Howard had three bags of brown rice, M, N and P and the mass of each bag of brown rice was in the ratio 2 : 5 : 1. Howard decided to transfer 10% of brown rice from Bag M into Bag N and 90% of brown rice from Bag P into Bag N. Given that the mass of Bag N was 24.4 kg in the end, how many kilograms of brown rice were transferred into Bag N?

	M	N	P
Before	2 u	5 u	1 u
Change 1	- 0.2 u	+ 0.2 u
Change 2		+ 0.9 u	- 0.9 u
After	1.8 u	6.1 u	0.1 u

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

Mass of brown rice transferred from Bag P into Bag N
= 90% x 1 u
= ⁹⁰₁₀₀ x 1 u
= 0.9 u

Mass of brown rice in Bag N in the end
= 5 u + 0.2 u + 0.9 u
= 6.1 u

6.1 u = 24.4 

1 u = 24.4 ÷ 6.1 = 4

Mass of brown rice transferred into Bag N
= 0.2 u + 0.9 u
= 1.1 u
= 1.1 x 4
= 4.4 kg

Answer(s): 4.4 kg