There were gold, black and silver balls in a container. Sean put another 6 gold balls and 4 silver balls in the container, then the ratio of the number of black balls to the number of silver balls became 4 : 3. Then, he doubled the number of gold balls and removed 21 silver balls. The ratio of the number of gold balls to black balls became 2 : 1. He counted and found that there were a total of 54 balls left in the container. Find the number of gold balls that he had at first.

	Gold balls	Black balls	Silver balls	Total
Before	4 u - 6	4 u	3 u - 4
Change 1	+ 6		+ 4
After 1	1x4 = 4 u	4 u	3 u
Change 2	+ 1x4 = 4 u		- 21
After 2	2x4 = 8 u	1x4 = 4 u	3 u - 21	54

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

Total number of balls in the end
= 8 u + 4 u + 3 u - 21
= 15 u - 21

15 u - 21 = 54
15 u = 54 + 21
15 u = 75

1 u = 75 ÷ 15 = 5

Number of gold balls at first
= 4 u - 6
= 4 x 5 - 6
= 20 - 6
= 14

Answer(s): 14