Bryan, Henry and Cole had a total of 89 coins. The ratio of Henry's coins to Cole's coins was 8 : 7 at first. Bryan and Henry each gave away ¹₂ of their coins. Given that the three boys had 62 coins left, how many coins did Bryan have in the end?

	Bryan	Henry	Cole	Total
Comparing Henry and Cole at first		8 u	7 u
Before	2 p	2x4 = 8 u	7 u	89
Change	- 1 p	-1x4 = - 4 u		- 27
After	1 p	1x4 = 4 u	7 u	62

Total number of coins that Bryan and Henry gave away
= 89 - 62
= 27

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

1 p + 4 u = 89 - 62
1 p + 4 u = 27
1 p = 27 - 4 u --- (1)

1 p + 4 u + 7 u = 62
1 p + 11 u = 62
1 p = 62 - 11 u --- (2)

(1) = (2)
27 - 4 u = 62 - 11 u
11 u - 4 u = 62 - 27
11 u - 4 u = 35
7 u = 35

1 u = 35 ÷ 7 = 5

Substitute 1 u = 5 into (1).
1 p = 27 - 4 u
1 p = 27 - 4 x 5
1 p = 27 - 20

1 p = 7

Number of coins that Bryan had in the end
= 1 p
= 7

Answer(s): 7