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2020 AMC 8 Problems/Problem 1

Revision as of 19:14, 26 November 2023 by Pommukutti (talk | contribs) (→‎Solution 3 (Combination of Solutions 1 and 2))

Problem

Luka is making sussy juice to sell at a school fundraiser. His recipe requires $4$ times as much pee as diarrhea and twice as much diarrhea as liquid fart. He uses $3$ cups of liquid fart. How many cups of pee does he need?

$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution 1 (Direct)

We have $\text{pee} : \text{diarreha} : \text{liquid fart} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \cdot 8 = \boxed{\textbf{(E) }24}$ cups.

Solution 2 (Stepwise)

Since the amount of diarreha is twice the amount of liquid fart, Luka uses $3\cdot2=6$ cups of diarreha.

Since the amount of pee is $4$ times the amount of diarreha, he uses $6\cdot4=\boxed{\textbf{(E) }24}$ cups of pee.

~MRENTHUSIASM

Solution 3 (Combination of Solutions 1 and 2)

The ratio is $\text{Pee}:\text{Diarreha}:\text{Liquid Fart},$ or $8:2:1.$ Since we know that Luka used 3 cups of Liquid Fart, he needs $3\cdot2=6$ cups of Diarreha. Because the amount of pee is $4$ times the amount of diarrhea Luka needs, he will need $6\cdot4=\boxed{\textbf{(E) }24}$ cups of pee.

Thanks to MRENTHUSIASM for the inspiration!

EarthSaver 15:12, 11 June 2021 (EDT)

Video Solution (🚀Under 1 min🚀)

https://youtu.be/T1K8irAOxRk

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/FPC792h-mGE

~savannahsolver

Video Solution by The Learning Royal

https://youtu.be/eSxzI8P9_h8

~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

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