There were some peaches in 3 baskets, G, H and J. 25% of the number of peaches in Basket G was equal to 10% of the number of peaches in Basket H. The number of peaches in Basket J was 60% of the total number of peaches. After Billy removed 25% of the peaches in Basket J, there were 230 more peaches in Basket J than in Basket H. In the end, how many peaches should be transferred from Basket H to Basket J so that the number of peaches in Basket G would be the same as Basket H?

Basket G	Basket H	Basket J
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket G	Basket H	Basket J
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 G = ¹₁₀ Basket H

Basket G : Basket H
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of peaches in Basket G and Basket H is the combined repeated identity.  Make the total number of peaches in Box G and Basket H the same.  LCM of 7 and 2 = 14

Number of peaches removed from Basket J
= ²⁵₁₀₀ x 21 u
= 5.25 u

Number of peaches left in Basket J
= 21 u - 5.25 u
= 15.75 u

Number of more peaches in Basket J than Basket H
= 15.75 u - 10 u
= 5.75 u

5.75 u = 230

1 u = 230 ÷ 5.75 = 40

	Basket G	Basket H	Basket J
Before	4 u	10 u	15.75 u
Change		- 6 u	+ 6 u
After	4 u	4 u	21.75 u

Number of peaches to be transferred from Basket H to Basket J so that the number of peaches in Basket G would be the same as Basket H.
= 10 u - 4 u
= 6 u
= 6 x 40
= 240

Answer(s): 240