There were black, brown and silver balls in a box. Sean put another 2 black balls and 4 silver balls in the box, then the ratio of the number of brown balls to the number of silver balls became 5 : 2. Then, he doubled the number of black balls and removed 26 silver balls. The ratio of the number of black balls to brown balls became 2 : 1. He counted and found that there were a total of 59 balls left in the box. Find the number of black balls that he had at first.

	Black balls	Brown balls	Silver balls	Total
Before	5 u - 2	5 u	2 u - 4
Change 1	+ 2		+ 4
After 1	1x5 = 5 u	5 u	2 u
Change 2	+ 1x5 = 5 u		- 26
After 2	2x5 = 10 u	1x5 = 5 u	2 u - 26	59

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

Total number of balls in the end
= 10 u + 5 u + 2 u - 26
= 17 u - 26

17 u - 26 = 59
17 u = 59 + 26
17 u = 85

1 u = 85 ÷ 17 = 5

Number of black balls at first
= 5 u - 2
= 5 x 5 - 2
= 25 - 2
= 23

Answer(s): 23