Difference between revisions of "2020 AMC 8 Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
Let <math>W, S,</math> and <math>L</math> represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that <math>W=4S</math> and <math>S=2L</math>. Since <math>L=3</math>, it follows that <math>S=6</math>, which in turn implies that <math>W=24 \implies\boxed{\textbf{(E) }24}</math>.<br> | Let <math>W, S,</math> and <math>L</math> represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that <math>W=4S</math> and <math>S=2L</math>. Since <math>L=3</math>, it follows that <math>S=6</math>, which in turn implies that <math>W=24 \implies\boxed{\textbf{(E) }24}</math>.<br> | ||
| − | + | <a href="url">link text</a> | |
==Solution 3== | ==Solution 3== | ||
Revision as of 18:17, 18 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?
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 
.
<a href="url">link text</a>
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
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. 
