Crate H contains 9 black balls and 4 green balls. Crate J contains 69 black balls and 34 green balls. How many green balls and black balls must be removed from Crate J to put into Crate H so that 50% of the balls in Crate A are black and 75% of the balls in Crate J are black?

	Crate H		Crate J
	Black balls	Green balls	Black balls	Green balls
Before	9	4	69	34
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of black balls = 9 + 69 = 78 

Number of green balls = 4 + 34 = 38 

1 u + 3 p = 78 --- (1)

1 u + 1 p = 38 ---(2)

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 78 - 38
3 p - 1 p = 40
2 p = 40

1 p = 40 ÷ 2 = 20

From (2):
1 u + 1 p = 38
1 u + 1 x 20 = 38
1 u + 20 = 38

1 u = 38 - 20 = 18

Number of green balls to be removed from Crate J to Crate H
= 34 - 1 p
= 34 - 1 x 20
= 34 - 20
= 14

Number of black balls to be removed from Crate J to Crate H
= 1 u - 9
= 18 - 9
= 9

Total number of green and black balls to be removed from Crate J to Crate H
= 14 + 9
= 23

Answer(s): 23