Answer:
4 liters of 60% solution; 2 liters of 30% solution
Explanation:
I like to use a simple, but effective, tool for most mixture problems. It is a kind of "X" diagram as in the attachment.
The ratios of solution concentrations are 3:6:5, so I've used those numbers in the diagram. The constituent solutions are on the left; the desired mixture is in the middle, and the numbers on the other legs of the X are the differences along the diagonals: 6 - 5 = 1; 5 - 3 = 2. This tells you the ratio of 60% solution to 30% solution is 2 : 1.
These ratio units (2, 1) add to 3. We want 6 liters of mixture, so we need to multiply these ratio units by 2 liters to get the amounts of constituents needed. The result is 4 liters of 60% solution and 2 liters of 30% solution.
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If you're writing equations, it often works well to let the variable represent the quantity of the greatest contributor—the 60% solution. Let the volume of that (in liters) be represented by v. Then the total volume of iodine in the mixture is ...
... 0.60·v + 0.30·(6 -v) = 0.50·6
... 0.30v = 0.20·6 . . . . subtract 0.30·6, collect terms
... v = 6·(0.20/0.30) = 4 . . . . divide by the coefficient of v
4 liters of 60% solution are needed. The other 2 liters are 30% solution.