There were some guavas in 3 baskets, E, F and G. 25% of the number of guavas in Basket E was equal to 10% of the number of guavas in Basket F. The number of guavas in Basket G was 70% of the total number of guavas. After Bryan removed 20% of the guavas in Basket G, there were 484 more guavas in Basket G than in Basket F. In the end, how many guavas should be transferred from Basket F to Basket G so that the number of guavas 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			- 9.8 u
After	6 u	15 u	39.2 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 guavas in Basket E and Basket F is the combined repeated identity.  Make the total number of guavas in Box E and Basket F the same.  LCM of 7 and 3 = 21

Number of guavas removed from Basket G
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of guavas left in Basket G
= 49 u - 9.8 u
= 39.2 u

Number of more guavas in Basket G than Basket F
= 39.2 u - 15 u
= 24.2 u

24.2 u = 484

1 u = 484 ÷ 24.2 = 20

	Basket E	Basket F	Basket G
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of guavas to be transferred from Basket F to Basket G so that the number of guavas in Basket E would be the same as Basket F.
= 15 u - 6 u
= 9 u
= 9 x 20
= 180

Answer(s): 180