Difference between revisions of "2023 AMC 10A Problems/Problem 8"
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==See Also== | ==See Also== | ||
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Revision as of 14:08, 10 November 2023
Contents
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 the Fahrenheit and the is the Breadus. We have and . We want to find . 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 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 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 degBr
~Dilip
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=88mPIms6wdZ6-deq&t=1602 ~Math-X
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.