how many liters of 10% alcohol solution must be mixed with 40 liters of a 50% solution to get a 40% solution?

Question

need a math persons help!!!

Answer 1

We need to add x liters of 10% solution to 40 liters of 50% to get (40 + x) liters of 40%.

x*(10/100) + 40*(50/100) = (40 + x) * (40/100)
Turn the fractions into decimals:
.10x + 40(.5) = .40(40 + x)
Multiply through the parenthesis:
.10x + 20 = 16 + .40x
Subtract 16 from each side:
.10x + 4 = .40x
Subtract .10x from each side:
4 = .30x
Divide each side by .30:
4/.30 = x
You can solve right from here, but I would multiply the top and bottom by 10 to get a cleaner fraction:
40/3 = x
x = 13 1/3 liters (or 13.33 if you need a decimal answer)


To check this out, 13 1/3 liters of 10% solution added to the 40 liters of 50% solution should make 53 1/3 liters of 40% solution.
13 1/3 * 10/100 = 4 / 3 = 1 1/3
40 * 50/100 = 20
So you have a total of 21 1/3 liters of alcohol

53 1/3 * 40/100 = 64/3 = 21 1/3

This can be done using decimals also, but I find a fraction of 1/3 to be much cleaner than .3333333333

Answer 2

let amount of10% be x
then total solution is 40+x
so
(10/100)*x+(50/100)*40
=(40/100)*(40+x)
solve this for x

Answer 3

this would actually be more of a chemistry question. you have to use a certain formula for it. and i dont know it off the top of my head.