There were some passion fruits in 3 baskets, C, D and E. 25% of the number of passion fruits in Basket C was equal to 10% of the number of passion fruits in Basket D. The number of passion fruits in Basket E was 60% of the total number of passion fruits. After Howard removed 20% of the passion fruits in Basket E, there were 340 more passion fruits in Basket E than in Basket D. In the end, how many passion fruits should be transferred from Basket D to Basket E so that the number of passion fruits in Basket C would be the same as Basket D?

Basket C	Basket D	Basket E
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket C	Basket D	Basket E
Before	4 u	10 u	21 u
Change			- 4.2 u
After	4 u	10 u	16.8 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Basket C = ¹₁₀ Basket D

Basket C : Basket D
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of passion fruits in Basket C and Basket D is the combined repeated identity.  Make the total number of passion fruits in Box C and Basket D the same.  LCM of 7 and 2 = 14

Number of passion fruits removed from Basket E
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of passion fruits left in Basket E
= 21 u - 4.2 u
= 16.8 u

Number of more passion fruits in Basket E than Basket D
= 16.8 u - 10 u
= 6.8 u

6.8 u = 340

1 u = 340 ÷ 6.8 = 50

	Basket C	Basket D	Basket E
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of passion fruits to be transferred from Basket D to Basket E so that the number of passion fruits in Basket C would be the same as Basket D.
= 10 u - 4 u
= 6 u
= 6 x 50
= 300

Answer(s): 300