Container A contains 8 red beads and 15 brown beads. Container B contains 26 red beads and 7 brown beads. How many brown beads and red beads must be transferred from Container B to put into Container A so that 50% of the beads in Container A are red and 80% of the beads in Container B are red?

	Container A		Container B
	Red beads	Brown beads	Red beads	Brown beads
Before	8	15	26	7
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of red beads = 8 + 26 = 34 

Number of brown beads = 15 + 7 = 22 

1 u + 4 p = 34 --- (1)

1 u + 1 p = 22 ---(2)

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 34 - 22
4 p - 1 p = 12
3 p = 12

1 p = 12 ÷ 3 = 4

From (2):
1 u + 1 p = 22
1 u + 1 x 4 = 22
1 u + 4 = 22

1 u = 22 - 4 = 18

Number of brown beads to be transferred from Container B to Container A
= 7 - 1 p
= 7 - 1 x 4
= 7 - 4
= 3

Number of red beads to be transferred from Container B to Container A
= 1 u - 8
= 18 - 8
= 10

Total number of brown and red beads to be transferred from Container B to Container A
= 3 + 10
= 13

Answer(s): 13