Question

Please help!!! Will get brainliest!!! A 20% solution of fertilizer is to be mixed with a 80% solution of fertilizer in order to get 30 gallons of a 60% solution. How many gallons of the 20% solution and 80% solution should be mixed?

Answer 1

X:  volume of the 20% sol'n to be used in the mixture
y:  vol. of the 80% sol'n to be used in the mixture

Then x+y must equal 30 gallons.

.20x:  amt. of pure fertizer in the 20% sol'n
.80x:  amt of pure fert. in the 80% sol'n
.60(x+y) = amt of pure fert in the 60% sol'n

Then:

.20x + .80y = .60(x+y) = .60(30)
Since x+y must equal 30, y=30-x, and
.20(x) + .80(30-x) = 18
.20x + 24 - .80x = 18
6 = .60x
x = 10
Use 10 gallons of the 20% solution and 20 gallons of the 80% solution.
The result will be 30 gallons of a 60% solution.