Asher had three bags of black beans, A, B and C and the mass of each bag of black beans was in the ratio 5 : 4 : 4. Asher decided to transfer 40% of black beans from Bag A into Bag B and 70% of black beans from Bag C into Bag B. Given that the mass of Bag B was 26.4 kg in the end, how many kilograms of black beans were transferred into Bag B?

	A	B	C
Before	5 u	4 u	4 u
Change 1	- 2 u	+ 2 u
Change 2		+ 2.8 u	- 2.8 u
After	3 u	8.8 u	1.2 u

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

Mass of black beans transferred from Bag C into Bag B
= 70% x 4 u
= ⁷⁰₁₀₀ x 4 u
= 2.8 u

Mass of black beans in Bag B in the end
= 4 u + 2 u + 2.8 u
= 8.8 u

8.8 u = 26.4 

1 u = 26.4 ÷ 8.8 = 3

Mass of black beans transferred into Bag B
= 2 u + 2.8 u
= 4.8 u
= 4.8 x 3
= 14.4 kg

Answer(s): 14.4 kg