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Difference between revisions of "2020 AMC 8 Problems/Problem 1"

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~Education, the Study of Everything
 
  
 
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Revision as of 16:52, 29 July 2023

Problem

Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water 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 $\text{water} : \text{sugar} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \cdot 8 = \boxed{\textbf{(E) }24}$ cups.

Solution 2 (Stepwise)

Since the amount of sugar is twice the amount of lemon juice, Luka uses $3\cdot2=6$ cups of sugar.

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

~MRENTHUSIASM

Solution 3 (Combination of Solutions 1 and 2)

The ratio is $\text{Water}:\text{Sugar}:\text{Lemon Juice},$ or $8:2:1.$ Since we know that Luka used 3 cups of lemon juice, he needs $3\cdot2=6$ cups of sugar. Because the amount of water is $4$ times the amount of sugar Luka needs, he will need $6\cdot4=\boxed{\textbf{(E) }24}$ cups of water.

Thanks to MRENTHUSIASM for the inspiration!

EarthSaver 15:12, 11 June 2021 (EDT)

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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