Jack had three bags of potato starch, T, U and V and the mass of each bag of potato starch was in the ratio 2 : 5 : 5. Jack decided to transfer 20% of potato starch from Bag T into Bag U and 90% of potato starch from Bag V into Bag U. Given that the mass of Bag U was 99 kg in the end, how many kilograms of potato starch were transferred into Bag U?

	T	U	V
Before	2 u	5 u	5 u
Change 1	- 0.4 u	+ 0.4 u
Change 2		+ 4.5 u	- 4.5 u
After	1.6 u	9.9 u	0.5 u

Mass of potato starch transferred from Bag T into Bag U
= 20% x 2 u
= ²⁰₁₀₀ x 2 u
= 1.6 u

Mass of potato starch transferred from Bag V into Bag U
= 90% x 5 u
= ⁹⁰₁₀₀ x 5 u
= 4.5 u

Mass of potato starch in Bag U in the end
= 5 u + 0.4 u + 4.5 u
= 9.9 u

9.9 u = 99 

1 u = 99 ÷ 9.9 = 10

Mass of potato starch transferred into Bag U
= 0.4 u + 4.5 u
= 4.9 u
= 4.9 x 10
= 49 kg

Answer(s): 49 kg