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==
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==Solution 1==
 
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.
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==Solution 2 (Stepwise)==
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Since the amount of sugar is twice the amount of lemon juice, Luka uses <math>3\cdot2=6</math> cups of sugar.
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Since the amount of water is <math>4</math> times the amount of sugar, he uses <math>6\cdot4=\boxed{\textbf{(E) }24}</math> cups of water.
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~MRENTHUSIASM
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==

Revision as of 22:45, 4 March 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 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

Video Solution by WhyMath

https://youtu.be/FPC792h-mGE

~savannahsolver

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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