Peter had three bags of brown rice, L, M and N and the mass of each bag of brown rice was in the ratio 4 : 3 : 4. Peter decided to transfer 20% of brown rice from Bag L into Bag M and 90% of brown rice from Bag N into Bag M. Given that the mass of Bag M was 29.6 kg in the end, how many kilograms of brown rice were transferred into Bag M?

	L	M	N
Before	4 u	3 u	4 u
Change 1	- 0.8 u	+ 0.8 u
Change 2		+ 3.6 u	- 3.6 u
After	3.2 u	7.4 u	0.4 u

Mass of brown rice transferred from Bag L into Bag M
= 20% x 4 u
= ²⁰₁₀₀ x 4 u
= 3.2 u

Mass of brown rice transferred from Bag N into Bag M
= 90% x 4 u
= ⁹⁰₁₀₀ x 4 u
= 3.6 u

Mass of brown rice in Bag M in the end
= 3 u + 0.8 u + 3.6 u
= 7.4 u

7.4 u = 29.6 

1 u = 29.6 ÷ 7.4 = 4

Mass of brown rice transferred into Bag M
= 0.8 u + 3.6 u
= 4.4 u
= 4.4 x 4
= 17.6 kg

Answer(s): 17.6 kg