Container P contains 4 green balls and 7 brown balls. Container Q contains 25 green balls and 12 brown balls. How many brown balls and green balls must be moved from Container Q to put into Container P so that 50% of the balls in Container A are green and 75% of the balls in Container Q are green?

	Container P		Container Q
	Green balls	Brown balls	Green balls	Brown balls
Before	4	7	25	12
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of green balls = 4 + 25 = 29 

Number of brown balls = 7 + 12 = 19 

1 u + 3 p = 29 --- (1)

1 u + 1 p = 19 ---(2)

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 29 - 19
3 p - 1 p = 10
2 p = 10

1 p = 10 ÷ 2 = 5

From (2):
1 u + 1 p = 19
1 u + 1 x 5 = 19
1 u + 5 = 19

1 u = 19 - 5 = 14

Number of brown balls to be moved from Container Q to Container P
= 12 - 1 p
= 12 - 1 x 5
= 12 - 5
= 7

Number of green balls to be moved from Container Q to Container P
= 1 u - 4
= 14 - 4
= 10

Total number of brown and green balls to be moved from Container Q to Container P
= 7 + 10
= 17

Answer(s): 17