Vincent, George and Bryan had a total of 60 coins. The ratio of George's coins to Bryan's coins was 2 : 3 at first. Vincent and George each gave away ¹₂ of their coins. Given that the three boys had 45 coins left, how many coins did Vincent have at first?

	Vincent	George	Bryan	Total
Comparing George and Bryan at first		2 u	3 u
Before	2 p	2x1 = 2 u	3 u	60
Change	- 1 p	-1x1 = - 1 u		- 15
After	1 p	1x1 = 1 u	3 u	45

Total number of coins that Vincent and George gave away
= 60 - 45
= 15

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

1 p + 1 u = 60 - 45
1 p + 1 u = 15
1 p = 15 - 1 u --- (1)

1 p + 1 u + 3 u = 45
1 p + 4 u = 45
1 p = 45 - 4 u --- (2)

(1) = (2)
15 - 1 u = 45 - 4 u
4 u - 1 u = 45 - 15
4 u - 1 u = 30
3 u = 30

1 u = 30 ÷ 3 = 10

Substitute 1 u = 10 into (1).
1 p = 15 - 1 u
1 p = 15 - 1 x 10
1 p = 15 - 10

1 p = 5

Number of coins that Vincent had at first
= 2 p
= 2 x 5
= 10

Answer(s): 10