Box S contains 7 silver balls and 6 green balls. Box T contains 55 silver balls and 28 green balls. How many green balls and silver balls must be removed from Box T to put into Box S so that 50% of the balls in Box A are silver and 70% of the balls in Box T are silver?

	Box S		Box T
	Silver balls	Green balls	Silver balls	Green balls
Before	7	6	55	28
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	7 p	3 p

50% = ⁵⁰₁₀₀ = 12

70% = ⁷⁰₁₀₀ = 710

Number of silver balls = 7 + 55 = 62 

Number of green balls = 6 + 28 = 34 

1 u + 7 p = 62 --- (1)

1 u + 3 p = 34 ---(2)

(1) - (2)
(1 u + 7 p) - (1 u + 3 p) = 62 - 34
7 p - 3 p = 28
4 p = 28

1 p = 28 ÷ 4 = 7

From (2):
1 u + 3 p = 34
1 u + 3 x 7 = 34
1 u + 21 = 34

1 u = 34 - 21 = 13

Number of green balls to be removed from Box T to Box S
= 28 - 3 p
= 28 - 3 x 7
= 28 - 21
= 7

Number of silver balls to be removed from Box T to Box S
= 1 u - 7
= 13 - 7
= 6

Total number of green and silver balls to be removed from Box T to Box S
= 7 + 6
= 13

Answer(s): 13