Elijah had three bags of black beans, S, T and U and the mass of each bag of black beans was in the ratio 5 : 2 : 3. Elijah decided to transfer 10% of black beans from Bag S into Bag T and 90% of black beans from Bag U into Bag T. Given that the mass of Bag T was 52 kg in the end, how many kilograms of black beans were transferred into Bag T?

	S	T	U
Before	5 u	2 u	3 u
Change 1	- 0.5 u	+ 0.5 u
Change 2		+ 2.7 u	- 2.7 u
After	4.5 u	5.2 u	0.3 u

Mass of black beans transferred from Bag S into Bag T
= 10% x 5 u
= ¹⁰₁₀₀ x 5 u
= 4.5 u

Mass of black beans transferred from Bag U into Bag T
= 90% x 3 u
= ⁹⁰₁₀₀ x 3 u
= 2.7 u

Mass of black beans in Bag T in the end
= 2 u + 0.5 u + 2.7 u
= 5.2 u

5.2 u = 52 

1 u = 52 ÷ 5.2 = 10

Mass of black beans transferred into Bag T
= 0.5 u + 2.7 u
= 3.2 u
= 3.2 x 10
= 32 kg

Answer(s): 32 kg