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

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

	Bag L	Bag M	Bag 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 Bag L = ¹₁₀ Bag M

Bag L : Bag M
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

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

Number of guavas removed from Bag N
= ²⁵₁₀₀ x 49 u
= 12.25 u

Number of guavas left in Bag N
= 49 u - 12.25 u
= 36.75 u

Number of more guavas in Bag N than Bag M
= 36.75 u - 15 u
= 21.75 u

21.75 u = 870

1 u = 870 ÷ 21.75 = 40

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

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

Answer(s): 360