David, Japheth and Luis had a total of 44 beads. The ratio of Japheth's beads to Luis's beads was 4 : 3 at first. David and Japheth each gave away ¹₂ of their beads. Given that the three boys had 28 beads left, how many beads did David have in the end?

	David	Japheth	Luis	Total
Comparing Japheth and Luis at first		4 u	3 u
Before	2 p	2x2 = 4 u	3 u	44
Change	- 1 p	-1x2 = - 2 u		- 16
After	1 p	1x2 = 2 u	3 u	28

Total number of beads that David and Japheth gave away
= 44 - 28
= 16

The number of beads that Japheth had at first is repeated.  Make the number of beads that Japheth had at first the same.  LCM of 4 and 2 is 4.

1 p + 2 u = 44 - 28
1 p + 2 u = 16
1 p = 16 - 2 u --- (1)

1 p + 2 u + 3 u = 28
1 p + 5 u = 28
1 p = 28 - 5 u --- (2)

(1) = (2)
16 - 2 u = 28 - 5 u
5 u - 2 u = 28 - 16
5 u - 2 u = 12
3 u = 12

1 u = 12 ÷ 3 = 4

Substitute 1 u = 4 into (1).
1 p = 16 - 2 u
1 p = 16 - 2 x 4
1 p = 16 - 8

1 p = 8

Number of beads that David had in the end
= 1 p
= 8

Answer(s): 8