Howard, Will and Zeph had a total of 82 beads. The ratio of Will's beads to Zeph's beads was 6 : 7 at first. Howard and Will each gave away ¹₂ of their beads. Given that the three boys had 62 beads left, how many beads did Howard have in the end?

	Howard	Will	Zeph	Total
Comparing Will and Zeph at first		6 u	7 u
Before	2 p	2x3 = 6 u	7 u	82
Change	- 1 p	-1x3 = - 3 u		- 20
After	1 p	1x3 = 3 u	7 u	62

Total number of beads that Howard and Will gave away
= 82 - 62
= 20

The number of beads that Will had at first is repeated.  Make the number of beads that Will had at first the same.  LCM of 6 and 2 is 6.

1 p + 3 u = 82 - 62
1 p + 3 u = 20
1 p = 20 - 3 u --- (1)

1 p + 3 u + 7 u = 62
1 p + 10 u = 62
1 p = 62 - 10 u --- (2)

(1) = (2)
20 - 3 u = 62 - 10 u
10 u - 3 u = 62 - 20
10 u - 3 u = 42
7 u = 42

1 u = 42 ÷ 7 = 6

Substitute 1 u = 6 into (1).
1 p = 20 - 3 u
1 p = 20 - 3 x 6
1 p = 20 - 18

1 p = 2

Number of beads that Howard had in the end
= 1 p
= 2

Answer(s): 2