There were some peaches in 3 bags, F, G and H. 25% of the number of peaches in Bag F was equal to 10% of the number of peaches in Bag G. The number of peaches in Bag H was 70% of the total number of peaches. After Ben removed 20% of the peaches in Bag H, there were 1210 more peaches in Bag H than in Bag G. In the end, how many peaches should be transferred from Bag G to Bag H so that the number of peaches in Bag F would be the same as Bag G?

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

	Bag F	Bag G	Bag H
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 Bag F = ¹₁₀ Bag G

Bag F : Bag G
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

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

Number of peaches removed from Bag H
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of peaches left in Bag H
= 49 u - 9.8 u
= 39.2 u

Number of more peaches in Bag H than Bag G
= 39.2 u - 15 u
= 24.2 u

24.2 u = 1210

1 u = 1210 ÷ 24.2 = 50

	Bag F	Bag G	Bag H
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of peaches to be transferred from Bag G to Bag H so that the number of peaches in Bag F would be the same as Bag G.
= 15 u - 6 u
= 9 u
= 9 x 50
= 450

Answer(s): 450