Ivan, Neave and Julian had a total of 40 coins. The ratio of Neave's coins to Julian's coins was 8 : 3 at first. Ivan and Neave each gave away ¹₂ of their coins. Given that the three boys had 23 coins left, how many coins did Ivan have at first?

	Ivan	Neave	Julian	Total
Comparing Neave and Julian at first		8 u	3 u
Before	2 p	2x4 = 8 u	3 u	40
Change	- 1 p	-1x4 = - 4 u		- 17
After	1 p	1x4 = 4 u	3 u	23

Total number of coins that Ivan and Neave gave away
= 40 - 23
= 17

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

1 p + 4 u = 40 - 23
1 p + 4 u = 17
1 p = 17 - 4 u --- (1)

1 p + 4 u + 3 u = 23
1 p + 7 u = 23
1 p = 23 - 7 u --- (2)

(1) = (2)
17 - 4 u = 23 - 7 u
7 u - 4 u = 23 - 17
7 u - 4 u = 6
3 u = 6

1 u = 6 ÷ 3 = 2

Substitute 1 u = 2 into (1).
1 p = 17 - 4 u
1 p = 17 - 4 x 2
1 p = 17 - 8

1 p = 9

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

Answer(s): 18