Alcohol mixture problem

Question

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

Answer 1

Given 900ml of a 30% alcohol solution.

Therefore we have .30(900)ml or 270ml pure alcohol.

Let x = ml of the 20% alcohol mixture and

y = ml of the 80% alcohol mixture

So, therefore our first equation is x + y = 900

The amount of pure alcohol in the 20% mixture is .20(x) and

the amount in the 80% mixture is .80(y)

Because the amount of pure alcohol in the 30% solution (270ml) must equal the amount of pure alcohol in both the 20% and 80% mixtures we are able to write our second equation.

270 = .20x + .80y or .2x + .8y = 270

So we have x + y = 900 (1st equation)

and .2x +.8y = 270 (2nd equation)

To solve, multiply the first equation by -2 and the second by 10 producing

-2x - 2y = -1800

and 2x + 8y = 2700

Adding both we get, 6y = 900 or y = 150

To find x replace, y with 150 in the first equation and we get

x + 150 = 900

Subtract 150 from both sides and we have x = 750

Therefore we need 750ml of the 20% mixture and 150ml of the 80% mixture.

Answer 2

900(.3) = .2x + .8(900-x) = .2x + 720 -.8x

-.6x = 270-720

.6x = 450

x = 450/.6 = 4500/6 = 750 mL of 20%

900-x = 150 mL of 80%

30 is within 10 of 20

30 is within 50 of 80

50 is 5 times 10

you need 5 times as much of 20 as 80

divide 900 into 6 = 150 of 80% and 750 mL of 20% to get 900 mL of 30%