There were white, red and silver balls in a container. Sean put another 8 white balls and 2 silver balls in the container, then the ratio of the number of red balls to the number of silver balls became 6 : 5. Then, he doubled the number of white balls and removed 16 silver balls. The ratio of the number of white balls to red balls became 2 : 1. He counted and found that there were a total of 53 balls left in the container. Find the number of white balls that he had at first.

	White balls	Red balls	Silver balls	Total
Before	6 u - 8	6 u	5 u - 2
Change 1	+ 8		+ 2
After 1	1x6 = 6 u	6 u	5 u
Change 2	+ 1x6 = 6 u		- 16
After 2	2x6 = 12 u	1x6 = 6 u	5 u - 16	53

The number of red balls remains unchanged. Make the number of red balls the same. LCM of 1 and 6 is 6.

Total number of balls in the end
= 12 u + 6 u + 5 u - 16
= 23 u - 16

23 u - 16 = 53
23 u = 53 + 16
23 u = 69

1 u = 69 ÷ 23 = 3

Number of white balls at first
= 6 u - 8
= 6 x 3 - 8
= 18 - 8
= 10

Answer(s): 10