Difference between revisions of "2020 AMC 8 Problems/Problem 1"
MRENTHUSIASM (talk | contribs) m (→Solution 1: The ratio should be water : sugar : lemonjuice) |
MRENTHUSIASM (talk | contribs) m (→Solution 3: Thank you for giving me credit. Note that Water:Sugar:Lemon juice should be 8:2:1, as shown in Solution 1.) |
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==Solution 3== | ==Solution 3== | ||
− | The ratio is Water:Sugar:Lemon juice, or <math> | + | The ratio is Water:Sugar:Lemon juice, or <math>8:2:1</math>. Since we know that Luka used 3 cups of lemon juice, he needs <math>3\cdot2=6</math> cups of sugar. Because the amount of water is <math>4</math> times the amount of sugar Luka needs, he will need <math>6\cdot4=\boxed{\textbf{(E) }24}</math> cups of water. |
Thanks to <math>MRENTHUSIASM</math> for the inspiration! | Thanks to <math>MRENTHUSIASM</math> for the inspiration! |
Revision as of 14:43, 9 June 2021
Contents
Problem
Luka is making lemonade to sell at a school fundraiser. His recipe requires times as much water as sugar and twice as much sugar as lemon juice. He uses cups of lemon juice. How many cups of water does he need?
Solution 1
We have that , so Luka needs cups.
Solution 2 (Stepwise)
Since the amount of sugar is twice the amount of lemon juice, Luka uses cups of sugar.
Since the amount of water is times the amount of sugar, he uses cups of water.
~MRENTHUSIASM
Solution 3
The ratio is Water:Sugar:Lemon juice, or . Since we know that Luka used 3 cups of lemon juice, he needs cups of sugar. Because the amount of water is times the amount of sugar Luka needs, he will need cups of water.
Thanks to for the inspiration!
~
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=34
~Interstigation
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.