Crate A contains 12 yellow balls and 8 gold balls. Crate B contains 69 yellow balls and 37 gold balls. How many gold balls and yellow balls must be removed from Crate B to put into Crate A so that 50% of the balls in Crate A are yellow and 70% of the balls in Crate B are yellow?

	Crate A		Crate B
	Yellow balls	Gold balls	Yellow balls	Gold balls
Before	12	8	69	37
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	7 p	3 p

50% = ⁵⁰₁₀₀ = 12

70% = ⁷⁰₁₀₀ = 710

Number of yellow balls = 12 + 69 = 81 

Number of gold balls = 8 + 37 = 45 

1 u + 7 p = 81 --- (1)

1 u + 3 p = 45 ---(2)

(1) - (2)
(1 u + 7 p) - (1 u + 3 p) = 81 - 45
7 p - 3 p = 36
4 p = 36

1 p = 36 ÷ 4 = 9

From (2):
1 u + 3 p = 45
1 u + 3 x 9 = 45
1 u + 27 = 45

1 u = 45 - 27 = 18

Number of gold balls to be removed from Crate B to Crate A
= 37 - 3 p
= 37 - 3 x 9
= 37 - 27
= 10

Number of yellow balls to be removed from Crate B to Crate A
= 1 u - 12
= 18 - 12
= 6

Total number of gold and yellow balls to be removed from Crate B to Crate A
= 10 + 6
= 16

Answer(s): 16