Wesley had three bags of corn starch, D, E and F and the mass of each bag of corn starch was in the ratio 1 : 1 : 3. Wesley decided to transfer 10% of corn starch from Bag D into Bag E and 60% of corn starch from Bag F into Bag E. Given that the mass of Bag E was 29 kg in the end, how many kilograms of corn starch were transferred into Bag E?

	D	E	F
Before	1 u	1 u	3 u
Change 1	- 0.1 u	+ 0.1 u
Change 2		+ 1.8 u	- 1.8 u
After	0.9 u	2.9 u	1.2 u

Mass of corn starch transferred from Bag D into Bag E
= 10% x 1 u
= ¹⁰₁₀₀ x 1 u
= 0.9 u

Mass of corn starch transferred from Bag F into Bag E
= 60% x 3 u
= ⁶⁰₁₀₀ x 3 u
= 1.8 u

Mass of corn starch in Bag E in the end
= 1 u + 0.1 u + 1.8 u
= 2.9 u

2.9 u = 29 

1 u = 29 ÷ 2.9 = 10

Mass of corn starch transferred into Bag E
= 0.1 u + 1.8 u
= 1.9 u
= 1.9 x 10
= 19 kg

Answer(s): 19 kg