Ryan had three bags of green beans, X, Y and Z and the mass of each bag of green beans was in the ratio 1 : 5 : 4. Ryan decided to transfer 20% of green beans from Bag X into Bag Y and 70% of green beans from Bag Z into Bag Y. Given that the mass of Bag Y was 56 kg in the end, how many kilograms of green beans were transferred into Bag Y?

	X	Y	Z
Before	1 u	5 u	4 u
Change 1	- 0.2 u	+ 0.2 u
Change 2		+ 2.8 u	- 2.8 u
After	0.8 u	8 u	1.2 u

Mass of green beans transferred from Bag X into Bag Y
= 20% x 1 u
= ²⁰₁₀₀ x 1 u
= 0.8 u

Mass of green beans transferred from Bag Z into Bag Y
= 70% x 4 u
= ⁷⁰₁₀₀ x 4 u
= 2.8 u

Mass of green beans in Bag Y in the end
= 5 u + 0.2 u + 2.8 u
= 8 u

8 u = 56 

1 u = 56 ÷ 8 = 7

Mass of green beans transferred into Bag Y
= 0.2 u + 2.8 u
= 3 u
= 3 x 7
= 21 kg

Answer(s): 21 kg