Zane, Luis and Xavier had some beads. Luis had 25% less beads than Zane. Luis had ⁴₇ of Xavier's. After Zane gave 81 beads to Luis, he had ¹₃ of what Luis had. How many more beads did Xavier have than Zane in the end?

Zane	Luis	Xavier
4x4	3x4
	4x3	7x3
16	12	21

	Xavier	Zane	Luis	Total beads of  Zane and Luis
Before	21x1 = 21 u	16x1 = 16 u	12x1 = 12 u	28x1 = 28 u
Change		- 81	+ 81
After	21 u	1x7 = 7 u	3x7 = 21 u	4x7 = 28 u

Number of beads that Luis had less than Zane at first in percent
= 100% - 25%
= 75%

75% = ⁷⁵₁₀₀ = ³₄

Zane : Luis = 4 : 3

The number of beads that Luis had at first is repeated.  Make the number of beads that Luis had at first the same.  LCM of 3 and 4 is 12.

When Zane gave 81 beads to Luis, the total number of beads that Luis and Zane had at first and in the end remains the same.  Make the total number of beads that Luis and Zane had the same.  LCM of 28 and 4 is 28.

Number of beads that Zane gave to Luis
= 16 u - 7 u
= 9 u

9 u = 81

1 u = 81 ÷ 9 = 9

Number of beads that Xavier had more than Zane in the end
= 21 u - 7 u
= 14 u
= 14 x 9
= 126

Answer(s): 126