There were some persimmons in 3 bags, L, M and N. 25% of the number of persimmons in Bag L was equal to 10% of the number of persimmons in Bag M. The number of persimmons in Bag N was 60% of the total number of persimmons. After Ryan removed 10% of the persimmons in Bag N, there were 356 more persimmons in Bag N than in Bag M. In the end, how many persimmons should be transferred from Bag M to Bag N so that the number of persimmons in Bag L would be the same as Bag M?

Bag L	Bag M	Bag N
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Bag L	Bag M	Bag N
Before	4 u	10 u	21 u
Change			- 2.1 u
After	4 u	10 u	18.9 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag L = ¹₁₀ Bag M

Bag L : Bag M
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of persimmons in Bag L and Bag M is the combined repeated identity.  Make the total number of persimmons in Box L and Bag M the same.  LCM of 7 and 2 = 14

Number of persimmons removed from Bag N
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of persimmons left in Bag N
= 21 u - 2.1 u
= 18.9 u

Number of more persimmons in Bag N than Bag M
= 18.9 u - 10 u
= 8.9 u

8.9 u = 356

1 u = 356 ÷ 8.9 = 40

	Bag L	Bag M	Bag N
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of persimmons to be transferred from Bag M to Bag N so that the number of persimmons in Bag L would be the same as Bag M.
= 10 u - 4 u
= 6 u
= 6 x 40
= 240

Answer(s): 240