2023 AMC 10A Problem 8

"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 $\boxed{\textbf{(D)}\ 37.5}$.

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.

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.

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 $\boxed{\textbf{(D)}\ 37.5}$.