Tim had three bags of corn starch, E, F and G and the mass of each bag of corn starch was in the ratio 5 : 1 : 4. Tim decided to transfer 10% of corn starch from Bag E into Bag F and 70% of corn starch from Bag G into Bag F. Given that the mass of Bag F was 30.1 kg in the end, how many kilograms of corn starch were transferred into Bag F?

	E	F	G
Before	5 u	1 u	4 u
Change 1	- 0.5 u	+ 0.5 u
Change 2		+ 2.8 u	- 2.8 u
After	4.5 u	4.3 u	1.2 u

Mass of corn starch transferred from Bag E into Bag F
= 10% x 5 u
= ¹⁰₁₀₀ x 5 u
= 4.5 u

Mass of corn starch transferred from Bag G into Bag F
= 70% x 4 u
= ⁷⁰₁₀₀ x 4 u
= 2.8 u

Mass of corn starch in Bag F in the end
= 1 u + 0.5 u + 2.8 u
= 4.3 u

4.3 u = 30.1 

1 u = 30.1 ÷ 4.3 = 7

Mass of corn starch transferred into Bag F
= 0.5 u + 2.8 u
= 3.3 u
= 3.3 x 7
= 23.1 kg

Answer(s): 23.1 kg