SOLUTION: You need 390 mL of a 80% alcohol solution. On hand, you have a 70% alcohol mixture. How much of the 70% alcohol mixture and pure alcohol will you need to obtain the desired solutio

Question 806217: You need 390 mL of a 80% alcohol solution. On hand, you have a 70% alcohol mixture. How much of the 70% alcohol mixture and pure alcohol will you need to obtain the desired solution?

Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
x+y=390,
.7x+y=.8*390
x=260, y=130
check
.7*260+130=.8*390
.7*260+130=312
182+130=312
ok
Let A denote the 100% solution and B denote the 70% solution which we have.
Let C denote the 80% solution which we want.
Using the method of alligation.
10= 80 - 70
20= 100 - 80
E.g. There are 10 parts of A plus 20 parts of B
Then 10+20=30
0.33333333*390=130
130 mL of 100% solution.
390-130=260
260 mL of 70% solution.
check
0.7 * 260 + 1 * 130 = 0.8*390
182.0 + 130 = 312.0
312.0 = 312.0
ok

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