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 150 mL from the beaker, and adds 150 mL of water. How many mL of water are now in the beaker?

bader Dec 12, 2023

#1**0 **

Here's how to solve the problem:

Initial amount of water: There are 200 mL of water in the beaker initially.

Amount of water added after each drain: After each drain, 150 mL of water is added to the beaker.

Number of times the solution is drained: The solution is drained twice.

Therefore, the total amount of water in the beaker after both drains and additions is:

Total water = Initial water + Water added after drains - Water drained

Total water = 200 mL + 2 * 150 mL - 2 * 150 mL

Total water = 200 mL + 300 mL - 300 mL

Total water = 200 mL

Therefore, there are 200 mL of water in the beaker now.

BuiIderBoi Dec 12, 2023