Crate Y contains 3 brown balls and 7 red balls. Crate Z contains 134 brown balls and 58 red balls. How many red balls and brown balls must be removed from Crate Z to put into Crate Y so that 50% of the balls in Crate A are brown and 70% of the balls in Crate Z are brown?

	Crate Y		Crate Z
	Brown balls	Red balls	Brown balls	Red balls
Before	3	7	134	58
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	7 p	3 p

50% = ⁵⁰₁₀₀ = 12

70% = ⁷⁰₁₀₀ = 710

Number of brown balls = 3 + 134 = 137 

Number of red balls = 7 + 58 = 65 

1 u + 7 p = 137 --- (1)

1 u + 3 p = 65 ---(2)

(1) - (2)
(1 u + 7 p) - (1 u + 3 p) = 137 - 65
7 p - 3 p = 72
4 p = 72

1 p = 72 ÷ 4 = 18

From (2):
1 u + 3 p = 65
1 u + 3 x 18 = 65
1 u + 54 = 65

1 u = 65 - 54 = 11

Number of red balls to be removed from Crate Z to Crate Y
= 58 - 3 p
= 58 - 3 x 18
= 58 - 54
= 4

Number of brown balls to be removed from Crate Z to Crate Y
= 1 u - 3
= 11 - 3
= 8

Total number of red and brown balls to be removed from Crate Z to Crate Y
= 4 + 8
= 12

Answer(s): 12