Lee, Tom and Nick had some balls. Tom had 20% less balls than Lee. Tom had ²₇ of Nick's. After Lee gave 132 balls to Tom, he had ¹₃ of what Tom had. How many more balls did Nick have than Lee in the end?

Lee	Tom	Nick
5x1	4x1
	2x2	7x2
5	4	14

	Nick	Lee	Tom	Total balls of  Lee and Tom
Before	14x4 = 56 u	5x4 = 20 u	4x4 = 16 u	9x4 = 36 u
Change		- 132	+ 132
After	56 u	1x9 = 9 u	3x9 = 27 u	4x9 = 36 u

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

80% = ⁸⁰₁₀₀ = ⁴₅

Lee : Tom = 5 : 4

The number of balls that Tom had at first is repeated.  Make the number of balls that Tom had at first the same.  LCM of 4 and 2 is 4.

When Lee gave 132 balls to Tom, the total number of balls that Tom and Lee had at first and in the end remains the same.  Make the total number of balls that Tom and Lee had the same.  LCM of 9 and 4 is 36.

Number of balls that Lee gave to Tom
= 20 u - 9 u
= 11 u

11 u = 132

1 u = 132 ÷ 11 = 12

Number of balls that Nick had more than Lee in the end
= 56 u - 9 u
= 47 u
= 47 x 12
= 564

Answer(s): 564