Elijah, Xavier and Cody had some balls. Xavier had 60% less balls than Elijah. Xavier had ⁵₈ of Cody's. After Elijah gave 72 balls to Xavier, he had ¹₄ of what Xavier had. How many more balls did Cody have than Elijah in the end?

Elijah	Xavier	Cody
5x5	2x5
	5x2	8x2
25	10	16

	Cody	Elijah	Xavier	Total balls of  Elijah and Xavier
Before	16x1 = 16 u	25x1 = 25 u	10x1 = 10 u	35x1 = 35 u
Change		- 72	+ 72
After	16 u	1x7 = 7 u	4x7 = 28 u	5x7 = 35 u

Number of balls that Xavier had less than Elijah at first in percent
= 100% - 60%
= 40%

40% = ⁴⁰₁₀₀ = ²₅

Elijah : Xavier = 5 : 2

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

When Elijah gave 72 balls to Xavier, the total number of balls that Xavier and Elijah had at first and in the end remains the same.  Make the total number of balls that Xavier and Elijah had the same.  LCM of 35 and 5 is 35.

Number of balls that Elijah gave to Xavier
= 25 u - 7 u
= 18 u

18 u = 72

1 u = 72 ÷ 18 = 4

Number of balls that Cody had more than Elijah in the end
= 16 u - 7 u
= 9 u
= 9 x 4
= 36

Answer(s): 36