Pierre, Simon and Ian had a total of 134 coins. The ratio of Simon's coins to Ian's coins was 4 : 7 at first. Pierre and Simon each gave away ¹₂ of their coins. Given that the three boys had 102 coins left, how many coins did Pierre have in the end?

	Pierre	Simon	Ian	Total
Comparing Simon and Ian at first		4 u	7 u
Before	2 p	2x2 = 4 u	7 u	134
Change	- 1 p	-1x2 = - 2 u		- 32
After	1 p	1x2 = 2 u	7 u	102

Total number of coins that Pierre and Simon gave away
= 134 - 102
= 32

The number of coins that Simon had at first is repeated.  Make the number of coins that Simon had at first the same.  LCM of 4 and 2 is 4.

1 p + 2 u = 134 - 102
1 p + 2 u = 32
1 p = 32 - 2 u --- (1)

1 p + 2 u + 7 u = 102
1 p + 9 u = 102
1 p = 102 - 9 u --- (2)

(1) = (2)
32 - 2 u = 102 - 9 u
9 u - 2 u = 102 - 32
9 u - 2 u = 70
7 u = 70

1 u = 70 ÷ 7 = 10

Substitute 1 u = 10 into (1).
1 p = 32 - 2 u
1 p = 32 - 2 x 10
1 p = 32 - 20

1 p = 12

Number of coins that Pierre had in the end
= 1 p
= 12

Answer(s): 12