There were some mangosteens in 3 bags, T, U and V. 20% of the number of mangosteens in Bag T was equal to 10% of the number of mangosteens in Bag U. The number of mangosteens in Bag V was 60% of the total number of mangosteens. After Brandon removed 10% of the mangosteens in Bag V, there were 205 more mangosteens in Bag V than in Bag U. In the end, how many mangosteens should be transferred from Bag U to Bag V so that the number of mangosteens in Bag T would be the same as Bag U?

Bag T	Bag U	Bag V
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Bag T	Bag U	Bag V
Before	2 u	4 u	9 u
Change			- 0.9 u
After	2 u	4 u	8.1 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Bag T = ¹₁₀ Bag U

Bag T : Bag U
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of mangosteens in Bag T and Bag U is the combined repeated identity.  Make the total number of mangosteens in Box T and Bag U the same.  LCM of 3 and 2 = 6

Number of mangosteens removed from Bag V
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of mangosteens left in Bag V
= 9 u - 0.9 u
= 8.1 u

Number of more mangosteens in Bag V than Bag U
= 8.1 u - 4 u
= 4.1 u

4.1 u = 205

1 u = 205 ÷ 4.1 = 50

	Bag T	Bag U	Bag V
Before	2 u	4 u	8.1 u
Change		- 2 u	+ 2 u
After	2 u	2 u	10.1 u

Number of mangosteens to be transferred from Bag U to Bag V so that the number of mangosteens in Bag T would be the same as Bag U.
= 4 u - 2 u
= 2 u
= 2 x 50
= 100

Answer(s): 100