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

(Solution 1)
(Problem)
Line 1: Line 1:
==Problem==
 
Luka is making lemonade to sell at a school fundraiser. His recipe requires <math>4</math> times as much water as sugar and twice as much sugar as lemon juice. He uses <math>3</math> cups of lemon juice. How many cups of water does he need?
 
 
<math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math>
 
 
 
==Solution 2==
 
==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.

Revision as of 12:30, 5 December 2020

Solution 2

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

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

Invalid username
Login to AoPS