There were blue, silver and green balls in a jar. Sean put another 8 blue balls and 13 green balls in the jar, then the ratio of the number of silver balls to the number of green balls became 5 : 3. Then, he doubled the number of blue balls and removed 10 green balls. The ratio of the number of blue balls to silver balls became 2 : 1. He counted and found that there were a total of 152 balls left in the jar. Find the number of blue balls that he had at first.

	Blue balls	Silver balls	Green balls	Total
Before	5 u - 8	5 u	3 u - 13
Change 1	+ 8		+ 13
After 1	1x5 = 5 u	5 u	3 u
Change 2	+ 1x5 = 5 u		- 10
After 2	2x5 = 10 u	1x5 = 5 u	3 u - 10	152

The number of silver balls remains unchanged. Make the number of silver balls the same. LCM of 1 and 5 is 5.

Total number of balls in the end
= 10 u + 5 u + 3 u - 10
= 18 u - 10

18 u - 10 = 152
18 u = 152 + 10
18 u = 162

1 u = 162 ÷ 18 = 9

Number of blue balls at first
= 5 u - 8
= 5 x 9 - 8
= 45 - 8
= 37

Answer(s): 37