Ken had three bags of red beans, X, Y and Z and the mass of each bag of red beans was in the ratio 3 : 5 : 1. Ken decided to transfer 30% of red beans from Bag X into Bag Y and 80% of red beans from Bag Z into Bag Y. Given that the mass of Bag Y was 40.2 kg in the end, how many kilograms of red beans were transferred into Bag Y?

	X	Y	Z
Before	3 u	5 u	1 u
Change 1	- 0.9 u	+ 0.9 u
Change 2		+ 0.8 u	- 0.8 u
After	2.1 u	6.7 u	0.2 u

Mass of red beans transferred from Bag X into Bag Y
= 30% x 3 u
= ³⁰₁₀₀ x 3 u
= 2.1 u

Mass of red beans transferred from Bag Z into Bag Y
= 80% x 1 u
= ⁸⁰₁₀₀ x 1 u
= 0.8 u

Mass of red beans in Bag Y in the end
= 5 u + 0.9 u + 0.8 u
= 6.7 u

6.7 u = 40.2 

1 u = 40.2 ÷ 6.7 = 6

Mass of red beans transferred into Bag Y
= 0.9 u + 0.8 u
= 1.7 u
= 1.7 x 6
= 10.2 kg

Answer(s): 10.2 kg