Daniel had three bags of soya beans, M, N and P and the mass of each bag of soya beans was in the ratio 2 : 1 : 1. Daniel decided to transfer 10% of soya beans from Bag M into Bag N and 80% of soya beans from Bag P into Bag N. Given that the mass of Bag N was 8 kg in the end, how many kilograms of soya beans were transferred into Bag N?

	M	N	P
Before	2 u	1 u	1 u
Change 1	- 0.2 u	+ 0.2 u
Change 2		+ 0.8 u	- 0.8 u
After	1.8 u	2 u	0.2 u

Mass of soya beans transferred from Bag M into Bag N
= 10% x 2 u
= ¹⁰₁₀₀ x 2 u
= 1.8 u

Mass of soya beans transferred from Bag P into Bag N
= 80% x 1 u
= ⁸⁰₁₀₀ x 1 u
= 0.8 u

Mass of soya beans in Bag N in the end
= 1 u + 0.2 u + 0.8 u
= 2 u

2 u = 8 

1 u = 8 ÷ 2 = 4

Mass of soya beans transferred into Bag N
= 0.2 u + 0.8 u
= 1 u
= 1 x 4
= 4 kg

Answer(s): 4 kg