Container E contains 7 yellow balls and 14 red balls. Container F contains 71 yellow balls and 24 red balls. How many red balls and yellow balls must be removed from Container F to put into Container E so that 50% of the balls in Container A are yellow and 75% of the balls in Container F are yellow?

	Container E		Container F
	Yellow balls	Red balls	Yellow balls	Red balls
Before	7	14	71	24
Change	+ ?	+ ?	- ?	- ?
After	1 u	1 u	3 p	1 p

50% = ⁵⁰₁₀₀ = 12

75% = ⁷⁵₁₀₀ = 34

Number of yellow balls = 7 + 71 = 78 

Number of red balls = 14 + 24 = 38 

1 u + 3 p = 78 --- (1)

1 u + 1 p = 38 ---(2)

(1) - (2)
(1 u + 3 p) - (1 u + 1 p) = 78 - 38
3 p - 1 p = 40
2 p = 40

1 p = 40 ÷ 2 = 20

From (2):
1 u + 1 p = 38
1 u + 1 x 20 = 38
1 u + 20 = 38

1 u = 38 - 20 = 18

Number of red balls to be removed from Container F to Container E
= 24 - 1 p
= 24 - 1 x 20
= 24 - 20
= 4

Number of yellow balls to be removed from Container F to Container E
= 1 u - 7
= 18 - 7
= 11

Total number of red and yellow balls to be removed from Container F to Container E
= 4 + 11
= 15

Answer(s): 15