2023 AMC 10A Problem 8

P1-10~2 minPrint

Linear Equations & Word Problems Functions Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

200 is 90 of the 240 degrees from 110 to 350, which is 3/8 of the way. On the new scale that is 3/8 of 100, or 37.5.

Solution

"Linear" means the two scales stretch evenly. If a Fahrenheit temperature is a certain fraction of the way from $110$ to $350$ , then its Breadus reading is the same fraction of the way from $0$ to $100$ .

1. How long is the Fahrenheit stretch? From $110$ to $350$ is $350 - 110 = 240$ degrees.
2. How far along is $200$ ? It is $200 - 110 = 90$ degrees past $110$ .
3. As a fraction of the stretch:
$\frac{90}{240} = \frac{3}{8}.$
4. On the Breadus scale the same stretch goes from $0$ to $100$ . Three eighths of the way is
$\frac38 \times 100 = 37.5.$

The answer is $\boxed{\textbf{(D)}\ 37.5}$ .

Why this works

Two matching pairs of readings fix a straight-line scale completely. The easiest way to use them is to ask "what fraction of the way from the low mark to the high mark is this?", because that fraction is the same on both scales. This is exactly how Celsius and Fahrenheit are converted.

Alternative approach

Find the rate. $240$ Fahrenheit degrees match $100$ Breadus degrees, so each Fahrenheit degree is worth $\frac{100}{240} = \frac{5}{12}$ of a Breadus degree. Starting from $110^\circ$ F (Breadus $0$ ) and going up $90$ Fahrenheit degrees adds $90 \times \frac{5}{12} = 37.5$ Breadus degrees.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The two scales are related linearly, so equal fractions of the corresponding intervals match. In Fahrenheit, the interval from $110$ to $350$ has length $240$ , and $200$ sits
$\frac{200 - 110}{240} = \frac{90}{240} = \frac{3}{8}$
of the way along it. On the Breadus scale the same interval runs from $0$ to $100$ , so the reading is $\frac38 \cdot 100 = 37.5$ .

Equivalently, the slope is $\frac{100}{240} = \frac{5}{12}$ Breadus degrees per Fahrenheit degree, and $B = \frac{5}{12}(F - 110)$ gives $B = \frac{5}{12} \cdot 90 = 37.5$ .

The answer is $\boxed{\textbf{(D)}\ 37.5}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 16: Functions
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 7: Proportion (§7.4)