There were some persimmons in 3 baskets, L, M and N. 25% of the number of persimmons in Basket L was equal to 10% of the number of persimmons in Basket M. The number of persimmons in Basket N was 70% of the total number of persimmons. After Cole removed 10% of the persimmons in Basket N, there were 1455 more persimmons in Basket N than in Basket M. In the end, how many persimmons should be transferred from Basket M to Basket N so that the number of persimmons 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			- 4.9 u
After	6 u	15 u	44.1 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 persimmons in Basket L and Basket M is the combined repeated identity.  Make the total number of persimmons in Box L and Basket M the same.  LCM of 7 and 3 = 21

Number of persimmons removed from Basket N
= ¹⁰₁₀₀ x 49 u
= 4.9 u

Number of persimmons left in Basket N
= 49 u - 4.9 u
= 44.1 u

Number of more persimmons in Basket N than Basket M
= 44.1 u - 15 u
= 29.1 u

29.1 u = 1455

1 u = 1455 ÷ 29.1 = 50

	Basket L	Basket M	Basket N
Before	6 u	15 u	44.1 u
Change		- 9 u	+ 9 u
After	6 u	6 u	53.1 u

Number of persimmons to be transferred from Basket M to Basket N so that the number of persimmons in Basket L would be the same as Basket M.
= 15 u - 6 u
= 9 u
= 9 x 50
= 450

Answer(s): 450