There were some mangoes in 3 baskets, E, F and G. 25% of the number of mangoes in Basket E was equal to 10% of the number of mangoes in Basket F. The number of mangoes in Basket G was 70% of the total number of mangoes. After Ian removed 10% of the mangoes in Basket G, there were 291 more mangoes in Basket G than in Basket F. In the end, how many mangoes should be transferred from Basket F to Basket G so that the number of mangoes in Basket E would be the same as Basket F?

Basket E	Basket F	Basket G
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Basket E	Basket F	Basket G
Before	6 u	15 u	49 u
Change			- 4.9 u
After	6 u	15 u	44.1 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Basket E = ¹₁₀ Basket F

Basket E : Basket F
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of mangoes in Basket E and Basket F is the combined repeated identity.  Make the total number of mangoes in Box E and Basket F the same.  LCM of 7 and 3 = 21

Number of mangoes removed from Basket G
= ¹⁰₁₀₀ x 49 u
= 4.9 u

Number of mangoes left in Basket G
= 49 u - 4.9 u
= 44.1 u

Number of more mangoes in Basket G than Basket F
= 44.1 u - 15 u
= 29.1 u

29.1 u = 291

1 u = 291 ÷ 29.1 = 10

	Basket E	Basket F	Basket G
Before	6 u	15 u	44.1 u
Change		- 9 u	+ 9 u
After	6 u	6 u	53.1 u

Number of mangoes to be transferred from Basket F to Basket G so that the number of mangoes in Basket E would be the same as Basket F.
= 15 u - 6 u
= 9 u
= 9 x 10
= 90

Answer(s): 90