Container B contains 7 silver balls and 10 blue balls. Container C contains 44 silver balls and 17 blue balls. How many blue balls and silver balls must be moved from Container C to put into Container B so that 50% of the balls in Container A are silver and 75% of the balls in Container C are silver?

	Container B		Container C
	Silver balls	Blue balls	Silver balls	Blue balls
Before	7	10	44	17
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of silver balls = 7 + 44 = 51 

Number of blue balls = 10 + 17 = 27 

1 u + 3 p = 51 --- (1)

1 u + 1 p = 27 ---(2)

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 51 - 27
3 p - 1 p = 24
2 p = 24

1 p = 24 ÷ 2 = 12

From (2):
1 u + 1 p = 27
1 u + 1 x 12 = 27
1 u + 12 = 27

1 u = 27 - 12 = 15

Number of blue balls to be moved from Container C to Container B
= 17 - 1 p
= 17 - 1 x 12
= 17 - 12
= 5

Number of silver balls to be moved from Container C to Container B
= 1 u - 7
= 15 - 7
= 8

Total number of blue and silver balls to be moved from Container C to Container B
= 5 + 8
= 13

Answer(s): 13