Container K contains 6 green balls and 5 black balls. Container L contains 54 green balls and 25 black balls. How many black balls and green balls must be moved from Container L to put into Container K so that 50% of the balls in Container A are green and 80% of the balls in Container L are green?

	Container K		Container L
	Green balls	Black balls	Green balls	Black balls
Before	6	5	54	25
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of green balls = 6 + 54 = 60 

Number of black balls = 5 + 25 = 30 

1 u + 4 p = 60 --- (1)

1 u + 1 p = 30 ---(2)

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 60 - 30
4 p - 1 p = 30
3 p = 30

1 p = 30 ÷ 3 = 10

From (2):
1 u + 1 p = 30
1 u + 1 x 10 = 30
1 u + 10 = 30

1 u = 30 - 10 = 20

Number of black balls to be moved from Container L to Container K
= 25 - 1 p
= 25 - 1 x 10
= 25 - 10
= 15

Number of green balls to be moved from Container L to Container K
= 1 u - 6
= 20 - 6
= 14

Total number of black and green balls to be moved from Container L to Container K
= 15 + 14
= 29

Answer(s): 29