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

Pick an answer.

(A)

(B)

34.5

(C)

(D)

37.5

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Two parallel number lines — one labeled ° F, the other ° 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° F sits 90 above the left anchor. Tool #8 (Analyze the Units) is the verification companion — track the unit ratio ° B / ° F through the calculation so the final number is in ° B. Algebra (Tool #13) would also work but the picture-plus-ratio path is faster and stays inside Grade 6 unit-rate thinking.

1STEP 1

Draw two aligned number lines: 110 and 350 on the F-line, 0 and 100 below on the B-line, and the target 200° F between the anchors.

110 ° F & 200 ° F & 350 ° F ; ↓ & ↓ & ↓ ; 0 ° B & ? ° B & 100 ° B

2STEP 2

Read the two spans between the anchors: the F-range is 350 - 110 = 240°, the B-range is 100 - 0 = 100°.

Δ F = 240, Δ B = 100

3STEP 3

On the F-line, the target sits 200 - 110 = 90° above the left anchor.

200 - 110 = 90 ° F

4STEP 4

Since it's linear, that same fractional position $\frac{90}{240}$ = $\frac{3}{8}$ carries over to the B-line.

\frac{90}{240}

\frac{9}{24}

\frac{3}{8}

5STEP 5

Take that $\frac{3}{8}$ of the B-range from 0° B: $\frac{3}{8}$ · 100 = 37.5° B, which is (D).

? = 0 +

\frac{3}{8}

· 100 =

\frac{300}{8}

= 37.5 ° B → (D)

Answer

37.5

Three quick checks. (1) Magnitude: 200 is roughly

\frac{3}{8}

of the way from 110 to 350, so the answer should be roughly

\frac{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° F into the formula gives

\frac{3}{8}

· 0 = 0° B ✓; plugging 350° F gives

\frac{3}{8}

· 240 ·

\frac{100}{240}

= 100° 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

\frac{1}{3}

of 100, ignoring the 110° F shift).

💡Key takeaway

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