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

Basket L	Basket M	Basket N
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Basket L	Basket M	Basket N
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 L = ¹₁₀ Basket M

Basket L : Basket M
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of peaches in Basket L and Basket M is the combined repeated identity.  Make the total number of peaches in Box L and Basket M the same.  LCM of 7 and 3 = 21

Number of peaches removed from Basket N
= ²⁵₁₀₀ x 49 u
= 12.25 u

Number of peaches left in Basket N
= 49 u - 12.25 u
= 36.75 u

Number of more peaches in Basket N than Basket M
= 36.75 u - 15 u
= 21.75 u

21.75 u = 870

1 u = 870 ÷ 21.75 = 40

	Basket L	Basket M	Basket N
Before	6 u	15 u	36.75 u
Change		- 9 u	+ 9 u
After	6 u	6 u	45.75 u

Number of peaches to be transferred from Basket M to Basket N so that the number of peaches in Basket L would be the same as Basket M.
= 15 u - 6 u
= 9 u
= 9 x 40
= 360

Answer(s): 360