Tom had three bags of green beans, J, K and L and the mass of each bag of green beans was in the ratio 4 : 5 : 2. Tom decided to transfer 10% of green beans from Bag J into Bag K and 80% of green beans from Bag L into Bag K. Given that the mass of Bag K was 14 kg in the end, how many kilograms of green beans were transferred into Bag K?

	J	K	L
Before	4 u	5 u	2 u
Change 1	- 0.4 u	+ 0.4 u
Change 2		+ 1.6 u	- 1.6 u
After	3.6 u	7 u	0.4 u

Mass of green beans transferred from Bag J into Bag K
= 10% x 4 u
= ¹⁰₁₀₀ x 4 u
= 3.6 u

Mass of green beans transferred from Bag L into Bag K
= 80% x 2 u
= ⁸⁰₁₀₀ x 2 u
= 1.6 u

Mass of green beans in Bag K in the end
= 5 u + 0.4 u + 1.6 u
= 7 u

7 u = 14 

1 u = 14 ÷ 7 = 2

Mass of green beans transferred into Bag K
= 0.4 u + 1.6 u
= 2 u
= 2 x 2
= 4 kg

Answer(s): 4 kg