How many liters of a 10% silver iodide solution must be mixed with 3 L of a 4% silver iodine solution to get s 6% solution?

To answer this problem, you should just substitute some figures in order to get the required amount.

So, let a be the amount of 10% iodide solution

So ...

4 (3) + 10a = 6 (3+a)

[4 represents the silver iodine solution, the 3 is the liters, while we are looking for the 6% solution]

12 + 10a = 18 + 6a

10a - 6a = 18 - 12

4a = 6

a = 6/4

a = 3/2

So you will need 3/2 or 1.5 liters of 10% iodide solution.