There were some starfruits in 3 baskets, U, V and W. 20% of the number of starfruits in Basket U was equal to 10% of the number of starfruits in Basket V. The number of starfruits in Basket W was 60% of the total number of starfruits. After Glen removed 10% of the starfruits in Basket W, there were 82 more starfruits in Basket W than in Basket V. In the end, how many starfruits should be transferred from Basket V to Basket W so that the number of starfruits in Basket U would be the same as Basket V?

Basket U	Basket V	Basket W
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Basket U	Basket V	Basket W
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 Basket U = ¹₁₀ Basket V

Basket U : Basket V
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of starfruits in Basket U and Basket V is the combined repeated identity.  Make the total number of starfruits in Box U and Basket V the same.  LCM of 3 and 2 = 6

Number of starfruits removed from Basket W
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of starfruits left in Basket W
= 9 u - 0.9 u
= 8.1 u

Number of more starfruits in Basket W than Basket V
= 8.1 u - 4 u
= 4.1 u

4.1 u = 82

1 u = 82 ÷ 4.1 = 20

	Basket U	Basket V	Basket W
Before	2 u	4 u	8.1 u
Change		- 2 u	+ 2 u
After	2 u	2 u	10.1 u

Number of starfruits to be transferred from Basket V to Basket W so that the number of starfruits in Basket U would be the same as Basket V.
= 4 u - 2 u
= 2 u
= 2 x 20
= 40

Answer(s): 40