Apharmacist has 30% and 60% iodine solutions on hand. how many liters of each iodine solution will be required to produce 6 liters of a 50% iodine.

4 liters of 60% solution; 2 liters of 30% solution

Step-by-step 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.

l.ln.lln.k.nkl

step-by-step explanation:

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.