consider the **last hop** into $Z$ — there are exactly three options ($M\to 1.NBT.B.3 Match $28$ to the answer choices. Of $28, 29, 30, 31, 32$, our value $28$ is exa Review
Reasonableness: Re-add the optimal route $A \to X \to M \to Y \to C \to Z$ in one shot: $5 + 2 + 6 + 5 + 10 = 28$, matching the step-by-step accumulation exactly. The route starts on the cheapest arrow out of $A$ ($A\to X = 5$), visits every intermediate town along the way, and ends on the cheapest arrow into $Z$ ($C\to Z = 10$) — so $28$ also feels right by size. The fact that $C\to Z = 10$ (the shortest single arrow into $Z$) appears in our winning route is a nice sanity check.
Alternative: An alternative is to use Tool #2 alone and list **every** simple $A\to Z$ path from scratch. That works but produces 7-8 paths, with the same partial sums recomputed many times. Combining Tool #7 (Subproblems) with the diagram, as we did, gets the same answer $28$ with far fewer additions.
CCSS standards used (min grade 2)
K.G.A.1Describe positions of objects using above, below, beside, in front of (Reading the diagram to see which arrows leave $A$ and which arrows enter $Z$.)1.NBT.B.3Compare two two-digit numbers using symbols (Comparing the three route totals $32, 30, 28$ to pick the smallest and match it to a choice.)2.NBT.B.5Fluently add and subtract within 100 (Two-digit sums like $5+2=7$, $5+10=15$, and the comparison of $7$ vs $8$.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Accumulating distances in two steps ($7+6=13$, $13+5=18$, $18+10=28$, etc.) and choosing the shorter running total.)
⭐ This AMC 8 problem only needs Grade 2 addition within 100 and two-digit comparison you already know!
⭐ This AMC 8 problem only needs Grade 2 addition within 100 and two-digit comparison you already know!
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