Box M contains 11 grey balls and 4 black balls. Box N contains 43 grey balls and 23 black balls. How many black balls and grey balls must be moved from Box N to put into Box M so that 50% of the balls in Box A are grey and 80% of the balls in Box N are grey?

	Box M		Box N
	Grey balls	Black balls	Grey balls	Black balls
Before	11	4	43	23
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of grey balls = 11 + 43 = 54 

Number of black balls = 4 + 23 = 27 

1 u + 4 p = 54 --- (1)

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

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 54 - 27
4 p - 1 p = 27
3 p = 27

1 p = 27 ÷ 3 = 9

From (2):
1 u + 1 p = 27
1 u + 1 x 9 = 27
1 u + 9 = 27

1 u = 27 - 9 = 18

Number of black balls to be moved from Box N to Box M
= 23 - 1 p
= 23 - 1 x 9
= 23 - 9
= 14

Number of grey balls to be moved from Box N to Box M
= 1 u - 11
= 18 - 11
= 7

Total number of black and grey balls to be moved from Box N to Box M
= 14 + 7
= 21

Answer(s): 21