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

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(Solution 1 (Substitution))
 
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or every real number x, define bxc to be equal to the greatest integer less than
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==Problem==
or equal to x. (We call this the “floor” of x.) For example, b4.2c = 4, b5.7c = 5,
+
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?
b−3.4c = −4, b0.4c = 0, and b2c = 2.
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(a) Determine the integer equal to �
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<math>\textbf{(A) }33\qquad\textbf{(B) }34.5\qquad\textbf{(C) }36\qquad\textbf{(D) }37.5\qquad\textbf{(E) }39</math>
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==Solution 1 (Substitution)==
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+
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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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2
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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>
3
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3
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~Technodoggo
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+ . . . +
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==Solution 3 (Intuitive)==
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59
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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>
3
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~MercilessAnimations
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==Solution 4==
60
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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>.
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.
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~Dilip
(The sum has 60 terms.)
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-missmango
(b) Determine a polynomial p(x) so that for every positive integer m > 4,
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~Minor edits by FutureSphinx
bp(m)c =
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==Solution 5==
1
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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>.
3
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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
3
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\end{bmatrix}, let \: B = \begin{bmatrix}
3
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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 \
m − 2
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b
3
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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}
m − 1
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-1/240 & 1/240 \
3
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35/24 & -11/24
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\end{bmatrix}\cdot\begin{bmatrix}
(The sum has m − 1 terms.)
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0 \
A polynomial f(x) is an algebraic expression of the form
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100
f(x) = anx
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\end{bmatrix}=\begin{bmatrix}
n + an−1x
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5/12 \
n−1 + · · · + a1x + a0 for some integer n ≥ 0 and for
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-275/6
some real numbers an, an−1, . . . , a1, a0.
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\end{bmatrix}.</math>
(c) For each integer n ≥ 1, define f(n) to be equal to an infinite sum:
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Therefore, <math>m = 5/12 \: and \: b = -275/6</math>.
f(n) =
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Plugging in <math>t = 200</math>, we get <math>(5/12)200+(-275/6) = \boxed{\textbf{(D) }37.5}</math>.
n
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1
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~Loquacious Autodidact
2 + 1�
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+
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==Video Solution by Little Fermat==
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https://youtu.be/h2Pf2hvF1wE?si=rxjQH1lLtTftMj9a&t=1428
2n
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~little-fermat
2
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==Video Solution by Math-X (First understand the problem!!!)==
2 + 1�
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https://youtu.be/GP-DYudh5qU?si=prG8ONR_AgTR4HkL&t=1683 ~Math-X
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
3n
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3
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https://www.youtube.com/watch?v=dfrF_P-FIEA
2 + 1�
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==Video Solution==
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4n
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https://youtu.be/bYzV5B425V4
4
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2 + 1�
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution (easy to digest) by Power Solve==
5n
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https://www.youtube.com/watch?v=Yi5p3_x9iU8
5
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2 + 1�
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==See Also==
+ · · ·
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{{AMC10 box|year=2023|ab=A|num-b=7|num-a=9}}
(The sum contains the terms �
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{{MAA Notice}}
kn
 
k
 
2 + 1�
 
for all positive integers k, and no other
 
terms.)
 
Suppose f(t + 1) − f(t) = 2 for some odd positive integer t. Prove that t is a
 
prime number.
 

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 (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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