AMC 10 · 2023 · #1

Grade 6 rate-ratio

Archetype Meeting Time with Piecewise Rates

ratelinear-equations-one-varunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: ratemulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet?

$\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Two bikers start at the same time from cities $A$ and $B$, which are $45$ miles apart, and pedal toward each other. Alicia (from $A$) goes $18$ mph; Beth (from $B$) goes $12$ mph. How far from $A$ are they when they meet?

Givens: Distance between $A$ and $B$ is $45$ miles; Alicia leaves $A$ at $18$ mph heading toward $B$; Beth leaves $B$ at $12$ mph heading toward $A$; They start at the same time; Answer choices: (A) $20$, (B) $24$, (C) $25$, (D) $26$, (E) $27$

Unknowns: Distance from city $A$ to the meeting point (in miles)

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #8 Analyze the Units

The problem describes positions on a road and motion along that road — Tool #1 (Draw a Diagram) is the natural lead. A simple labeled segment from $A$ to $B$ with arrows showing the two bikers makes the structure obvious: the gap of $45$ miles closes at the combined speed $18 + 12 = 30$ mph. Tool #8 (Analyze the Units) is the verification companion — tracking miles, mph, and hours through every multiplication makes sure the final number is in miles, which is what the problem asks for. Algebra (Tool #13) would also work but the diagram-plus-units path is faster and more visual for a Grade 6 rate concept.

Execute — Answer: E

#1 Draw a Diagram 6.RP.A.2 Step 1

Draw a $45$-mile segment with $A$ on the left and $B$ on the right.
Put Alicia at $A$ with a rightward arrow labeled $18$ mph; put Beth at $B$ with a leftward arrow labeled $12$ mph.
The picture shows the gap between them shrinking from both ends at the same time.

$$A \xrightarrow{18\,\text{mph}} \;\cdots\; \xleftarrow{12\,\text{mph}} B, \quad |AB| = 45$$

💡 Drawing the road with two arrows turns a word problem into a closing-gap picture — the Grade 6 "unit rate" idea that miles-per-hour is how fast the gap shrinks.

#1 Draw a Diagram 4.OA.A.3 Step 2

From the diagram, the two arrows close the gap at the same time, so their speeds add.
Compute the combined closing speed.

$$18 + 12 = 30 \text{ mph}$$

💡 When two things move toward each other, the gap shrinks at the sum of their speeds — a Grade 4 multi-step word-problem addition.

#8 Analyze the Units 6.RP.A.3 Step 3

Find the time until they meet by dividing the $45$-mile gap by the $30$-mph closing speed.
Track units: miles divided by miles-per-hour leaves hours.

$$t = \dfrac{45 \text{ mi}}{30 \text{ mi/hr}} = 1.5 \text{ hr}$$

💡 Distance $\div$ rate $=$ time — the Grade 6 rate-reasoning that miles cancel and leave hours, exactly the unit the next step needs.

#8 Analyze the Units 6.RP.A.3 Step 4

The meeting point is wherever Alicia has biked to in $1.5$ hours.
Multiply her speed by the time.
The question asks for distance from $A$, which is exactly Alicia's traveled distance.

$$d_A = 18 \text{ mi/hr} \times 1.5 \text{ hr} = 27 \text{ mi} \;\Rightarrow\; \textbf{(E)}$$

💡 Rate $\times$ time $=$ distance — the same Grade 6 unit-rate move in reverse; mph $\times$ hr cancels to mi, the requested unit.

[1] #1 6.RP.A.2 Draw a $45$-mile segment with $A$ on the left and $B$ on the right. Put Alicia a

[2] #1 4.OA.A.3 From the diagram, the two arrows close the gap at the same time, so their speeds

[3] #8 6.RP.A.3 Find the time until they meet by dividing the $45$-mile gap by the $30$-mph clos

[4] #8 6.RP.A.3 The meeting point is wherever Alicia has biked to in $1.5$ hours. Multiply her s

Review

Reasonableness: Cross-check from Beth's side. In $1.5$ hours Beth covers $12 \times 1.5 = 18$ miles from $B$ toward $A$. Alicia's $27$ miles plus Beth's $18$ miles is $27 + 18 = 45$ miles — the full distance, so they really do meet at that spot. Also, since Alicia is the faster rider, the meeting point should sit past the midpoint ($22.5$ mi) on the $B$ side; $27 > 22.5$ confirms this.

Alternative: Tool #13 (Convert to Algebra): let $t$ be the meeting time. Alicia's position from $A$ is $18t$ and Beth's position from $A$ is $45 - 12t$. Setting them equal gives $18t = 45 - 12t \Rightarrow 30t = 45 \Rightarrow t = 1.5$, then $18 \cdot 1.5 = 27$ miles. Same answer (E), more symbol-pushing.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Adding the two speeds $18 + 12 = 30$ to get the combined closing speed at which the gap between Alicia and Beth shrinks.)
6.RP.A.2 Understand the concept of a unit rate and use rate language (Interpreting "$18$ miles per hour" as a unit rate — the number of miles Alicia covers each hour — so the diagram has a precise meaning.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Using distance $=$ rate $\times$ time twice: once to get $t = 45/30 = 1.5$ hr, then $d_A = 18 \times 1.5 = 27$ mi.)

⭐ This AMC 10 problem only needs Grade 6 unit rates you already know — speeds add when two riders close in, then distance equals speed times time!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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