AMC 10 · 2023 · #8

Grade 6 algebra

linear-equations-two-varslope-interceptratio-proportion identify-subproblemsconvert-to-algebra ↑ Prerequisites: linear-equations-one-varratio-proportion

📏 Medium solution 💡 2 insights

Problem

Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. Bread is done when its internal temperature is $200$ degrees Fahrenheit. What is this in degrees on the Breadus scale?

$\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39$

Pick an answer.

(A)

(B)

34.5

(C)

(D)

37.5

(E)

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Toolkit + CCSS Solution

Understand

Restated: Barb's Breadus scale ($^\circ B$) is a linear function of Fahrenheit ($^\circ F$). Two anchor points are given: $110^\circ F = 0^\circ B$ (rising) and $350^\circ F = 100^\circ B$ (baking). Find the Breadus reading at $200^\circ F$ (done).

Givens: Breadus scale is linear in Fahrenheit; Anchor 1: $110^\circ F \leftrightarrow 0^\circ B$; Anchor 2: $350^\circ F \leftrightarrow 100^\circ B$; Target: $200^\circ F \leftrightarrow ?\,^\circ B$; Answer choices: (A) $33$, (B) $34.5$, (C) $36$, (D) $37.5$, (E) $39$

Unknowns: The Breadus temperature corresponding to $200^\circ F$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #8 Analyze the Units

Two parallel number lines — one labeled $^\circ F$, the other $^\circ B$ — make the linear correspondence visible: $110$ on top sits above $0$ on bottom, and $350$ on top sits above $100$ on bottom. Tool #1 (Draw a Diagram) is the natural lead because the trigger is "positions on a scale". From the picture you read off two facts: the $F$-line covers $240$ degrees while the $B$-line covers $100$, and the target $200^\circ F$ sits $90$ above the left anchor. Tool #8 (Analyze the Units) is the verification companion — track the unit ratio $^\circ B / ^\circ F$ through the calculation so the final number is in $^\circ B$. Algebra (Tool #13) would also work but the picture-plus-ratio path is faster and stays inside Grade 6 unit-rate thinking.

Execute — Answer: D

#1 Draw a Diagram 5.G.A.1 Step 1

Draw two horizontal number lines, one above the other.
Mark $110$ and $350$ on the top $F$-line; mark $0$ and $100$ on the bottom $B$-line, lined up vertically with their partners.
Add the target tick at $200^\circ F$ between them.

$$\begin{array}{ccc} 110\,^\circ F & 200\,^\circ F & 350\,^\circ F \\ \downarrow & \downarrow & \downarrow \\ 0\,^\circ B & ?\,^\circ B & 100\,^\circ B \end{array}$$

💡 Two aligned number lines are the linear-relationship picture — Grade 5 coordinate-system intuition without needing the word "slope".

#1 Draw a Diagram 4.NBT.B.4 Step 2

Read off the two range lengths between the anchors.
The top $F$-range is $350 - 110 = 240$ degrees; the bottom $B$-range is $100 - 0 = 100$ degrees.
The same physical interval looks like $240$ on top and $100$ on bottom.

$$\Delta F = 240, \quad \Delta B = 100$$

💡 Two subtractions to find the lengths of the two segments — Grade 4 fluency with multi-digit subtraction.

#1 Draw a Diagram 4.NBT.B.4 Step 3

Measure how far the target $200^\circ F$ sits from the left anchor on the $F$-line.

$$200 - 110 = 90\,^\circ F$$

💡 Subtraction along the number line — Grade 4 multi-digit subtraction.

#1 Draw a Diagram 4.NF.A.1 Step 4

Because the relationship is linear, the same fractional position $\tfrac{90}{240}$ of the way across also gives the answer on the $B$-line.
Simplify the fraction.

$$\dfrac{90}{240} = \dfrac{9}{24} = \dfrac{3}{8}$$

💡 Equal-step scales share the same fractional landmarks — Grade 4 equivalent-fraction reasoning.

