Question

Determine the amounts of 20% and 50% salt solutions that should be mixed to obtain 300 gallons of 41% salt solution

Please answer

Answer 1

Let x be the number of gallons of 20% solution to be mixed.
Then the amount of 50% solution to be mixed must be 300-x.
.20 (x) + .50 (300-x) = .41 (300)
.2x + 150 - .5x = 123
-.3x = - 27
x = 90 gallons of 20%
300-x = 210 gallons of 50%

Hope this helps!! (If not I'm sorry!)