Fred, Archie and Peter had a total of 81 stickers. The ratio of Archie's stickers to Peter's stickers was 8 : 7 at first. Fred and Archie each gave away ¹₂ of their stickers. Given that the three boys had 58 stickers left, how many stickers did Fred have in the end?

	Fred	Archie	Peter	Total
Comparing Archie and Peter at first		8 u	7 u
Before	2 p	2x4 = 8 u	7 u	81
Change	- 1 p	-1x4 = - 4 u		- 23
After	1 p	1x4 = 4 u	7 u	58

Total number of stickers that Fred and Archie gave away
= 81 - 58
= 23

The number of stickers that Archie had at first is repeated.  Make the number of stickers that Archie had at first the same.  LCM of 8 and 2 is 8.

1 p + 4 u = 81 - 58
1 p + 4 u = 23
1 p = 23 - 4 u --- (1)

1 p + 4 u + 7 u = 58
1 p + 11 u = 58
1 p = 58 - 11 u --- (2)

(1) = (2)
23 - 4 u = 58 - 11 u
11 u - 4 u = 58 - 23
11 u - 4 u = 35
7 u = 35

1 u = 35 ÷ 7 = 5

Substitute 1 u = 5 into (1).
1 p = 23 - 4 u
1 p = 23 - 4 x 5
1 p = 23 - 20

1 p = 3

Number of stickers that Fred had in the end
= 1 p
= 3

Answer(s): 3