#8 Analyze the Units 6.RP.A.3 Step 5

Apply that fractional position to the $B$-range of length $100$, starting from $0^\circ B$.
Units check: a $\tfrac{^\circ B}{^\circ F}$ ratio multiplied by $^\circ F$ correctly returns $^\circ B$.

$$? = 0 + \dfrac{3}{8} \cdot 100 = \dfrac{300}{8} = 37.5\,^\circ B \;\Rightarrow\; \textbf{(D)}$$

💡 Same fraction of the way across both scales — Grade 6 ratio/proportion reasoning, with units carried through to confirm the answer is in $^\circ B$.

[1] #1 5.G.A.1 Draw two horizontal number lines, one above the other. Mark $110$ and $350$ on t

[2] #1 4.NBT.B.4 Read off the two range lengths between the anchors. The top $F$-range is $350 -

[3] #1 4.NBT.B.4 Measure how far the target $200^\circ F$ sits from the left anchor on the $F$-li

[4] #1 4.NF.A.1 Because the relationship is linear, the same fractional position $\tfrac{90}{240

[5] #8 6.RP.A.3 Apply that fractional position to the $B$-range of length $100$, starting from $

Review

Reasonableness: Three quick checks. (1) Magnitude: $200$ is roughly $\tfrac{3}{8}$ of the way from $110$ to $350$, so the answer should be roughly $\tfrac{3}{8}$ of the way from $0$ to $100$, i.e. somewhere around $37$ or $38$ — and (D) $37.5$ fits perfectly. (2) Endpoints: plugging $110^\circ F$ into the formula gives $\tfrac{3}{8} \cdot 0 = 0^\circ B$ ✓; plugging $350^\circ F$ gives $\tfrac{3}{8} \cdot 240 \cdot \tfrac{100}{240} = 100^\circ B$ ✓. (3) Eliminate: the other choices $33, 34.5, 36, 39$ each correspond to wrong anchor pairs (e.g. $33$ comes from treating $200$ as $\tfrac{1}{3}$ of $100$, ignoring the $110^\circ F$ shift).

Alternative: Tool #13 (Convert to Algebra): write $B = m F + b$ as a linear function. The two anchors give $0 = 110m + b$ and $100 = 350 m + b$, so subtracting yields $100 = 240 m$, i.e. $m = \tfrac{5}{12}$, and $b = -\tfrac{5}{12} \cdot 110 = -\tfrac{275}{6}$. Then $B(200) = \tfrac{5}{12} \cdot 200 - \tfrac{275}{6} = \tfrac{1000}{12} - \tfrac{550}{12} = \tfrac{450}{12} = 37.5$. Same answer, more work — exactly the case where the diagram-plus-ratio path wins.

CCSS standards used (min grade 6)

5.G.A.1 Use a pair of perpendicular number lines to form a coordinate system (Drawing two stacked number lines with the $F$- and $B$-scales aligned so that the linear correspondence is read off geometrically.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Subtractions $350 - 110 = 240$, $100 - 0 = 100$, and $200 - 110 = 90$ that find the lengths and the target's offset along the $F$-line.)
4.NF.A.1 Explain why a fraction $\tfrac{a}{b}$ is equivalent to $\tfrac{n a}{n b}$ (Simplifying $\tfrac{90}{240}$ to $\tfrac{9}{24}$ to $\tfrac{3}{8}$ — equivalent-fraction reduction.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Mapping the fractional position $\tfrac{3}{8}$ on the $F$-line to the same fractional position on the $B$-line, scaling the $B$-range $100$ by $\tfrac{3}{8}$ to get $37.5^\circ B$.)

⭐ This AMC 10 problem only needs Grade 6 ratio reasoning you already know — line up the two scales, see that $200^\circ F$ is $\tfrac{3}{8}$ of the way along, and take that same $\tfrac{3}{8}$ of $100$ on the Breadus scale.