Difference between revisions of "2023 AMC 10A Problems/Problem 8"
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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{\text{(D) }37.5.}</math> | 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{\text{(D) }37.5.}</math> | ||
Revision as of 20:49, 9 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 
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
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. 
~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
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. 
