Ruth has a beaker containing a solution of 800 mL of acid and 200 mL of water. She thinks the solution is a little strong, so she drains 100 mL from the beaker, adds 100 mL of water, and stirs the solution. Ruth thinks the solution is still too strong, so again she drains 100 mL from the beaker, and adds 100 mL of water. How many mL of water are now in the beaker?

Since there is 4 times as much acid as there is water in the original solution, there will be 80 ml of acid and 20 ml of water in the 100 ml drained so the new amounts are 720 ml acid and 180 ml water. Adding 100 ml water makes it 720 ml acid and 280 ml water. When she again drains 100 ml of solution there will be 76 and 2/3 ml acid and 23 and 1/3 ml water so the new amounts are 693 and 1/3 ml acid and 256 and 2/3 ml water. Adding 100 ml of water makes it 693 and 1/3 ml acid and 356 and 2/3 ml water.

 At the end, 352mL water

Step-by-step explanation:  The original solution is 80% acid, 20% water. The first 100 mL drain leaves 720 acid, 180 water.

Adding 100mL water results in 1000mL with 720acid, 280 water, now 72% acid, 28% water.  

The second drain takes 72mL acid and 28mL water leaving 648mL acid and 252mL water (900 mL)

The second Addition of 100mL water results in a 1000mL solution with 648mL acid and 352mL water.  ( 64.8% acid,  35.2% water)

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(a) the volume of the cone is 20/3 in3

hence, option a is correct.