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Difference between revisions of "2023 AMC 10A Problems/Problem 8"
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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?
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==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 <math>110</math> degrees Fahrenheit, which is <math>0</math> degrees on the Breadus scale. Bread is baked at <math>350</math> degrees Fahrenheit, which is <math>100</math> degrees on the Breadus scale. Bread is done when its internal temperature is <math>200</math> degrees Fahrenheit. What is this in degrees on the Breadus scale?
  
 
<math>\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39</math>
 
<math>\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39</math>
  
 +
==Solution 1 (Substitution)==
  
 +
To solve this question, you can use <math>f(x) = mx + b</math> where the <math>x</math> is Fahrenheit and the <math>y</math> is Breadus. We have <math>(110,0)</math> and <math>(350,100)</math>. We want to find the value of <math>y</math> in <math>(200,y)</math> that falls on this line. 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>
  
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==Solution 1==
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~walmartbrian
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==Solution 2 (Faster)==
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 +
Let <math>^\circ B</math> denote degrees Breadus. We notice that <math>200^\circ F</math> is <math>90^\circ F</math> degrees to <math>0^\circ B</math>, and <math>150^\circ F</math> to <math>100^\circ B</math>. This ratio is <math>90:150=3:5</math>; therefore, <math>200^\circ F</math> will be <math>\dfrac3{3+5}=\dfrac38</math> of the way from <math>0</math> to <math>100</math>, which is <math>\boxed{\textbf{(D) }37.5}</math>
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 +
~Technodoggo
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 +
==Solution 3 (Intuitive)==
 +
 
 +
From <math>110</math> to <math>350</math> degrees Fahrenheit, the Breadus scale goes from <math>1</math> to <math>100</math>. <math>110</math> to <math>350</math> degrees is a span of <math>240</math>, and we can use this to determine how many Fahrenheit each Breadus unit is worth. <math>240</math> divided by <math>100</math> is <math>2.4</math>, so each Breadus unit is <math>2.4</math> Fahrenheit, starting at <math>110</math> Fahrenheit. For example, <math>1</math> degree on the Breadus scale is <math>110 + 2.4</math>, or <math>112.4</math> Fahrenheit. Using this information, we can figure out how many Breadus degrees <math>200</math> Fahrenheit is. <math>200-110</math> is <math>90</math>, so we divide <math>90</math> by <math>2.4</math> to find the answer, which is <math>\boxed{\textbf{(D) }37.5}</math>
  
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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. <math>\boxed{(D)}</math>
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~MercilessAnimations
  
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~walmartbrian
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==Solution 4==
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We note that the range of F temperatures that <math>0-100</math> <math>\text{Br}^\circ</math> represents is <math>350-110 = 240</math> <math>\text{F}^\circ</math>.
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<math>200</math> <math>\text{F}^\circ</math> is <math>(200-110) = 90</math> <math>\text{F}^\circ</math> along the way to getting to <math>240</math> <math>\text{F}^\circ</math>, the end of this range, or <math>90/240 = 9/24 = 3/8 = 0.375</math> of the way. Therefore if we switch to the Br scale, we are <math>0.375</math> of the way to <math>100</math> from <math>0</math>, or at <math>\boxed{\textbf{(D) }37.5}</math> <math>\text{Br}^\circ</math>.
 +
 
