2020 AMC 8 Problems/Problem 1: Difference between revisions
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==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)}\ 24</math>. | |||
-franzliszt | |||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|before=First problem|num-a=2}} | {{AMC8 box|year=2020|before=First problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:57, 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 
.
~ 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
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 | ||
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