Riordan, Ivan and Seth had some balls. Ivan had 25% more balls than Riordan. Ivan had ²₅ of Seth's. After Riordan gave 20 balls to Ivan, he had ¹₅ of what Ivan had. How many more balls did Seth have than Ivan in the end?

Riordan	Ivan	Seth
4x2	5x2
	2x5	5x5
8	10	25

	Seth	Riordan	Ivan	Total balls of  Riordan and Ivan
Before	25x1 = 25 u	8x1 = 8 u	10x1 = 10 u	18x1 = 18 u
Change		- 20	+ 20
After	25 u	1x3 = 3 u	5x3 = 15 u	6x3 = 18 u

Number of balls that Ivan had more than Riordan at first in percent
= 100%+ 25%
= 125%

125% = ¹²⁵₁₀₀ = ⁵₄

Riordan : Ivan = 4 : 5

The number of balls that Ivan had at first is repeated.  Make the number of balls that Ivan had at first the same.  LCM of 5 and 2 is 10.

When Riordan gave 20 balls to Ivan, the total number of balls that Ivan and Riordan had at first and in the end remains the same.  Make the total number of balls that Ivan and Riordan had the same.  LCM of 18 and 6 is 18.

Number of balls that Riordan gave to Ivan
= 8 u - 3 u
= 5 u

5 u = 20

1 u = 20 ÷ 5 = 4

Number of balls that Seth had more than Ivan in the end
= 25 u - 15 u
= 10 u
= 10 x 4
= 40

Answer(s): 40