There were some persimmons in 3 bags, M, N and P. 20% of the number of persimmons in Bag M was equal to 10% of the number of persimmons in Bag N. The number of persimmons in Bag P was 60% of the total number of persimmons. After Tommy removed 20% of the persimmons in Bag P, there were 160 more persimmons in Bag P than in Bag N. In the end, how many persimmons should be transferred from Bag N to Bag P so that the number of persimmons in Bag M would be the same as Bag N?

Bag M	Bag N	Bag P
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Bag M	Bag N	Bag P
Before	2 u	4 u	9 u
Change			- 1.8 u
After	2 u	4 u	7.2 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Bag M = ¹₁₀ Bag N

Bag M : Bag N
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of persimmons in Bag M and Bag N is the combined repeated identity.  Make the total number of persimmons in Box M and Bag N the same.  LCM of 3 and 2 = 6

Number of persimmons removed from Bag P
= ²⁰₁₀₀ x 9 u
= 1.8 u

Number of persimmons left in Bag P
= 9 u - 1.8 u
= 7.2 u

Number of more persimmons in Bag P than Bag N
= 7.2 u - 4 u
= 3.2 u

3.2 u = 160

1 u = 160 ÷ 3.2 = 50

	Bag M	Bag N	Bag P
Before	2 u	4 u	7.2 u
Change		- 2 u	+ 2 u
After	2 u	2 u	9.2 u

Number of persimmons to be transferred from Bag N to Bag P so that the number of persimmons in Bag M would be the same as Bag N.
= 4 u - 2 u
= 2 u
= 2 x 50
= 100

Answer(s): 100