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 $150$ mL from the beaker, adds $150$ mL of water, and stirs the solution. Ruth thinks the solution is still too strong, so again she drains $250$ mL from the beaker, and adds $250$ mL of water. How many mL of water are now in the beaker?

Answer 1

After the first step, the solution is $200-150=50$ mL too strong, so there are $200-150=50$ mL too many mL of acid. Since $200$ mL of acid diluted in $1000$ mL of total solution is $20\%$ acid, each $1$ mL of acid contributes $\frac{20}{200}=0.1\%$ acid. Thus, $50$ mL contributes $0.1\times 50=5\%$ too much acid. Thus, the solution is now $(100-5)\%=95\%$ acid and $(100-95)\%=5\%$ water, and so has $\frac{95}{5}=19$ times as many mL of acid as water.

The second time, Ruth drains $250$ mL, which is $19\times 250=4750$ mL of acid, and adds $250$ mL, or $5\%$ of the water volume. Say the new mixture has $V$ mL of water, so $\frac{5}{V}=\frac{4750}{1000-V}$. If we cross-multiply, we find that $\frac{5}{V}=\frac{475}{100-V}=\frac{100(4.75)}{100-V}=\frac{\frac{100}{V}(P)}{\frac{V-100}{V}(Q)}=\frac{P}{Q}$ Solving, we find $P=\frac{475}{4.75}=\frac{500}{5}=100 = Q$. Therefore, $Q=V-100$ so $V-100=100$ and $V=\boxed{200}$.