Crate V contains 15 gold balls and 10 green balls. Crate W contains 36 gold balls and 19 green balls. How many green balls and gold balls must be transferred from Crate W to put into Crate V so that 50% of the balls in Crate A are gold and 75% of the balls in Crate W are gold?

	Crate V		Crate W
	Gold balls	Green balls	Gold balls	Green balls
Before	15	10	36	19
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of gold balls = 15 + 36 = 51 

Number of green balls = 10 + 19 = 29 

1 u + 3 p = 51 --- (1)

1 u + 1 p = 29 ---(2)

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 51 - 29
3 p - 1 p = 22
2 p = 22

1 p = 22 ÷ 2 = 11

From (2):
1 u + 1 p = 29
1 u + 1 x 11 = 29
1 u + 11 = 29

1 u = 29 - 11 = 18

Number of green balls to be transferred from Crate W to Crate V
= 19 - 1 p
= 19 - 1 x 11
= 19 - 11
= 8

Number of gold balls to be transferred from Crate W to Crate V
= 1 u - 15
= 18 - 15
= 3

Total number of green and gold balls to be transferred from Crate W to Crate V
= 8 + 3
= 11

Answer(s): 11