Michael, Liam and Zane had some balls. Liam had 80% less balls than Michael. Liam had ⁴₇ of Zane's. After Michael gave 32 balls to Liam, he had ¹₅ of what Liam had. How many more balls did Zane have than Michael in the end?

Michael	Liam	Zane
5x4	1x4
	4x1	7x1
20	4	7

	Zane	Michael	Liam	Total balls of  Michael and Liam
Before	7x1 = 7 u	20x1 = 20 u	4x1 = 4 u	24x1 = 24 u
Change		- 32	+ 32
After	7 u	1x4 = 4 u	5x4 = 20 u	6x4 = 24 u

Number of balls that Liam had less than Michael at first in percent
= 100% - 80%
= 20%

20% = ²⁰₁₀₀ = ¹₅

Michael : Liam = 5 : 1

The number of balls that Liam had at first is repeated.  Make the number of balls that Liam had at first the same.  LCM of 1 and 4 is 4.

When Michael gave 32 balls to Liam, the total number of balls that Liam and Michael had at first and in the end remains the same.  Make the total number of balls that Liam and Michael had the same.  LCM of 24 and 6 is 24.

Number of balls that Michael gave to Liam
= 20 u - 4 u
= 16 u

16 u = 32

1 u = 32 ÷ 16 = 2

Number of balls that Zane had more than Michael in the end
= 7 u - 4 u
= 3 u
= 3 x 2
= 6

Answer(s): 6