Crate B contains 9 green beads and 10 white beads. Crate C contains 21 green beads and 12 white beads. How many white beads and green beads must be removed from Crate C to put into Crate B so that 50% of the beads in Crate A are green and 70% of the beads in Crate C are green?

	Crate B		Crate C
	Green beads	White beads	Green beads	White beads
Before	9	10	21	12
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	7 p	3 p

50% = ⁵⁰₁₀₀ = 12

70% = ⁷⁰₁₀₀ = 710

Number of green beads = 9 + 21 = 30 

Number of white beads = 10 + 12 = 22 

1 u + 7 p = 30 --- (1)

1 u + 3 p = 22 ---(2)

(1) - (2)
(1 u + 7 p) - (1 u + 3 p) = 30 - 22
7 p - 3 p = 8
4 p = 8

1 p = 8 ÷ 4 = 2

From (2):
1 u + 3 p = 22
1 u + 3 x 2 = 22
1 u + 6 = 22

1 u = 22 - 6 = 16

Number of white beads to be removed from Crate C to Crate B
= 12 - 3 p
= 12 - 3 x 2
= 12 - 6
= 6

Number of green beads to be removed from Crate C to Crate B
= 1 u - 9
= 16 - 9
= 7

Total number of white and green beads to be removed from Crate C to Crate B
= 6 + 7
= 13

Answer(s): 13