Sensim Math Original · sm-7

Easy mode Grade 2

Problem

Picture a summer camp on a hillside with six cabins. Hana is a counselor who carries the mail from cabin $H$ (Headquarters, at the top) down to cabin $R$ (Riverside, at the bottom).

The six cabins are $H$ , $P$ (Pinewood), $L$ (Lakeview), $B$ (Birch), $M$ (Maple), and $R$ (Riverside). Trails connect some of the cabins. Hana is only allowed to walk downhill, so each trail can only be used in one direction (always away from $H$ , toward $R$ ).

Here are the trails and their lengths in meters:

$H \to P$ : $4$ m
$H \to L$ : $9$ m
$P \to L$ : $4$ m
$P \to B$ : $11$ m
$L \to B$ : $6$ m
$L \to M$ : $12$ m
$B \to M$ : $5$ m
$B \to R$ : $18$ m
$M \to R$ : $7$ m

Hana wants to walk from $H$ to $R$ using these trails, going only in the allowed direction. What is the shortest total distance, in meters, she can walk?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A summer-camp counselor, Hana, must carry mail from cabin $H$ to cabin $R$ along nine one-way (downhill-only) trails connecting six cabins $H, P, L, B, M, R$. The trail lengths in meters are fixed and listed in the problem. We must find the smallest total length of any allowed $H \to R$ trail sequence.

Givens: Six cabins: $H, P, L, B, M, R$; Nine directed trails with lengths (meters): $H\to P = 4$, $H\to L = 9$, $P\to L = 4$, $P\to B = 11$, $L\to B = 6$, $L\to M = 12$, $B\to M = 5$, $B\to R = 18$, $M\to R = 7$; Hana may only walk in the listed direction (downhill); Five answer choices: (A) 24, (B) 25, (C) 26, (D) 27, (E) 28

Unknowns: The minimum total distance, in meters, of any allowed sequence of trails from $H$ to $R$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #3 Eliminate Possibilities

The trails form a small map, so first use Tool #1 to sketch the cabins and arrows so we can see which trails leave $H$ and which trails enter $R$. Because every path from $H$ to $R$ must pass through one of the intermediate cabins $P, L, B, M$, we use Tool #7 to break the big question into the easier subproblems "shortest distance from $H$ to each intermediate cabin," growing the partial answers outward layer by layer. Tool #2 (systematic list) handles the final hop into $R$ — there are exactly two trails entering $R$, so we list both options and add. Tool #3 then matches the minimum to one of the five offered choices.

Execute — Answer: C

#1 Draw a Diagram K.G.A.1 Step 1

Sketch the cabins and arrows.
Only two trails leave $H$ directly: $H\to P = 4$ and $H\to L = 9$.
Only two trails enter $R$: $B\to R = 18$ and $M\to R = 7$.
So every valid route must start with $H\to P$ or $H\to L$ and finish with $B\to R$ or $M\to R$.
This framing is the whole solution's backbone.

$$H \xrightarrow{4} P,\ H \xrightarrow{9} L,\quad B \xrightarrow{18} R,\ M \xrightarrow{7} R$$

💡 Reading where each arrow starts and ends is the same kind of left-right-above-below position vocabulary kids learn in Kindergarten geometry.

#7 Identify Subproblems 2.NBT.B.5 Step 2

Break the journey into small pieces.
First find the shortest distance from $H$ to the two nearest cabins, $P$ and $L$.
There is only one way into $P$, so the shortest distance is $H\to P = 4$.
There are two ways into $L$: directly $H\to L = 9$, or via $P$ as $H\to P\to L = 4 + 4 = 8$.
Since $8 < 9$, the shortest distance from $H$ to $L$ is $8$.

$$d(P) = 4.\quad d(L) = \min(9,\ 4 + 4) = \min(9, 8) = 8$$

💡 Adding $4 + 4 = 8$ and comparing $8$ with $9$ is exactly Grade 2 fluent addition and comparison inside 100.

#7 Identify Subproblems 2.OA.A.1 Step 3

Push outward one more layer to $B$ and $M$.
Cabin $B$ is reached by $P\to B = 11$ or $L\to B = 6$, giving candidates $d(P) + 11 = 4 + 11 = 15$ and $d(L) + 6 = 8 + 6 = 14$, so $d(B) = 14$.
Cabin $M$ is reached by $L\to M = 12$ or $B\to M = 5$, giving $d(L) + 12 = 8 + 12 = 20$ and $d(B) + 5 = 14 + 5 = 19$, so $d(M) = 19$.
At each cabin we keep only the shorter running total.

$$d(B) = \min(4+11,\ 8+6) = \min(15, 14) = 14.\quad d(M) = \min(8+12,\ 14+5) = \min(20, 19) = 19$$

💡 Carrying two small running totals forward and keeping the smaller one is exactly the Grade 2 two-step addition word-problem move.

