Mark had three bags of soya beans, A, B and C and the mass of each bag of soya beans was in the ratio 4 : 5 : 3. Mark decided to transfer 30% of soya beans from Bag A into Bag B and 80% of soya beans from Bag C into Bag B. Given that the mass of Bag B was 51.6 kg in the end, how many kilograms of soya beans were transferred into Bag B?

	A	B	C
Before	4 u	5 u	3 u
Change 1	- 1.2 u	+ 1.2 u
Change 2		+ 2.4 u	- 2.4 u
After	2.8 u	8.6 u	0.6 u

Mass of soya beans transferred from Bag A into Bag B
= 30% x 4 u
= ³⁰₁₀₀ x 4 u
= 2.8 u

Mass of soya beans transferred from Bag C into Bag B
= 80% x 3 u
= ⁸⁰₁₀₀ x 3 u
= 2.4 u

Mass of soya beans in Bag B in the end
= 5 u + 1.2 u + 2.4 u
= 8.6 u

8.6 u = 51.6 

1 u = 51.6 ÷ 8.6 = 6

Mass of soya beans transferred into Bag B
= 1.2 u + 2.4 u
= 3.6 u
= 3.6 x 6
= 21.6 kg

Answer(s): 21.6 kg