Simon had some red and blue balls. There were 7 times as many red as blue balls. 60% of the balls in the box were red balls and the rest were blue balls. After he bought equal number of red and blue balls, of his balls were red. If he had 368 balls at first, how many balls did he buy?

Let's break down the problem step by step:

1. We know that Simon initially had 368 balls. According to the problem, 60% of these balls were red. So, we calculate 60% of 368 to find out the number of red balls.

   0.60 * 368 = 220.8

   Since the number of balls can't be a fraction, we round this to the nearest whole number, which is 221. So, Simon initially had 221 red balls.

2. The rest of the balls were blue. So, we subtract the number of red balls from the total number of balls to find out the number of blue balls.

   368 - 221 = 147

   So, Simon initially had 147 blue balls.

3. The problem states that Simon bought an equal number of red and blue balls. After he did this, 7/8 of his balls were red. So, we can set up an equation to find out how many balls Simon bought. Let's let x represent the number of balls Simon bought.

   (221 + x) / (368 + 2x) = 7/8

4. To solve for x, we first cross-multiply:

   8 * (221 + x) = 7 * (368 + 2x)

   1768 + 8x = 2576 + 14x

5. Then, we subtract 8x from both sides of the equation:

   1768 = 2576 + 6x

6. Next, we subtract 2576 from both sides:

   -808 = 6x

7. Finally, we divide both sides by 6 to solve for x:

   x = -808 / 6 = -134.67

   Since the number of balls can't be a fraction or negative, we round this to the nearest whole number, which is -135.

   However, it doesn't make sense for Simon to buy a negative number of balls. This suggests that there may be a mistake in the problem. The initial conditions (that there were 7 times as many red balls as blue balls and that 60% of the balls were red) seem to be inconsistent. Please check the problem and try again.