There were some passion fruits in 3 baskets, N, P and Q. 25% of the number of passion fruits in Basket N was equal to 10% of the number of passion fruits in Basket P. The number of passion fruits in Basket Q was 60% of the total number of passion fruits. After Luis removed 25% of the passion fruits in Basket Q, there were 230 more passion fruits in Basket Q than in Basket P. In the end, how many passion fruits should be transferred from Basket P to Basket Q so that the number of passion fruits in Basket N would be the same as Basket P?

Basket N	Basket P	Basket Q
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket N	Basket P	Basket Q
Before	4 u	10 u	21 u
Change			- 5.25 u
After	4 u	10 u	15.75 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Basket N = ¹₁₀ Basket P

Basket N : Basket P
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

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

Number of passion fruits removed from Basket Q
= ²⁵₁₀₀ x 21 u
= 5.25 u

Number of passion fruits left in Basket Q
= 21 u - 5.25 u
= 15.75 u

Number of more passion fruits in Basket Q than Basket P
= 15.75 u - 10 u
= 5.75 u

5.75 u = 230

1 u = 230 ÷ 5.75 = 40

	Basket N	Basket P	Basket Q
Before	4 u	10 u	15.75 u
Change		- 6 u	+ 6 u
After	4 u	4 u	21.75 u

Number of passion fruits to be transferred from Basket P to Basket Q so that the number of passion fruits in Basket N would be the same as Basket P.
= 10 u - 4 u
= 6 u
= 6 x 40
= 240

Answer(s): 240