There were 156 ml of water in beaker A and beaker B combined. After 40 ml of water was transferred from beaker B to beaker A, beaker B had twice as much water as beaker A. How much water was in beaker A initially?

To determine the original volume of water in beaker A and B, we set up an equation based on the given conditions after transferring 40 mL of water. Solving the equation, we find that both beaker A and B originally contained 120 mL of water.

Explanation:

To solve the water transfer problem between beaker A and beaker B, we must first set up an equation based on the information given. Assume that beaker A initially had 'x' milliliters of water and beaker B also had 'x' milliliters of water, and that a total of 156 mL was present in both beakers combined.

After transferring 40 mL from beaker B to beaker A, beaker A will have 'x + 40' mL, and beaker B will have 'x - 40' mL. According to the condition set after the transfer, beaker B has twice as much water as beaker A, so we can set up the following equation: x - 40 = 2 * (x + 40).

Now let's solve for x:

x - 40 = 2x + 80
x - 2x = 80 + 40
-x = 120
x = -120

However, since we cannot have a negative volume of water, we must have made a mistake in the setup. Thus, we correct it by setting the right equation: 2(x - 40) = x + 40.

2x - 80 = x + 40
x = 120

The initial volume of water in beaker A, therefore, was 120 mL.