Michael, Elijah and Fred had some balls. Elijah had 20% more balls than Michael. Elijah had ³₇ of Fred's. After Michael gave 36 balls to Elijah, he had ¹₂ of what Elijah had. How many more balls did Fred have than Elijah in the end?

Michael	Elijah	Fred
5x1	6x1
	3x2	7x2
5	6	14

	Fred	Michael	Elijah	Total balls of  Michael and Elijah
Before	14x3 = 42 u	5x3 = 15 u	6x3 = 18 u	11x3 = 33 u
Change		- 36	+ 36
After	42 u	1x11 = 11 u	2x11 = 22 u	3x11 = 33 u

Number of balls that Elijah had more than Michael at first in percent
= 100%+ 20%
= 120%

120% = ¹²⁰₁₀₀ = ⁶₅

Michael : Elijah = 5 : 6

The number of balls that Elijah had at first is repeated.  Make the number of balls that Elijah had at first the same.  LCM of 6 and 3 is 6.

When Michael gave 36 balls to Elijah, the total number of balls that Elijah and Michael had at first and in the end remains the same.  Make the total number of balls that Elijah and Michael had the same.  LCM of 11 and 3 is 33.

Number of balls that Michael gave to Elijah
= 15 u - 11 u
= 4 u

4 u = 36

1 u = 36 ÷ 4 = 9

Number of balls that Fred had more than Elijah in the end
= 42 u - 22 u
= 20 u
= 20 x 9
= 180

Answer(s): 180