Vincent, John and Gabriel had a total of 139 coins. The ratio of John's coins to Gabriel's coins was 8 : 9 at first. Vincent and John each gave away ¹₂ of their coins. Given that the three boys had 101 coins left, how many coins did Vincent have in the end?

	Vincent	John	Gabriel	Total
Comparing John and Gabriel at first		8 u	9 u
Before	2 p	2x4 = 8 u	9 u	139
Change	- 1 p	-1x4 = - 4 u		- 38
After	1 p	1x4 = 4 u	9 u	101

Total number of coins that Vincent and John gave away
= 139 - 101
= 38

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

1 p + 4 u = 139 - 101
1 p + 4 u = 38
1 p = 38 - 4 u --- (1)

1 p + 4 u + 9 u = 101
1 p + 13 u = 101
1 p = 101 - 13 u --- (2)

(1) = (2)
38 - 4 u = 101 - 13 u
13 u - 4 u = 101 - 38
13 u - 4 u = 63
9 u = 63

1 u = 63 ÷ 9 = 7

Substitute 1 u = 7 into (1).
1 p = 38 - 4 u
1 p = 38 - 4 x 7
1 p = 38 - 28

1 p = 10

Number of coins that Vincent had in the end
= 1 p
= 10

Answer(s): 10