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? | + | ==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> | ||
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− | Solution 1 | + | ==Solution 1 (Substitution)== |
− | To solve this question, you can use y = mx + b where the x is | + | To solve this question, you can use <math>y = 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> |
~walmartbrian | ~walmartbrian | ||
+ | |||
+ | ==Solution 2 (Faster)== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~Technodoggo | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ~MercilessAnimations | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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>. | ||
+ | <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 | ||
+ | -missmango | ||
+ | ~Minor edits by FutureSphinx | ||
+ | |||
+ | ==Solution 5== | ||
+ | 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>. | ||
+ | ~MC413551 | ||
+ | ==Solution 6 (extremely simple)== | ||
+ | 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. | ||
+ | From the problem, <math>110m + 1b = 0</math> and <math>350m + 1b = 100.</math> | ||
+ | We can rewrite this problem in terms of linear algebra to solve it. | ||
+ | |||
+ | <math>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}.</math> | ||
+ | We can write the system of equations as Ax = B. | ||
+ | We can solve for x using the expression x = <math>A^{-1}B</math>. | ||
+ | Calculating this value we get <math>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}.</math> | ||
+ | Therefore, <math>m = 5/12 \: and \: b = -275/6</math>. | ||
+ | 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 Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/cMgngeSmFCY?si=88mPIms6wdZ6-deq&t=1602 ~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 09:14, 17 August 2024
Contents
- 1 Problem
- 2 Solution 1 (Substitution)
- 3 Solution 2 (Faster)
- 4 Solution 3 (Intuitive)
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (extremely simple)
- 8 Video Solution by Math-X (First understand the problem!!!)
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 Video Solution
- 11 Video Solution (easy to digest) by Power Solve
- 12 See Also
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 degrees Fahrenheit, which is
degrees on the Breadus scale. Bread is baked at
degrees Fahrenheit, which is
degrees on the Breadus scale. Bread is done when its internal temperature is
degrees Fahrenheit. What is this in degrees on the Breadus scale?
Solution 1 (Substitution)
To solve this question, you can use where the
is Fahrenheit and the
is Breadus. We have
and
. We want to find the value of
in
that falls on this line. The slope for these two points is
;
. Solving for
using
,
. We get
. Plugging in
. Simplifying,
~walmartbrian
Solution 2 (Faster)
Let denote degrees Breadus. We notice that
is
degrees to
, and
to
. This ratio is
; therefore,
will be
of the way from
to
, which is
~Technodoggo
Solution 3 (Intuitive)
From to
degrees Fahrenheit, the Breadus scale goes from
to
.
to
degrees is a span of
, and we can use this to determine how many Fahrenheit each Breadus unit is worth.
divided by
is
, so each Breadus unit is
Fahrenheit, starting at
Fahrenheit. For example,
degree on the Breadus scale is
, or
Fahrenheit. Using this information, we can figure out how many Breadus degrees
Fahrenheit is.
is
, so we divide
by
to find the answer, which is
~MercilessAnimations
Solution 4
We note that the range of F temperatures that
represents is
.
is
along the way to getting to
, the end of this range, or
of the way. Therefore if we switch to the Br scale, we are
of the way to
from
, or at
.
~Dilip -missmango ~Minor edits by FutureSphinx
Solution 5
We have the points and
. We want to find
. The equation of the line is
. We use this to find
, or
.
~MC413551
Solution 6 (extremely simple)
We can write the value on the Breadus scale as
, where
is the temperature in Fahrenheit.
From the problem,
and
We can rewrite this problem in terms of linear algebra to solve it.
We can write the system of equations as Ax = B.
We can solve for x using the expression x =
.
Calculating this value we get
Therefore,
.
Plugging in
, we get
.
~Loquacious Autodidact
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=88mPIms6wdZ6-deq&t=1602 ~Math-X
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=dfrF_P-FIEA
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.