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?
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.
