Ken, Dylan and Jack had a total of 67 pens. The ratio of Dylan's pens to Jack's pens was 8 : 7 at first. Ken and Dylan each gave away ¹₂ of their pens. Given that the three boys had 44 pens left, how many pens did Ken have in the end?

	Ken	Dylan	Jack	Total
Comparing Dylan and Jack at first		8 u	7 u
Before	2 p	2x4 = 8 u	7 u	67
Change	- 1 p	-1x4 = - 4 u		- 23
After	1 p	1x4 = 4 u	7 u	44

Total number of pens that Ken and Dylan gave away
= 67 - 44
= 23

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

1 p + 4 u = 67 - 44
1 p + 4 u = 23
1 p = 23 - 4 u --- (1)

1 p + 4 u + 7 u = 44
1 p + 11 u = 44
1 p = 44 - 11 u --- (2)

(1) = (2)
23 - 4 u = 44 - 11 u
11 u - 4 u = 44 - 23
11 u - 4 u = 21
7 u = 21

1 u = 21 ÷ 7 = 3

Substitute 1 u = 3 into (1).
1 p = 23 - 4 u
1 p = 23 - 4 x 3
1 p = 23 - 12

1 p = 11

Number of pens that Ken had in the end
= 1 p
= 11

Answer(s): 11