Box M contains 2 blue balls and 4 green balls. Box N contains 44 blue balls and 16 green balls. How many green balls and blue balls must be removed from Box N to put into Box M so that 50% of the balls in Box A are blue and 75% of the balls in Box N are blue?

	Box M		Box N
	Blue balls	Green balls	Blue balls	Green balls
Before	2	4	44	16
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of blue balls = 2 + 44 = 46 

Number of green balls = 4 + 16 = 20 

1 u + 3 p = 46 --- (1)

1 u + 1 p = 20 ---(2)

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 46 - 20
3 p - 1 p = 26
2 p = 26

1 p = 26 ÷ 2 = 13

From (2):
1 u + 1 p = 20
1 u + 1 x 13 = 20
1 u + 13 = 20

1 u = 20 - 13 = 7

Number of green balls to be removed from Box N to Box M
= 16 - 1 p
= 16 - 1 x 13
= 16 - 13
= 3

Number of blue balls to be removed from Box N to Box M
= 1 u - 2
= 7 - 2
= 5

Total number of green and blue balls to be removed from Box N to Box M
= 3 + 5
= 8

Answer(s): 8