Alcohol solution problem

Question

You need 900 mL of a 30% alcohol solution. On hand, you have a 20% alcohol mixture and a 80% alcohol mixture. How much of each mixture will you need to obtain the desired solution?

Answer 1

20% alcohol mixture = 0.2

80% alcohol mixture = 0.8

30% alcohol mixture = 0.3

total volume = 900 ml

Let x represent the volume of 20% alcohol mixture

Let (900ml - x) represent the volume of the 80% mixture

0.2x + [0.8(900-x)] = 0.3(900)

0.2x + 720 - 0.8x = 270

-0.6x = -450

X = 750

So you need 750ml of 20% alcohol mixture

900 - 750 = 150ml of the 80% alcohol mixture

to get 900ml of 30% alcohol mixture

Answer 2

You will need 750 mL of 20% solution plus 150 mL of 80% solution.

Let X = mL needed of 20% solution

Let Y = mL needed of 80% solution

X + Y = 900

.2X + .8Y = .3(900)

solve simultaneously to obtain X = 750 and Y = 150