Brandon had three bags of black beans, K, L and M and the mass of each bag of black beans was in the ratio 5 : 4 : 4. Brandon decided to transfer 40% of black beans from Bag K into Bag L and 80% of black beans from Bag M into Bag L. Given that the mass of Bag L was 18.4 kg in the end, how many kilograms of black beans were transferred into Bag L?

	K	L	M
Before	5 u	4 u	4 u
Change 1	- 2 u	+ 2 u
Change 2		+ 3.2 u	- 3.2 u
After	3 u	9.2 u	0.8 u

Mass of black beans transferred from Bag K into Bag L
= 40% x 5 u
= ⁴⁰₁₀₀ x 5 u
= 3 u

Mass of black beans transferred from Bag M into Bag L
= 80% x 4 u
= ⁸⁰₁₀₀ x 4 u
= 3.2 u

Mass of black beans in Bag L in the end
= 4 u + 2 u + 3.2 u
= 9.2 u

9.2 u = 18.4 

1 u = 18.4 ÷ 9.2 = 2

Mass of black beans transferred into Bag L
= 2 u + 3.2 u
= 5.2 u
= 5.2 x 2
= 10.4 kg

Answer(s): 10.4 kg