Crate R contains 6 silver balls and 4 gold balls. Crate S contains 47 silver balls and 25 gold balls. How many gold balls and silver balls must be removed from Crate S to put into Crate R so that 50% of the balls in Crate A are silver and 70% of the balls in Crate S are silver?

	Crate R		Crate S
	Silver balls	Gold balls	Silver balls	Gold balls
Before	6	4	47	25
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	7 p	3 p

50% = ⁵⁰₁₀₀ = 12

70% = ⁷⁰₁₀₀ = 710

Number of silver balls = 6 + 47 = 53 

Number of gold balls = 4 + 25 = 29 

1 u + 7 p = 53 --- (1)

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

(1) - (2)
(1 u + 7 p) - (1 u + 3 p) = 53 - 29
7 p - 3 p = 24
4 p = 24

1 p = 24 ÷ 4 = 6

From (2):
1 u + 3 p = 29
1 u + 3 x 6 = 29
1 u + 18 = 29

1 u = 29 - 18 = 11

Number of gold balls to be removed from Crate S to Crate R
= 25 - 3 p
= 25 - 3 x 6
= 25 - 18
= 7

Number of silver balls to be removed from Crate S to Crate R
= 1 u - 6
= 11 - 6
= 5

Total number of gold and silver balls to be removed from Crate S to Crate R
= 7 + 5
= 12

Answer(s): 12