Container K contains 2 brown beads and 10 yellow beads. Container L contains 81 brown beads and 22 yellow beads. How many yellow beads and brown beads must be removed from Container L to put into Container K so that 50% of the beads in Container A are brown and 80% of the beads in Container L are brown?

	Container K		Container L
	Brown beads	Yellow beads	Brown beads	Yellow beads
Before	2	10	81	22
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of brown beads = 2 + 81 = 83 

Number of yellow beads = 10 + 22 = 32 

1 u + 4 p = 83 --- (1)

1 u + 1 p = 32 ---(2)

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 83 - 32
4 p - 1 p = 51
3 p = 51

1 p = 51 ÷ 3 = 17

From (2):
1 u + 1 p = 32
1 u + 1 x 17 = 32
1 u + 17 = 32

1 u = 32 - 17 = 15

Number of yellow beads to be removed from Container L to Container K
= 22 - 1 p
= 22 - 1 x 17
= 22 - 17
= 5

Number of brown beads to be removed from Container L to Container K
= 1 u - 2
= 15 - 2
= 13

Total number of yellow and brown beads to be removed from Container L to Container K
= 5 + 13
= 18

Answer(s): 18