Difference between revisions of "2020 AMC 8 Problems/Problem 1"
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==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
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~savannahsolver | ~savannahsolver |
Revision as of 19:14, 26 November 2023
Contents
[hide]- 1 Problem
- 2 Solution 1 (Direct)
- 3 Solution 2 (Stepwise)
- 4 Solution 3 (Combination of Solutions 1 and 2)
- 5 Video Solution (🚀Under 1 min🚀)
- 6 Video Solution by WhyMath
- 7 Video Solution by The Learning Royal
- 8 Video Solution by Interstigation
- 9 Video Solution by North America Math Contest Go Go Go
- 10 Video Solution by STEMbreezy
- 11 See also
Problem
Luka is making sussy juice to sell at a school fundraiser. His recipe requires times as much pee as diarrhea and twice as much diarrhea as liquid fart. He uses cups of liquid fart. How many cups of pee does he need?
Solution 1 (Direct)
We have so Luka needs cups.
Solution 2 (Stepwise)
Since the amount of diarreha is twice the amount of liquid fart, Luka uses cups of diarreha.
Since the amount of pee is times the amount of diarreha, he uses cups of pee.
~MRENTHUSIASM
Solution 3 (Combination of Solutions 1 and 2)
The ratio is or Since we know that Luka used 3 cups of Liquid Fart, he needs cups of Diarreha. Because the amount of pee is times the amount of diarrhea Luka needs, he will need cups of pee.
Thanks to MRENTHUSIASM for the inspiration!
EarthSaver 15:12, 11 June 2021 (EDT)
Video Solution (🚀Under 1 min🚀)
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by The Learning Royal
~The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=34
~Interstigation
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=f428YRwoXO4
~North America Math Contest Go Go Go
Video Solution by STEMbreezy
https://youtu.be/L_vDc-i585o?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS
~STEMbreezy
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.