Question 870136
One solution is 60% acid and another is 15% how much would be needed of each to make 300 L at 50% acid.
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let x=amt of 60% solution needed
300-x=amt of 15% solution needed
60%x+15%(300-x)=50%*300
.60x+45-.15x=150
.45x=105
x=105/.45
x≈233.33 L
300-x≈66.66 L
amt of 60% solution needed≈233.33 L
amt of 15% solution needed≈66.66 L