How much 10%alcohol solution be mixed w/ 14ounces of 18% solution to get a 12% solution

.1x +.18(14) = .12(14+x)

OK, we have an unknown amount of 10% solution, mixed with 14 ounces of 18% to give us 12% solution in the amount of the 14 ounces plus what we added.

x=42

So we must add 42 ounces of 10% solution to make this mix work.

Is your teacher a Douglas Adams fan? Does he or she know where his or her towel is? (See ref 3 if interested.)

Let the amount of 10% alcohol solution used be x ounces

0.10x + 0.18(14) = ( x+14) *0.12

0.1x + 2.52 = 0.12x + 1.68

0.02x = 0.84

x = 0.84 / 0.02 = 84/2 = 42 ounces ANSWER

.18(14)=2.52

Let x=ounces of solution for the 10% one

.1x+2.52=.12(x+14)
.1x+2.52=.12x+1.68
2.52=.02x+1.68
.84=.02x
42=x

42 ounces

x = amout of 10% alcohol
Then .1x +.18(14) = .12(14+x)
.1x + 2.52 = 1.68 + .12x
.02x = .84
x = 42 ounces

Let x = ounces of 10% solution

14 * 0.18 + x * 0.10 = (14 + x) * 0.12 <=== solve for x

42 ounces

(.1Va+.18Vb)/(Va+Vb)=.12
do the math and get a ratio.

Oh didn't see you gave the volume of one of them.

(0.10(x) + 2.52)/(x + 14) = 3/25

3x + 42 = 2.5x + 63

0.5x = 21

x = 42 oz