Container L contains 14 brown beads and 2 white beads. Container M contains 37 brown beads and 25 white beads. How many white beads and brown beads must be removed from Container M to put into Container L so that 50% of the beads in Container A are brown and 80% of the beads in Container M are brown?

	Container L		Container M
	Brown beads	White beads	Brown beads	White beads
Before	14	2	37	25
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of brown beads = 14 + 37 = 51 

Number of white beads = 2 + 25 = 27 

1 u + 4 p = 51 --- (1)

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

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 51 - 27
4 p - 1 p = 24
3 p = 24

1 p = 24 ÷ 3 = 8

From (2):
1 u + 1 p = 27
1 u + 1 x 8 = 27
1 u + 8 = 27

1 u = 27 - 8 = 19

Number of white beads to be removed from Container M to Container L
= 25 - 1 p
= 25 - 1 x 8
= 25 - 8
= 17

Number of brown beads to be removed from Container M to Container L
= 1 u - 14
= 19 - 14
= 5

Total number of white and brown beads to be removed from Container M to Container L
= 17 + 5
= 22

Answer(s): 22