AMC 8 · 2024 · #14

Easy mode Grade 2

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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

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

#1 Draw a Diagram 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.

$$A \xrightarrow{5} X,\ A \xrightarrow{8} M,\quad M \xrightarrow{25} Z,\ Y \xrightarrow{17} Z,\ C \xrightarrow{10} Z$$

💡 Describing how towns and arrows sit relative to each other (which arrow leaves which town) is the basic position-vocabulary skill from Kindergarten geometry.

#7 Identify Subproblems 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 $7$.

$$A\to X:\ 5.\quad A\to M:\ \min(8,\ 5+2) = \min(8, 7) = 7$$

💡 Adding two small two-digit numbers $5+2=7$ and comparing $7$ with $8$ is exactly Grade 2 fluent addition and comparison within 100.

#7 Identify Subproblems 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 $= 18$.

$$A\to Y:\ \min(5+10,\ 7+6) = \min(15, 13) = 13.\quad A\to C:\ \min(7+14,\ 13+5) = \min(21, 18) = 18$$

💡 Adding distances in two steps and keeping the smaller running total is a Grade 2 two-step word-problem move.

#2 Make a Systematic List 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 $28$.

$$\min(7+25,\ 13+17,\ 18+10) = \min(32, 30, 28) = 28$$

💡 Listing every way to reach $Z$ and adding the two-digit pieces is a Grade 2 two-step addition exercise within 100.

#3 Eliminate Possibilities 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.

$$28 \;\Rightarrow\; \textbf{(A)}$$

💡 Comparing a handful of two-digit numbers and picking the smallest is exactly the Grade 1 two-digit comparison skill.

[1] #1 K.G.A.1 Start by getting the arrows organized in a picture. Only two arrows leave $A$ di

[2] #7 2.NBT.B.5 Break the big problem into **smaller pieces**: first find the shortest distance

[3] #7 2.OA.A.1 Push the same idea one layer further to find the shortest distance to $Y$ and $C

[4] #2 2.OA.A.1 Now consider the **last hop** into $Z$ — there are exactly three options ($M\to

[5] #3 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.1 Describe 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.3 Compare 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.5 Fluently 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.1 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 2)

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