Crate S contains 9 grey balls and 2 blue balls. Crate T contains 26 grey balls and 21 blue balls. How many blue balls and grey balls must be moved from Crate T to put into Crate S so that 50% of the balls in Crate A are grey and 70% of the balls in Crate T are grey?

	Crate S		Crate T
	Grey balls	Blue balls	Grey balls	Blue balls
Before	9	2	26	21
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	7 p	3 p

50% = ⁵⁰₁₀₀ = 12

70% = ⁷⁰₁₀₀ = 710

Number of grey balls = 9 + 26 = 35 

Number of blue balls = 2 + 21 = 23 

1 u + 7 p = 35 --- (1)

1 u + 3 p = 23 ---(2)

(1) - (2)
(1 u + 7 p) - (1 u + 3 p) = 35 - 23
7 p - 3 p = 12
4 p = 12

1 p = 12 ÷ 4 = 3

From (2):
1 u + 3 p = 23
1 u + 3 x 3 = 23
1 u + 9 = 23

1 u = 23 - 9 = 14

Number of blue balls to be moved from Crate T to Crate S
= 21 - 3 p
= 21 - 3 x 3
= 21 - 9
= 12

Number of grey balls to be moved from Crate T to Crate S
= 1 u - 9
= 14 - 9
= 5

Total number of blue and grey balls to be moved from Crate T to Crate S
= 12 + 5
= 17

Answer(s): 17