Crate S contains 10 red balls and 7 brown balls. Crate T contains 107 red balls and 50 brown balls. How many brown balls and red balls must be transferred from Crate T to put into Crate S so that 50% of the balls in Crate A are red and 70% of the balls in Crate T are red?

	Crate S		Crate T
	Red balls	Brown balls	Red balls	Brown balls
Before	10	7	107	50
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	7 p	3 p

50% = ⁵⁰₁₀₀ = 12

70% = ⁷⁰₁₀₀ = 710

Number of red balls = 10 + 107 = 117 

Number of brown balls = 7 + 50 = 57 

1 u + 7 p = 117 --- (1)

1 u + 3 p = 57 ---(2)

(1) - (2)
(1 u + 7 p) - (1 u + 3 p) = 117 - 57
7 p - 3 p = 60
4 p = 60

1 p = 60 ÷ 4 = 15

From (2):
1 u + 3 p = 57
1 u + 3 x 15 = 57
1 u + 45 = 57

1 u = 57 - 45 = 12

Number of brown balls to be transferred from Crate T to Crate S
= 50 - 3 p
= 50 - 3 x 15
= 50 - 45
= 5

Number of red balls to be transferred from Crate T to Crate S
= 1 u - 10
= 12 - 10
= 2

Total number of brown and red balls to be transferred from Crate T to Crate S
= 5 + 2
= 7

Answer(s): 7