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

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Let <math>W, S,</math> and <math>L</math> represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that <math>W=4S</math> and <math>S=2L</math>. Since <math>L=3</math>, it follows that <math>S=6</math>, which in turn implies that <math>W=24 \implies\boxed{\textbf{(E) }24}</math>.<br>
 
Let <math>W, S,</math> and <math>L</math> represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that <math>W=4S</math> and <math>S=2L</math>. Since <math>L=3</math>, it follows that <math>S=6</math>, which in turn implies that <math>W=24 \implies\boxed{\textbf{(E) }24}</math>.<br>
 
~ junaidmansuri
 
~ junaidmansuri
 +
 +
==Solution 3==
 +
We have that <math>\textsl{lemonade} : \textsl{water} : \textsl{lemon juice} = 8 : 2 : 1,</math> so we have <math>3 \cdot 8 = \boxed{\textbf{(E) }24}.</math>
  
 
==See also==
 
==See also==
 
{{AMC8 box|year=2020|before=First problem|num-a=2}}
 
{{AMC8 box|year=2020|before=First problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:25, 18 November 2020

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

Luka will need $3\cdot 2=6$ cups of sugar and $6\cdot 4=24$ cups of water. The answer is $\boxed{\textbf{(E) } 24}$.

Solution 2

Let $W, S,$ and $L$ represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that $W=4S$ and $S=2L$. Since $L=3$, it follows that $S=6$, which in turn implies that $W=24 \implies\boxed{\textbf{(E) }24}$.
~ junaidmansuri

Solution 3

We have that $\textsl{lemonade} : \textsl{water} : \textsl{lemon juice} = 8 : 2 : 1,$ so we have $3 \cdot 8 = \boxed{\textbf{(E) }24}.$

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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