There were some pomegranates in 3 baskets, T, U and V. 20% of the number of pomegranates in Basket T was equal to 10% of the number of pomegranates in Basket U. The number of pomegranates in Basket V was 60% of the total number of pomegranates. After Warren removed 10% of the pomegranates in Basket V, there were 164 more pomegranates in Basket V than in Basket U. In the end, how many pomegranates should be transferred from Basket U to Basket V so that the number of pomegranates in Basket T would be the same as Basket U?

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

	Basket T	Basket U	Basket 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 Basket T = ¹₁₀ Basket U

Basket T : Basket U
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

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

Number of pomegranates removed from Basket V
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of pomegranates left in Basket V
= 9 u - 0.9 u
= 8.1 u

Number of more pomegranates in Basket V than Basket U
= 8.1 u - 4 u
= 4.1 u

4.1 u = 164

1 u = 164 ÷ 4.1 = 40

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

Number of pomegranates to be transferred from Basket U to Basket V so that the number of pomegranates in Basket T would be the same as Basket U.
= 4 u - 2 u
= 2 u
= 2 x 40
= 80

Answer(s): 80