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 $300$ mL from the beaker, adds $300$ mL of water, and stirs the solution. Ruth thinks the solution is still too strong, so again she drains $400$ mL from the beaker, and adds $400$ mL of water. How many mL of water are now in the beaker?

Question

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 $300$ mL from the beaker, adds $300$ mL of water, and stirs the solution.  Ruth thinks the solution is still too strong, so again she drains $400$ mL from the beaker, and adds $400$ mL of water.  How many mL of water are now in the beaker?

GPT-4o mini · Answer

Initially, Ruth has a solution composed of $800$ mL of acid and $200$ mL of water, making the total volume of the solution:

\[ 800 \text{ mL (acid)} + 200 \text{ mL (water)} = 1000 \text{ mL} \]

Next, Ruth drains $300$ mL from the beaker. This removed liquid consists of both acid and water in the same proportion as the original solution.

The proportion of acid and water in the original solution is:

Volume of acid: $ \frac{800}{1000} = 0.8 $
Volume of water: $ \frac{200}{1000} = 0.2 $

Calculating the amount of acid and water in the $300$ mL that Ruth drains:

Acid removed: $ 300 \text{ mL} \times 0.8 = 240 \text{ mL} $
Water removed: $ 300 \text{ mL} \times 0.2 = 60 \text{ mL} $

After draining $300$ mL, the quantities left in the beaker are:

Remaining acid: $ 800 \text{ mL} - 240 \text{ mL} = 560 \text{ mL} $
Remaining water: $ 200 \text{ mL} - 60 \text{ mL} = 140 \text{ mL} $

The new total volume of the solution after draining is:

\[ 560 \text{ mL (acid)} + 140 \text{ mL (water)} = 700 \text{ mL} \]

Next, she adds $300$ mL of water:

\[ 140 \text{ mL} + 300 \text{ mL} = 440 \text{ mL (water)} \]

Now the total volume of the solution becomes:

\[ 560 \text{ mL (acid)} + 440 \text{ mL (water)} = 1000 \text{ mL} \]

After this, Ruth thinks the solution is still too strong, so she drains $400$ mL. Again, this $400$ mL consists of acid and water in the same proportions as currently present.

The proportion of acid and water in the $1000$ mL solution is:

Current volume of acid: $ \frac{560}{1000} = 0.56 $
Current volume of water: $ \frac{440}{1000} = 0.44 $

Calculating the amount of acid and water in the $400$ mL that Ruth drains:

Acid removed: $ 400 \text{ mL} \times 0.56 = 224 \text{ mL} $
Water removed: $ 400 \text{ mL} \times 0.44 = 176 \text{ mL} $

After draining $400$ mL, the quantities left in the beaker become:

Remaining acid: $ 560 \text{ mL} - 224 \text{ mL} = 336 \text{ mL} $
Remaining water: $ 440 \text{ mL} - 176 \text{ mL} = 264 \text{ mL} $

Now she adds $400$ mL of water:

\[ 264 \text{ mL} + 400 \text{ mL} = 664 \text{ mL (water)} \]

Thus, the final amount of water in the beaker is:

\[ \boxed{664} \text{ mL} \]