Fred, Lee and Henry had some balls. Lee had 60% less balls than Fred. Lee had ³₇ of Henry's. After Fred gave 69 balls to Lee, he had ¹₅ of what Lee had. How many more balls did Henry have than Fred in the end?

Fred	Lee	Henry
5x3	2x3
	3x2	7x2
15	6	14

	Henry	Fred	Lee	Total balls of  Fred and Lee
Before	14x2 = 28 u	15x2 = 30 u	6x2 = 12 u	21x2 = 42 u
Change		- 69	+ 69
After	28 u	1x7 = 7 u	5x7 = 35 u	6x7 = 42 u

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

40% = ⁴⁰₁₀₀ = ²₅

Fred : Lee = 5 : 2

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

When Fred gave 69 balls to Lee, the total number of balls that Lee and Fred had at first and in the end remains the same.  Make the total number of balls that Lee and Fred had the same.  LCM of 21 and 6 is 42.

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

23 u = 69

1 u = 69 ÷ 23 = 3

Number of balls that Henry had more than Fred in the end
= 28 u - 7 u
= 21 u
= 21 x 3
= 63

Answer(s): 63