Crate R contains 8 brown balls and 4 green balls. Crate S contains 38 brown balls and 20 green balls. How many green balls and brown balls must be removed from Crate S to put into Crate R so that 50% of the balls in Crate A are brown and 75% of the balls in Crate S are brown?

	Crate R		Crate S
	Brown balls	Green balls	Brown balls	Green balls
Before	8	4	38	20
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of brown balls = 8 + 38 = 46 

Number of green balls = 4 + 20 = 24 

1 u + 3 p = 46 --- (1)

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

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 46 - 24
3 p - 1 p = 22
2 p = 22

1 p = 22 ÷ 2 = 11

From (2):
1 u + 1 p = 24
1 u + 1 x 11 = 24
1 u + 11 = 24

1 u = 24 - 11 = 13

Number of green balls to be removed from Crate S to Crate R
= 20 - 1 p
= 20 - 1 x 11
= 20 - 11
= 9

Number of brown balls to be removed from Crate S to Crate R
= 1 u - 8
= 13 - 8
= 5

Total number of green and brown balls to be removed from Crate S to Crate R
= 9 + 5
= 14

Answer(s): 14