75% of the balls in a bag were green.60% of the remainder were red and the rest were purple. There were 234 more green balls than purple balls. Some red balls were removed from the bag. Thereafter, 10% of the remaining balls in the bag were red. How many red balls were removed from the bag?

To solve this problem, we need to follow these steps:

1. Let's denote the total number of balls in the bag as X. According to the problem, 75% of the balls were green, so the number of green balls is 0.75X.

2. The remainder of the balls, which are either red or purple, is 25% of the total, so there are 0.25X red and purple balls.

3. Of these remaining balls, 60% were red, so the number of red balls is 0.60 * 0.25X = 0.15X.

4. The rest of the balls were purple, so the number of purple balls is 0.25X - 0.15X = 0.10X.

5. The problem states that there were 234 more green balls than purple balls, so we can set up the equation 0.75X = 0.10X + 234 and solve for X. This gives us X = 468.

6. So, initially there were 0.15 * 468 = 70.2, but since the number of balls must be a whole number, we round this to 70. So, there were initially 70 red balls in the bag.

7. After some red balls were removed, 10% of the remaining balls in the bag were red. Let's denote the number of remaining balls as Y and the number of removed balls as R. We know that Y = X - R and that 0.10Y = 70 - R.

8. Substituting the first equation into the second gives us 0.10(X - R) = 70 - R. Solving this equation for R gives us R = 42.

So, 42 red balls were removed from the bag.