Container K contains 11 blue beads and 17 silver beads. Container L contains 40 blue beads and 10 silver beads. How many silver beads and blue beads must be removed from Container L to put into Container K so that 50% of the beads in Container A are blue and 80% of the beads in Container L are blue?

	Container K		Container L
	Blue beads	Silver beads	Blue beads	Silver beads
Before	11	17	40	10
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of blue beads = 11 + 40 = 51 

Number of silver beads = 17 + 10 = 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 silver beads to be removed from Container L to Container K
= 10 - 1 p
= 10 - 1 x 8
= 10 - 8
= 2

Number of blue beads to be removed from Container L to Container K
= 1 u - 11
= 19 - 11
= 8

Total number of silver and blue beads to be removed from Container L to Container K
= 2 + 8
= 10

Answer(s): 10