Jenson, Xavier and Wesley had some beads. Xavier had 80% less beads than Jenson. Xavier had ²₇ of Wesley's. After Jenson gave 88 beads to Xavier, he had ¹₅ of what Xavier had. How many more beads did Wesley have than Jenson in the end?

Jenson	Xavier	Wesley
5x2	1x2
	2x1	7x1
10	2	7

	Wesley	Jenson	Xavier	Total beads of  Jenson and Xavier
Before	7x1 = 7 u	10x1 = 10 u	2x1 = 2 u	12x1 = 12 u
Change		- 88	+ 88
After	7 u	1x2 = 2 u	5x2 = 10 u	6x2 = 12 u

Number of beads that Xavier had less than Jenson at first in percent
= 100% - 80%
= 20%

20% = ²⁰₁₀₀ = ¹₅

Jenson : Xavier = 5 : 1

The number of beads that Xavier had at first is repeated.  Make the number of beads that Xavier had at first the same.  LCM of 1 and 2 is 2.

When Jenson gave 88 beads to Xavier, the total number of beads that Xavier and Jenson had at first and in the end remains the same.  Make the total number of beads that Xavier and Jenson had the same.  LCM of 12 and 6 is 12.

Number of beads that Jenson gave to Xavier
= 10 u - 2 u
= 8 u

8 u = 88

1 u = 88 ÷ 8 = 11

Number of beads that Wesley had more than Jenson in the end
= 7 u - 2 u
= 5 u
= 5 x 11
= 55

Answer(s): 55