There were some passion fruits in 3 baskets, T, U and V. 25% of the number of passion fruits in Basket T was equal to 10% of the number of passion fruits in Basket U. The number of passion fruits in Basket V was 70% of the total number of passion fruits. After Tom removed 20% of the passion fruits in Basket V, there were 484 more passion fruits in Basket V than in Basket U. In the end, how many passion fruits should be transferred from Basket U to Basket V so that the number of passion fruits in Basket T would be the same as Basket U?

Basket T	Basket U	Basket V
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Basket T	Basket U	Basket V
Before	6 u	15 u	49 u
Change			- 9.8 u
After	6 u	15 u	39.2 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Basket T = ¹₁₀ Basket U

Basket T : Basket U
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of passion fruits in Basket T and Basket U is the combined repeated identity.  Make the total number of passion fruits in Box T and Basket U the same.  LCM of 7 and 3 = 21

Number of passion fruits removed from Basket V
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of passion fruits left in Basket V
= 49 u - 9.8 u
= 39.2 u

Number of more passion fruits in Basket V than Basket U
= 39.2 u - 15 u
= 24.2 u

24.2 u = 484

1 u = 484 ÷ 24.2 = 20

	Basket T	Basket U	Basket V
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of passion fruits to be transferred from Basket U to Basket V so that the number of passion fruits in Basket T would be the same as Basket U.
= 15 u - 6 u
= 9 u
= 9 x 20
= 180

Answer(s): 180