Crate N contains 4 gold balls and 5 green balls. Crate P contains 64 gold balls and 25 green balls. How many green balls and gold balls must be transferred from Crate P to put into Crate N so that 50% of the balls in Crate A are gold and 75% of the balls in Crate P are gold?

	Crate N		Crate P
	Gold balls	Green balls	Gold balls	Green balls
Before	4	5	64	25
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of gold balls = 4 + 64 = 68 

Number of green balls = 5 + 25 = 30 

1 u + 3 p = 68 --- (1)

1 u + 1 p = 30 ---(2)

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 68 - 30
3 p - 1 p = 38
2 p = 38

1 p = 38 ÷ 2 = 19

From (2):
1 u + 1 p = 30
1 u + 1 x 19 = 30
1 u + 19 = 30

1 u = 30 - 19 = 11

Number of green balls to be transferred from Crate P to Crate N
= 25 - 1 p
= 25 - 1 x 19
= 25 - 19
= 6

Number of gold balls to be transferred from Crate P to Crate N
= 1 u - 4
= 11 - 4
= 7

Total number of green and gold balls to be transferred from Crate P to Crate N
= 6 + 7
= 13

Answer(s): 13