Harry, Fred and Andy had a total of 131 coins. The ratio of Fred's coins to Andy's coins was 2 : 9 at first. Harry and Fred each gave away ¹₂ of their coins. Given that the three boys had 115 coins left, how many coins did Harry have in the end?

	Harry	Fred	Andy	Total
Comparing Fred and Andy at first		2 u	9 u
Before	2 p	2x1 = 2 u	9 u	131
Change	- 1 p	-1x1 = - 1 u		- 16
After	1 p	1x1 = 1 u	9 u	115

Total number of coins that Harry and Fred gave away
= 131 - 115
= 16

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

1 p + 1 u = 131 - 115
1 p + 1 u = 16
1 p = 16 - 1 u --- (1)

1 p + 1 u + 9 u = 115
1 p + 10 u = 115
1 p = 115 - 10 u --- (2)

(1) = (2)
16 - 1 u = 115 - 10 u
10 u - 1 u = 115 - 16
10 u - 1 u = 99
9 u = 99

1 u = 99 ÷ 9 = 11

Substitute 1 u = 11 into (1).
1 p = 16 - 1 u
1 p = 16 - 1 x 11
1 p = 16 - 11

1 p = 5

Number of coins that Harry had in the end
= 1 p
= 5

Answer(s): 5