Difference between revisions of "2020 AMC 8 Problems/Problem 1"
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==Solution 4== | ==Solution 4== | ||
− | We are given that <math>4w:s</math> and <math>2s=l</math> which we combine to get <math>8w:2s:l</math>. Letting all the variables equal <math>3</math>, we find that the answer is <math>3\cdot 8=\textbf{(E)} | + | We are given that <math>4w:s</math> and <math>2s=l</math> which we combine to get <math>8w:2s:l</math>. Letting all the variables equal <math>3</math>, we find that the answer is <math>3\cdot 8=\boxed{\textbf{(E) }24} 24</math>. |
-franzliszt | -franzliszt | ||
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==Solution 5== | ==Solution 5== | ||
Revision as of 19:54, 19 November 2020
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?
Contents
Solution 1
Luka will need cups of sugar and cups of water. The answer is .
Solution 2
Let and represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that and . Since , it follows that , which in turn implies that .
~junaidmansuri
Solution 3
We have that so we have
[pog]
Solution 4
We are given that and which we combine to get . Letting all the variables equal , we find that the answer is .
-franzliszt
Solution 5
Put the numbers in ratios and when w = water, s = sugar, and lj = lemon juice. then since we know there is cups of lemon juice, do the math.
~ bsu1
Video Solution
~savannahsolver
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.
we just get 4*3*2=24$ which is E -oceanxia