Pierre, David and Riordan had some beads. David had 80% more beads than Pierre. David had ³₈ of Riordan's. After Pierre gave 40 beads to David, he had ¹₅ of what David had. How many more beads did Riordan have than David in the end?

Pierre	David	Riordan
5x1	9x1
	3x3	8x3
5	9	24

	Riordan	Pierre	David	Total beads of  Pierre and David
Before	24x3 = 72 u	5x3 = 15 u	9x3 = 27 u	14x3 = 42 u
Change		- 40	+ 40
After	72 u	1x7 = 7 u	5x7 = 35 u	6x7 = 42 u

Number of beads that David had more than Pierre at first in percent
= 100%+ 80%
= 180%

180% = ¹⁸⁰₁₀₀ = ⁹₅

Pierre : David = 5 : 9

The number of beads that David had at first is repeated.  Make the number of beads that David had at first the same.  LCM of 9 and 3 is 9.

When Pierre gave 40 beads to David, the total number of beads that David and Pierre had at first and in the end remains the same.  Make the total number of beads that David and Pierre had the same.  LCM of 14 and 6 is 42.

Number of beads that Pierre gave to David
= 15 u - 7 u
= 8 u

8 u = 40

1 u = 40 ÷ 8 = 5

Number of beads that Riordan had more than David in the end
= 72 u - 35 u
= 37 u
= 37 x 5
= 185

Answer(s): 185