Container M contains 3 blue beads and 4 black beads. Container N contains 12 blue beads and 7 black beads. How many black beads and blue beads must be removed from Container N to put into Container M so that 50% of the beads in Container A are blue and 75% of the beads in Container N are blue?

	Container M		Container N
	Blue beads	Black beads	Blue beads	Black beads
Before	3	4	12	7
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of blue beads = 3 + 12 = 15 

Number of black beads = 4 + 7 = 11 

1 u + 3 p = 15 --- (1)

1 u + 1 p = 11 ---(2)

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 15 - 11
3 p - 1 p = 4
2 p = 4

1 p = 4 ÷ 2 = 2

From (2):
1 u + 1 p = 11
1 u + 1 x 2 = 11
1 u + 2 = 11

1 u = 11 - 2 = 9

Number of black beads to be removed from Container N to Container M
= 7 - 1 p
= 7 - 1 x 2
= 7 - 2
= 5

Number of blue beads to be removed from Container N to Container M
= 1 u - 3
= 9 - 3
= 6

Total number of black and blue beads to be removed from Container N to Container M
= 5 + 6
= 11

Answer(s): 11