Riordan, Seth and Cody had some beads. Seth had 60% more beads than Riordan. Seth had ³₇ of Cody's. After Riordan gave 24 beads to Seth, he had ¹₂ of what Seth had. How many more beads did Cody have than Seth in the end?

Riordan	Seth	Cody
5x3	8x3
	3x8	7x8
15	24	56

	Cody	Riordan	Seth	Total beads of  Riordan and Seth
Before	56x1 = 56 u	15x1 = 15 u	24x1 = 24 u	39x1 = 39 u
Change		- 24	+ 24
After	56 u	1x13 = 13 u	2x13 = 26 u	3x13 = 39 u

Number of beads that Seth had more than Riordan at first in percent
= 100%+ 60%
= 160%

160% = ¹⁶⁰₁₀₀ = ⁸₅

Riordan : Seth = 5 : 8

The number of beads that Seth had at first is repeated.  Make the number of beads that Seth had at first the same.  LCM of 8 and 3 is 24.

When Riordan gave 24 beads to Seth, the total number of beads that Seth and Riordan had at first and in the end remains the same.  Make the total number of beads that Seth and Riordan had the same.  LCM of 39 and 3 is 39.

Number of beads that Riordan gave to Seth
= 15 u - 13 u
= 2 u

2 u = 24

1 u = 24 ÷ 2 = 12

Number of beads that Cody had more than Seth in the end
= 56 u - 26 u
= 30 u
= 30 x 12
= 360

Answer(s): 360