Question: The salt content of two cups has nothing to do with the result. So you can assume the most extreme situation: Cup A is 100g pure water, and Cup B is 60g pure salt (salt was also a liquid at that time). The result calculated in this way is also correct, and this simplification is only for the convenience of calculation.

Then, we can take a big barrel and pour100g water and 60g salt into it for dissolution (assuming that the salt can be dissolved indefinitely and the total volume remains unchanged). So, now the concentration of brine is 60/( 100+60)=3/8.

Divide this brine into cup A and cup B, and the original requirement will be met: the concentration of brine in the two cups is the same. Then we have to calculate how much water in A reaches B. (The same amount of salt reaches A)

At present, there is 100g salt water in A's cup, and the concentration is 3/8, that is, the water content is100 * (5/8) = 62.5g, so the water poured out by A is100-62.5 = 37.5g..

Similarly, the salt poured out by B is 37.5 grams.