Archie, Vaidev and Luis had a total of 110 coins. The ratio of Vaidev's coins to Luis's coins was 8 : 3 at first. Archie and Vaidev each gave away ¹₂ of their coins. Given that the three boys had 67 coins left, how many coins did Archie have at first?

	Archie	Vaidev	Luis	Total
Comparing Vaidev and Luis at first		8 u	3 u
Before	2 p	2x4 = 8 u	3 u	110
Change	- 1 p	-1x4 = - 4 u		- 43
After	1 p	1x4 = 4 u	3 u	67

Total number of coins that Archie and Vaidev gave away
= 110 - 67
= 43

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

1 p + 4 u = 110 - 67
1 p + 4 u = 43
1 p = 43 - 4 u --- (1)

1 p + 4 u + 3 u = 67
1 p + 7 u = 67
1 p = 67 - 7 u --- (2)

(1) = (2)
43 - 4 u = 67 - 7 u
7 u - 4 u = 67 - 43
7 u - 4 u = 24
3 u = 24

1 u = 24 ÷ 3 = 8

Substitute 1 u = 8 into (1).
1 p = 43 - 4 u
1 p = 43 - 4 x 8
1 p = 43 - 32

1 p = 11

Number of coins that Archie had at first
= 2 p
= 2 x 11
= 22

Answer(s): 22