There were some pears in 3 baskets, F, G and H. 25% of the number of pears in Basket F was equal to 10% of the number of pears in Basket G. The number of pears in Basket H was 70% of the total number of pears. After Michael removed 25% of the pears in Basket H, there were 870 more pears in Basket H than in Basket G. In the end, how many pears should be transferred from Basket G to Basket H so that the number of pears in Basket F would be the same as Basket G?

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

	Basket F	Basket G	Basket H
Before	6 u	15 u	49 u
Change			- 12.25 u
After	6 u	15 u	36.75 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Basket F = ¹₁₀ Basket G

Basket F : Basket G
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

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

Number of pears removed from Basket H
= ²⁵₁₀₀ x 49 u
= 12.25 u

Number of pears left in Basket H
= 49 u - 12.25 u
= 36.75 u

Number of more pears in Basket H than Basket G
= 36.75 u - 15 u
= 21.75 u

21.75 u = 870

1 u = 870 ÷ 21.75 = 40

	Basket F	Basket G	Basket H
Before	6 u	15 u	36.75 u
Change		- 9 u	+ 9 u
After	6 u	6 u	45.75 u

Number of pears to be transferred from Basket G to Basket H so that the number of pears in Basket F would be the same as Basket G.
= 15 u - 6 u
= 9 u
= 9 x 40
= 360

Answer(s): 360