There were gold, silver and green balls in a bottle. Sean put another 6 gold balls and 13 green balls in the bottle, then the ratio of the number of silver balls to the number of green balls became 4 : 3. Then, he doubled the number of gold balls and removed 14 green balls. The ratio of the number of gold balls to silver balls became 2 : 1. He counted and found that there were a total of 46 balls left in the bottle. Find the number of gold balls that he had at first.

	Gold balls	Silver balls	Green balls	Total
Before	4 u - 6	4 u	3 u - 13
Change 1	+ 6		+ 13
After 1	1x4 = 4 u	4 u	3 u
Change 2	+ 1x4 = 4 u		- 14
After 2	2x4 = 8 u	1x4 = 4 u	3 u - 14	46

The number of silver balls remains unchanged. Make the number of silver balls the same. LCM of 1 and 4 is 4.

Total number of balls in the end
= 8 u + 4 u + 3 u - 14
= 15 u - 14

15 u - 14 = 46
15 u = 46 + 14
15 u = 60

1 u = 60 ÷ 15 = 4

Number of gold balls at first
= 4 u - 6
= 4 x 4 - 6
= 16 - 6
= 10

Answer(s): 10