Container P contains 7 blue beads and 5 purple beads. Container Q contains 56 blue beads and 19 purple beads. How many purple beads and blue beads must be removed from Container Q to put into Container P so that 50% of the beads in Container A are blue and 80% of the beads in Container Q are blue?

	Container P		Container Q
	Blue beads	Purple beads	Blue beads	Purple beads
Before	7	5	56	19
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of blue beads = 7 + 56 = 63 

Number of purple beads = 5 + 19 = 24 

1 u + 4 p = 63 --- (1)

1 u + 1 p = 24 ---(2)

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 63 - 24
4 p - 1 p = 39
3 p = 39

1 p = 39 ÷ 3 = 13

From (2):
1 u + 1 p = 24
1 u + 1 x 13 = 24
1 u + 13 = 24

1 u = 24 - 13 = 11

Number of purple beads to be removed from Container Q to Container P
= 19 - 1 p
= 19 - 1 x 13
= 19 - 13
= 6

Number of blue beads to be removed from Container Q to Container P
= 1 u - 7
= 11 - 7
= 4

Total number of purple and blue beads to be removed from Container Q to Container P
= 6 + 4
= 10

Answer(s): 10