Question

Show an equation and a solution. 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.)

Answer 1

Number of liters of pure antifreeze to be added (x):
0.4(12) + x = 0.6(12 + x)
2(12) + 5x = 3(12 + x)
24 + 5x = 36 + 3x
2x = 12
x = 6

Answer: 6 liters of pure antifreeze must be added.

Answer 2

Proof new mixture is 60% solution:
= 100%([0.4{12 liters} + 6 liters]/[12 liters + 6 liters])
= 100%([4.8 liters + 6 liters]/18 liters)
= 100%(10.8 liters/18 liters)
= 100%(0.6) = 60%

Answer 3

at the moment you have 40% of 12 L = \(.4\times 12=4.8\) L of antifreeze
add say \(x\) amount of pure antifreeze,
then you have \(4.8+x\) liters of a antifreeze and \(12+x\) liters of solution
you want this \(12+x\) liters to be 60% antifreeze, so set
\(4.8+x=.6(12+x)\) and solve for \(x\)
probably easiest to get rid of the decimals by multiplying both sides by 10 to get
\[48+10x=6(12+x)\]
\[48+10x=72+6x\]
\[4x=24\]
\[x=6\]

Answer 4

same answer and almost same method as jackellyn wrote above, except for line number 2 where she went from
\[.4\times 12+x=.6(12+x\] to
\[2\times 12+5x=3(12+x)\] where the step was multiplication by 5

Answer 5

thnks!!!