2023 AMC 10A Problems/Problem 8

Revision as of 20:49, 9 November 2023 by Andyluo (talk | contribs) (Solution 2)

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?

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


Solution 1

To solve this question, you can use y = mx + b where the x is the Fahrenheit and the y is the Breadus. We have (110,0) and (350,100). We want to find (200,y). The slope for these two points is 5/12; y = 5x/12 + b. Solving for b using (110, 0), 550/12 = -b. We get b = -275/6. Plugging in (200, y), 1000/12-550/12=y. Simplifying, 450/12 = 37.5. $\boxed{(D)}$

~walmartbrian

Solution 2 (faster)

Let $^\circ B$ denote degrees Breadus. We notice that $200^\circ F$ is $90^\circ F$ degrees to $0^\circ B$, and $150^\circ F$ to $100^\circ B$. This ratio is $90:150=3:5$; therefore, $200^\circ F$ will be $\dfrac3{3+5}=\dfrac38$ of the way from $0$ to $100$, which is $\boxed{\text{(D) }37.5.}$

~Technodoggo

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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