How much liquid (in mL) was originally in the beaker with the 20% solution and how much liquid (in mL) was originally in the beaker with 15% solution?

Question

You have one beaker which contains an unknown amount of a 15% iodine solution and another beaker which contains an unknown amount of a 20% iodine solution. You add only half of the 20% solution to the 15% solution, and the mixture becomes an 18% solution. Then you add 50mL of the 10% solution to the whole mixture. After this, the whole entire mixture of the 15% solution, half of the 20% solution, and 50 mL of a 10% solution becomes a 14% solution.

Answer 1

Let x = the amount in ml of 15% iodine solution (containing 85% water)

Let y = the amount in ml of 20% iodine solution (containing 80% water)

0.15x + 1/2*(0.20y) = 0.18*(x + y/2) which simplifies to y = 3x

Then, when adding 50ml of the 10% solution to the 18% new mixture to get a 14% solution:

0.18*(x + y/2) + 0.10*50 = 0.14*(x + y/2 + 50)

0.18x + 0.09y + 5 = 0.14x + 0.07y + 7

0.04x + 0.02y = 2 and multiply by 50 to simplify:

2x + y = 100 and then substitute 3x for y to get:

2x + 3x = 100

5x = 100

x = 20

So y = 3x = 60

Now, to check you answer:

20*0.15 = 3 plus 17 water

(60/2)*0.20 =6 plus 24 water

gives you 9 iodine plus 41 water (which is 18% iodine)

then add 50 of the 10% solution (meaning 5 iodine and 45 water) to the above mixture to get:

14 iodine plus 86 water (which is 14% iodine)

Good luck.

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Thank you.

Ira S.