#2 Make a Systematic List 2.NBT.B.5 Step 4

Now use a systematic list for the final hop into $R$.
Only two trails enter $R$, so we list both: arriving via $B$ gives $d(B) + 18 = 14 + 18 = 32$, and arriving via $M$ gives $d(M) + 7 = 19 + 7 = 26$.
The smaller of $32$ and $26$ is $26$, so the shortest distance from $H$ to $R$ is $26$ m, achieved by the route $H \to P \to L \to B \to M \to R$.

$$d(R) = \min(d(B)+18,\ d(M)+7) = \min(32, 26) = 26$$

💡 Listing the last two ways into the goal and adding two two-digit numbers ($14+18$ and $19+7$) is Grade 2 fluent addition within 100.

#3 Eliminate Possibilities 1.NBT.B.3 Step 5

Compare $26$ to the five offered choices $24, 25, 26, 27, 28$.
Our value $26$ is exactly choice (C).
The neighbours $27$ and $28$ are real "near-miss" route totals — for example $H\to P\to L\to M\to R = 4+4+12+7 = 27$ matches (D), and $H\to L\to M\to R = 9+12+7 = 28$ matches (E) — but neither beats $26$.
The smaller candidates $24$ and $25$ are below the sum of the five edges along the optimal route, so they are impossible.

$$d(R) = 26 \;\Rightarrow\; \textbf{(C)}$$

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

[1] #1 K.G.A.1 Sketch the cabins and arrows. Only two trails leave $H$ directly: $H\to P = 4$ a

[2] #7 2.NBT.B.5 Break the journey into small pieces. First find the shortest distance from $H$ t

[3] #7 2.OA.A.1 Push outward one more layer to $B$ and $M$. Cabin $B$ is reached by $P\to B = 11

[4] #2 2.NBT.B.5 Now use a systematic list for the final hop into $R$. Only two trails enter $R$,

[5] #3 1.NBT.B.3 Compare $26$ to the five offered choices $24, 25, 26, 27, 28$. Our value $26$ is

Review

Reasonableness: Re-add the optimal route $H \to P \to L \to B \to M \to R$ in one shot: $4 + 4 + 6 + 5 + 7 = 26$, matching the layered totals exactly. The route uses the cheapest trail out of $H$ ($H\to P = 4$) and the cheapest trail into $R$ ($M\to R = 7$), and the three middle hops $4, 6, 5$ are each the small option at their respective cabin — so $26$ also looks right by feel. The two next-best totals ($27$ from $H\to L\to B\to M\to R = 9+6+5+7$, and $28$ from $H\to L\to M\to R = 9+12+7$) match choices (D) and (E), which confirms (C) is the only minimum.

Alternative: An alternative is to use Tool #2 alone and enumerate **every** simple $H\to R$ route from scratch. Counting by hand gives seven such routes — $H\!\to\! P\!\to\! L\!\to\! B\!\to\! M\!\to\! R$, $H\!\to\! P\!\to\! L\!\to\! B\!\to\! R$, $H\!\to\! P\!\to\! L\!\to\! M\!\to\! R$, $H\!\to\! P\!\to\! B\!\to\! M\!\to\! R$, $H\!\to\! P\!\to\! B\!\to\! R$, $H\!\to\! L\!\to\! B\!\to\! M\!\to\! R$, $H\!\to\! L\!\to\! B\!\to\! R$, $H\!\to\! L\!\to\! M\!\to\! R$ — and the smallest total is again $26$, but the subproblem approach we used reaches the same answer with far fewer additions because the partial sums $d(P), d(L), d(B), d(M)$ are reused.

CCSS standards used (min grade 2)

K.G.A.1 Describe positions of objects using above, below, beside, in front of (Reading the cabin map to see which trails leave $H$ and which trails enter $R$.)
1.NBT.B.3 Compare two two-digit numbers using symbols (Comparing the candidate route totals $26, 27, 28$ to pick the smallest and match it to a choice.)
2.NBT.B.5 Fluently add and subtract within 100 (Two-digit sums like $4+4=8$, $14+18=32$, and $19+7=26$ used to build and compare partial distances.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Accumulating trail distances in two steps ($4+11=15$ vs. $8+6=14$, $8+12=20$ vs. $14+5=19$) and keeping the shorter running total at each cabin.)

⭐ 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: C

Review

CCSS standards used (min grade 2)

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