AMC 10 · 2023 · #8
Grade 6 algebraProblem
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?
Pick an answer.
AMC 10 2023 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: 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
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) $39$
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
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.
💡 Two aligned number lines are the linear-relationship picture — Grade 5 coordinate-system intuition without needing the word "slope".
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.
💡 Two subtractions to find the lengths of the two segments — Grade 4 fluency with multi-digit subtraction.
4.NBT.B.4 Step 3 Measure how far the target $200^\circ F$ sits from the left anchor on the $F$-line.
💡 Subtraction along the number line — Grade 4 multi-digit subtraction.
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.
💡 Equal-step scales share the same fractional landmarks — Grade 4 equivalent-fraction reasoning.
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 $^\circ B$.
💡 Same fraction of the way across both scales — Grade 6 ratio/proportion reasoning, with units carried through to confirm the answer is in $^\circ B$.
5.G.A.1 Draw two horizontal number lines, one above the other. Mark $110$ and $350$ on t 4.NBT.B.4 Read off the two range lengths between the anchors. The top $F$-range is $350 - 4.NBT.B.4 Measure how far the target $200^\circ F$ sits from the left anchor on the $F$-li 4.NF.A.1 Because the relationship is linear, the same fractional position $\tfrac{90}{240 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.1Use 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.4Fluently 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.1Explain 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.3Use 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.
⭐ 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.