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

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<math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math>
 
<math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math>
  
==Solution 1==
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==Solution==
Luka will need <math>3\cdot 2=6</math> cups of sugar, and thus <math>6\cdot 4=24</math> cups of water. The answer is <math>\boxed{\textbf{(E) } 24}</math>.
 
 
 
==Solution 2==
 
 
We have that <math>\text{lemonade} : \text{water} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1</math>, so Luka needs <math>3 \cdot 8 = \boxed{\textbf{(E) }24}</math> cups.
 
We have that <math>\text{lemonade} : \text{water} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1</math>, so Luka needs <math>3 \cdot 8 = \boxed{\textbf{(E) }24}</math> cups.
  
 
==Video Solution==
 
==Video Solution==
https://youtu.be/FPC792h-mGE
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https://youtu.be/eSxzI8P9_h8
  
 
==See also==
 
==See also==
{{AMC8 box|year=2020|before=First problem|num-a=2}}
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{{AMC8 box|year=2020|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:20, 9 January 2021

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

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.

Video Solution

https://youtu.be/eSxzI8P9_h8

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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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