2020 AMC 8 Problems/Problem 1

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

We have that $\text{lemonade} : \text{water} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1$ , 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

$The ratio is Water:Sugar:Lemon juice, or$ 4:2:1 $. Since we know that Luka used 3 cups of lemon juice, he needs$ 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!

~ $PrincessBelle$

Video Solution by WhyMath

https://youtu.be/FPC792h-mGE

~savannahsolver

Video Solution

https://youtu.be/eSxzI8P9_h8

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=34

~Interstigation

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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