Crate K contains 6 brown beads and 7 green beads. Crate L contains 42 brown beads and 14 green beads. How many green beads and brown beads must be transferred from Crate L to put into Crate K so that 50% of the beads in Crate A are brown and 80% of the beads in Crate L are brown?

	Crate K		Crate L
	Brown beads	Green beads	Brown beads	Green beads
Before	6	7	42	14
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of brown beads = 6 + 42 = 48 

Number of green beads = 7 + 14 = 21 

1 u + 4 p = 48 --- (1)

1 u + 1 p = 21 ---(2)

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 48 - 21
4 p - 1 p = 27
3 p = 27

1 p = 27 ÷ 3 = 9

From (2):
1 u + 1 p = 21
1 u + 1 x 9 = 21
1 u + 9 = 21

1 u = 21 - 9 = 12

Number of green beads to be transferred from Crate L to Crate K
= 14 - 1 p
= 14 - 1 x 9
= 14 - 9
= 5

Number of brown beads to be transferred from Crate L to Crate K
= 1 u - 6
= 12 - 6
= 6

Total number of green and brown beads to be transferred from Crate L to Crate K
= 5 + 6
= 11

Answer(s): 11