Box U contains 5 red balls and 2 black balls. Box V contains 51 red balls and 27 black balls. How many black balls and red balls must be removed from Box V to put into Box U so that 50% of the balls in Box A are red and 80% of the balls in Box V are red?

	Box U		Box V
	Red balls	Black balls	Red balls	Black balls
Before	5	2	51	27
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	4 p	1 p

50% = ⁵⁰₁₀₀ = 12

80% = ⁸⁰₁₀₀ = 45

Number of red balls = 5 + 51 = 56 

Number of black balls = 2 + 27 = 29 

1 u + 4 p = 56 --- (1)

1 u + 1 p = 29 ---(2)

(1) - (2)
(1 u + 4 p) - (1 u + 1 p) = 56 - 29
4 p - 1 p = 27
3 p = 27

1 p = 27 ÷ 3 = 9

From (2):
1 u + 1 p = 29
1 u + 1 x 9 = 29
1 u + 9 = 29

1 u = 29 - 9 = 20

Number of black balls to be removed from Box V to Box U
= 27 - 1 p
= 27 - 1 x 9
= 27 - 9
= 18

Number of red balls to be removed from Box V to Box U
= 1 u - 5
= 20 - 5
= 15

Total number of black and red balls to be removed from Box V to Box U
= 18 + 15
= 33

Answer(s): 33