There were some lemons in 3 bags, L, M and N. 25% of the number of lemons in Bag L was equal to 10% of the number of lemons in Bag M. The number of lemons in Bag N was 60% of the total number of lemons. After Flynn removed 20% of the lemons in Bag N, there were 340 more lemons in Bag N than in Bag M. In the end, how many lemons should be transferred from Bag M to Bag N so that the number of lemons 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			- 4.2 u
After	4 u	10 u	16.8 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 lemons in Bag L and Bag M is the combined repeated identity.  Make the total number of lemons in Box L and Bag M the same.  LCM of 7 and 2 = 14

Number of lemons removed from Bag N
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of lemons left in Bag N
= 21 u - 4.2 u
= 16.8 u

Number of more lemons in Bag N than Bag M
= 16.8 u - 10 u
= 6.8 u

6.8 u = 340

1 u = 340 ÷ 6.8 = 50

	Bag L	Bag M	Bag N
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of lemons to be transferred from Bag M to Bag N so that the number of lemons in Bag L would be the same as Bag M.
= 10 u - 4 u
= 6 u
= 6 x 50
= 300

Answer(s): 300