Difference between revisions of "2023 AMC 10A Problems/Problem 8"

Revision as of 08:18, 10 November 2023

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 (Substitution)

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 $\frac{5}{12}$ ; $y = \frac{5}{12}x + b$ . Solving for $b$ using $(110, 0)$ , $\frac{550}{12} = -b$ . We get $b = \frac{-275}{6}$ . Plugging in $(200, y), \frac{1000}{12}-\frac{550}{12}=y$ . Simplifying, $\frac{450}{12} = \boxed{\textbf{(D) }37.5}$

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

~Technodoggo

Solution 3 (Intuitive)

From $110$ to $350$ degrees Fahrenheit, the Breadus scale goes from $1$ to $100$ . $110$ to $350$ degrees is a a span of $240$ , and we can use this to determine how many Fahrenheit each Breadus unit is worth. $240$ divided by $100$ is $2.4$ , so each Breadus unit is $2.4$ Fahrenheit, starting at $110$ Fahrenheit. For example, $1$ degree on the Breadus scale is $110 + 2.4$ , or $112.4$ Fahrenheit. Using this information, we can figure out how many Breadus degrees $200$ Fahrenheit is. $200-110$ is $90$ , so we divide $90$ by $2.4$ to find the answer, which is $\boxed{\textbf{(D) }37.5}$

~MercilessAnimations

Solution 4

We note that the range of F temperatures that 0-100 Br represents is 350-110 = 240 degF 200degF is (200-110) = 90 degF along the way to getting to 240 degF, the end of this range, or 90/240 = 9/24 = 3/8 = .375 of the way Therefore if we switch to the Br scale, we are .375 of the way to 100 from 0, or at $\boxed{\textbf{(D) 37.5}}$ degBr

~Dilip

Revision as of 03:18, 10 November 2023 (view source) Dilip (talk \| contribs) (→‎Solution 3 (Intuitive)) ← Older edit		Revision as of 08:18, 10 November 2023 (view source) Andyluo (talk \| contribs) m (→‎Solution 1) Newer edit →
Line 6:		Line 6:


−	==Solution 1==	+	==Solution 1 (Substitution)==

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

2023 AMC 10A (Problems • Answer Key • Resources)
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

Page

Toolbox

Search