Box K contains 4 black balls and 10 red balls. Box L contains 104 black balls and 46 red balls. How many red balls and black balls must be removed from Box L to put into Box K so that 50% of the balls in Box A are black and 70% of the balls in Box L are black?

	Box K		Box L
	Black balls	Red balls	Black balls	Red balls
Before	4	10	104	46
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	7 p	3 p

50% = ⁵⁰₁₀₀ = 12

70% = ⁷⁰₁₀₀ = 710

Number of black balls = 4 + 104 = 108 

Number of red balls = 10 + 46 = 56 

1 u + 7 p = 108 --- (1)

1 u + 3 p = 56 ---(2)

(1) - (2)
(1 u + 7 p) - (1 u + 3 p) = 108 - 56
7 p - 3 p = 52
4 p = 52

1 p = 52 ÷ 4 = 13

From (2):
1 u + 3 p = 56
1 u + 3 x 13 = 56
1 u + 39 = 56

1 u = 56 - 39 = 17

Number of red balls to be removed from Box L to Box K
= 46 - 3 p
= 46 - 3 x 13
= 46 - 39
= 7

Number of black balls to be removed from Box L to Box K
= 1 u - 4
= 17 - 4
= 13

Total number of red and black balls to be removed from Box L to Box K
= 7 + 13
= 20

Answer(s): 20