Neave, Riordan and Ian had some balls. Riordan had 80% less balls than Neave. Riordan had ⁵₈ of Ian's. After Neave gave 76 balls to Riordan, he had ¹₄ of what Riordan had. How many more balls did Ian have than Neave in the end?

Neave	Riordan	Ian
5x5	1x5
	5x1	8x1
25	5	8

	Ian	Neave	Riordan	Total balls of  Neave and Riordan
Before	8x1 = 8 u	25x1 = 25 u	5x1 = 5 u	30x1 = 30 u
Change		- 76	+ 76
After	8 u	1x6 = 6 u	4x6 = 24 u	5x6 = 30 u

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

20% = ²⁰₁₀₀ = ¹₅

Neave : Riordan = 5 : 1

The number of balls that Riordan had at first is repeated.  Make the number of balls that Riordan had at first the same.  LCM of 1 and 5 is 5.

When Neave gave 76 balls to Riordan, the total number of balls that Riordan and Neave had at first and in the end remains the same.  Make the total number of balls that Riordan and Neave had the same.  LCM of 30 and 5 is 30.

Number of balls that Neave gave to Riordan
= 25 u - 6 u
= 19 u

19 u = 76

1 u = 76 ÷ 19 = 4

Number of balls that Ian had more than Neave in the end
= 8 u - 6 u
= 2 u
= 2 x 4
= 8

Answer(s): 8