How much pure antifreeze must be added to 12 L of a 40% solution to obtain a 60% solution. (Remember that pure antifreeze is 100% = 1.) What is the correct question to solve this problem?

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How much pure antifreeze must be added to 12 L of a 40% solution to obtain a 60% solution. (Remember that pure antifreeze is 100% = 1.) What is the correct question to solve this problem?

Mathematics
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add 6 L
10.8/18
let the amount added be x litres 12 L of a 40 % solution has 4.8 litres of antifreeze (4.8+x)/ (12+x) *100=60 4.8+x=0.6(12+x) 4.8+x=7.2+0.6x 0.4x=2.4 x=6 litres

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Other answers:

6 litres
x + 12 = y x + .40(12) = .60y
Satellite, congratulations on getting 10,000 medals
\[.4\times 12+x=.6(12+x)\] solve for x thank you!!!
Wow, that's cool! 10000 medals!!!
kind of silly, huh?
Yup. You're officially the Godfather of OS
What's silly?
yeah, but haven't moved up a level since august
lol
You can't right?
really. i think it is impossible. no matter though.
For users who are below lvl 100, the bar on the right corner shows how far you are from the next lvl. What does it look like when you're lvl 100? Can you post the picture satellite, just curious.....
satellite do you work here? How long has it taken you to get so many medals?!
He's not a mod....
Satellite was here before OS was OS as we know it.
Satellite, can you post the pic please....
Where's muffin?
What muffin?

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