There were some chikoos in 3 boxes, L, M and N. 25% of the number of chikoos in Box L was equal to 10% of the number of chikoos in Box M. The number of chikoos in Box N was 70% of the total number of chikoos. After Will removed 10% of the chikoos in Box N, there were 873 more chikoos in Box N than in Box M. In the end, how many chikoos should be transferred from Box M to Box N so that the number of chikoos in Box L would be the same as Box M?

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

	Box L	Box M	Box 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 Box L = ¹₁₀ Box M

Box L : Box M
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

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

Number of chikoos removed from Box N
= ¹⁰₁₀₀ x 49 u
= 4.9 u

Number of chikoos left in Box N
= 49 u - 4.9 u
= 44.1 u

Number of more chikoos in Box N than Box M
= 44.1 u - 15 u
= 29.1 u

29.1 u = 873

1 u = 873 ÷ 29.1 = 30

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

Number of chikoos to be transferred from Box M to Box N so that the number of chikoos in Box L would be the same as Box M.
= 15 u - 6 u
= 9 u
= 9 x 30
= 270

Answer(s): 270