Container A contains 14 black balls and 18 gold balls. Container B contains 22 black balls and 6 gold balls. How many gold balls and black balls must be removed from Container B to put into Container A so that 50% of the balls in Container A are black and 80% of the balls in Container B are black?

	Container A		Container B
	Black balls	Gold balls	Black balls	Gold balls
Before	14	18	22	6
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of black balls = 14 + 22 = 36 

Number of gold balls = 18 + 6 = 24 

1 u + 4 p = 36 --- (1)

1 u + 1 p = 24 ---(2)

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 36 - 24
4 p - 1 p = 12
3 p = 12

1 p = 12 ÷ 3 = 4

From (2):
1 u + 1 p = 24
1 u + 1 x 4 = 24
1 u + 4 = 24

1 u = 24 - 4 = 20

Number of gold balls to be removed from Container B to Container A
= 6 - 1 p
= 6 - 1 x 4
= 6 - 4
= 2

Number of black balls to be removed from Container B to Container A
= 1 u - 14
= 20 - 14
= 6

Total number of gold and black balls to be removed from Container B to Container A
= 2 + 6
= 8

Answer(s): 8