 +
~Dilip
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-missmango
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~Minor edits by FutureSphinx
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==Solution 5==
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We have the points <math>(0, 110)</math> and <math>(100, 350)</math>. We want to find <math>(x, 200)</math>. The equation of the line is <math>y=\frac{12}{5}x+110</math>. We use this to find <math>x=\frac{75}{2}=37.5</math>, or <math>\boxed{D}</math>.
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~MC413551
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==Solution 6 (extremely simple)==
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We can write the value <math>y</math> on the Breadus scale as <math>y = mt + b</math>, where <math>t</math> is the temperature in Fahrenheit.
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From the problem, <math>110m + 1b = 0</math> and <math>350m + 1b = 100.</math>
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We can rewrite this problem in terms of linear algebra to solve it.
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<math>Let \: A =\begin{bmatrix}
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110 & 1 \\
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350 & 1
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\end{bmatrix}, let \: B = \begin{bmatrix}
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0 \\
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100
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\end{bmatrix}, and \: let \: x = \begin{bmatrix}
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m \\
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b
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\end{bmatrix}.</math>
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We can write the system of equations as Ax = B.
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We can solve for x using the expression x = <math>A^{-1}B</math>.
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Calculating this value we get <math>x = \begin{bmatrix}
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-1/240 & 1/240 \\
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35/24 & -11/24
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\end{bmatrix}\cdot\begin{bmatrix}
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0 \\
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100
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\end{bmatrix}=\begin{bmatrix}
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5/12 \\
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-275/6
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\end{bmatrix}.</math>
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Therefore, <math>m = 5/12 \: and \: b = -275/6</math>.
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Plugging in <math>t = 200</math>, we get <math>(5/12)200+(-275/6) = \boxed{\textbf{(D) }37.5}</math>.
 +
 
 +
~Loquacious Autodidact
 +
 
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=rxjQH1lLtTftMj9a&t=1428
 +
~little-fermat
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=prG8ONR_AgTR4HkL&t=1683 ~Math-X
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 
 +
https://www.youtube.com/watch?v=dfrF_P-FIEA
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/bYzV5B425V4
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Video Solution (easy to digest) by Power Solve==
 +
https://www.youtube.com/watch?v=Yi5p3_x9iU8
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2023|ab=A|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Latest revision as of 15:29, 13 November 2024

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 $f(x) = mx + b$ where the $x$ is Fahrenheit and the $y$ is Breadus. We have $(110,0)$ and $(350,100)$. We want to find the value of $y$ in $(200,y)$ that falls on this line. 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 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 $0-100$ $\text{Br}^\circ$ represents is $350-110 = 240$ $\text{F}^\circ$. $200$ $\text{F}^\circ$ is $(200-110) = 90$ $\text{F}^\circ$ along the way to getting to $240$ $\text{F}^\circ$, the end of this range, or $90/240 = 9/24 = 3/8 = 0.375$ of the way. Therefore if we switch to the Br scale, we are $0.375$ of the way to $100$ from $0$, or at $\boxed{\textbf{(D) }37.5}$ $\text{Br}^\circ$.

~Dilip -missmango ~Minor edits by FutureSphinx

Solution 5

We have the points $(0, 110)$ and $(100, 350)$. We want to find $(x, 200)$. The equation of the line is $y=\frac{12}{5}x+110$. We use this to find $x=\frac{75}{2}=37.5$, or $\boxed{D}$. ~MC413551

Solution 6 (extremely simple)

We can write the value $y$ on the Breadus scale as $y = mt + b$, where $t$ is the temperature in Fahrenheit. From the problem, $110m + 1b = 0$ and $350m + 1b = 100.$ We can rewrite this problem in terms of linear algebra to solve it.

$Let \: A =\begin{bmatrix} 110 & 1 \\ 350 & 1 \end{bmatrix}, let \: B = \begin{bmatrix} 0 \\ 100 \end{bmatrix}, and \: let \: x = \begin{bmatrix} m \\ b \end{bmatrix}.$ We can write the system of equations as Ax = B. We can solve for x using the expression x = $A^{-1}B$. Calculating this value we get $x = \begin{bmatrix} -1/240 & 1/240 \\ 35/24 & -11/24 \end{bmatrix}\cdot\begin{bmatrix} 0 \\ 100 \end{bmatrix}=\begin{bmatrix} 5/12 \\ -275/6 \end{bmatrix}.$ Therefore, $m = 5/12 \: and \: b = -275/6$. Plugging in $t = 200$, we get $(5/12)200+(-275/6) = \boxed{\textbf{(D) }37.5}$.

~Loquacious Autodidact

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=rxjQH1lLtTftMj9a&t=1428 ~little-fermat

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=prG8ONR_AgTR4HkL&t=1683 ~Math-X

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=dfrF_P-FIEA

Video Solution

https://youtu.be/bYzV5B425V4

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution (easy to digest) by Power Solve

https://www.youtube.com/watch?v=Yi5p3_x9iU8

See Also

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

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