Lee, George and Elijah had some balls. George had 60% less balls than Lee. George had ³₇ of Elijah's. After Lee gave 207 balls to George, he had ¹₅ of what George had. How many more balls did Elijah have than Lee in the end?

Lee	George	Elijah
5x3	2x3
	3x2	7x2
15	6	14

	Elijah	Lee	George	Total balls of  Lee and George
Before	14x2 = 28 u	15x2 = 30 u	6x2 = 12 u	21x2 = 42 u
Change		- 207	+ 207
After	28 u	1x7 = 7 u	5x7 = 35 u	6x7 = 42 u

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

40% = ⁴⁰₁₀₀ = ²₅

Lee : George = 5 : 2

The number of balls that George had at first is repeated.  Make the number of balls that George had at first the same.  LCM of 2 and 3 is 6.

When Lee gave 207 balls to George, the total number of balls that George and Lee had at first and in the end remains the same.  Make the total number of balls that George and Lee had the same.  LCM of 21 and 6 is 42.

Number of balls that Lee gave to George
= 30 u - 7 u
= 23 u

23 u = 207

1 u = 207 ÷ 23 = 9

Number of balls that Elijah had more than Lee in the end
= 28 u - 7 u
= 21 u
= 21 x 9
= 189

Answer(s): 189