Difference between revisions of "2020 AMC 8 Problems/Problem 1"
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+ | ==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? | 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> | <math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Direct)== |
− | + | We have <math>\text{water} : \text{sugar} : \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. | |
− | ==Solution 2== | + | ==Solution 2 (Stepwise)== |
− | + | Since the amount of sugar is twice the amount of lemon juice, Luka uses <math>3\cdot2=6</math> cups of sugar. | |
− | |||
− | + | 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. | |
− | |||
− | + | ~MRENTHUSIASM | |
− | ==Solution 4= | + | ==Solution 3 (Combination of Solutions 1 and 2)== |
+ | The ratio is <math>\text{Water}:\text{Sugar}:\text{Lemon Juice},</math> 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 MRENTHUSIASM for the inspiration! | ||
− | + | EarthSaver 15:12, 11 June 2021 (EDT) | |
− | + | ==Video Solution by WhyMath== | |
− | ==Solution | + | https://youtu.be/FPC792h-mGE |
+ | |||
+ | ~savannahsolver | ||
− | + | ==Video Solution by The Learning Royal== | |
+ | https://youtu.be/eSxzI8P9_h8 | ||
− | ~ | + | ~The Learning Royal |
− | ==Video Solution== | + | ==Video Solution by Interstigation== |
− | https://youtu.be/ | + | https://youtu.be/YnwkBZTv5Fw?t=34 |
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution by North America Math Contest Go Go Go== | ||
+ | https://www.youtube.com/watch?v=f428YRwoXO4 | ||
− | ~ | + | ~North America Math Contest Go Go Go |
==See also== | ==See also== | ||
− | {{AMC8 box|year=2020|before=First | + | {{AMC8 box|year=2020|before=First Problem|num-a=2}} |
{{MAA Notice}} | {{MAA Notice}} | ||
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− |
Latest revision as of 18:49, 11 January 2022
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 (Direct)
We have 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 (Combination of Solutions 1 and 2)
The ratio is 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 MRENTHUSIASM for the inspiration!
EarthSaver 15:12, 11 June 2021 (EDT)
Video Solution by WhyMath
~savannahsolver
Video Solution by The Learning Royal
~The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=34
~Interstigation
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=f428YRwoXO4
~North America Math Contest Go Go Go
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.