Lee, Jenson and Oscar had a total of 96 coins. The ratio of Jenson's coins to Oscar's coins was 4 : 9 at first. Lee and Jenson each gave away ¹₂ of their coins. Given that the three boys had 75 coins left, how many coins did Lee have at first?

	Lee	Jenson	Oscar	Total
Comparing Jenson and Oscar at first		4 u	9 u
Before	2 p	2x2 = 4 u	9 u	96
Change	- 1 p	-1x2 = - 2 u		- 21
After	1 p	1x2 = 2 u	9 u	75

Total number of coins that Lee and Jenson gave away
= 96 - 75
= 21

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

1 p + 2 u = 96 - 75
1 p + 2 u = 21
1 p = 21 - 2 u --- (1)

1 p + 2 u + 9 u = 75
1 p + 11 u = 75
1 p = 75 - 11 u --- (2)

(1) = (2)
21 - 2 u = 75 - 11 u
11 u - 2 u = 75 - 21
11 u - 2 u = 54
9 u = 54

1 u = 54 ÷ 9 = 6

Substitute 1 u = 6 into (1).
1 p = 21 - 2 u
1 p = 21 - 2 x 6
1 p = 21 - 12

1 p = 9

Number of coins that Lee had at first
= 2 p
= 2 x 9
= 18

Answer(s): 18