There were some chikoos in 3 boxes, N, P and Q. 25% of the number of chikoos in Box N was equal to 10% of the number of chikoos in Box P. The number of chikoos in Box Q was 60% of the total number of chikoos. After Paul removed 10% of the chikoos in Box Q, there were 267 more chikoos in Box Q than in Box P. In the end, how many chikoos should be transferred from Box P to Box Q so that the number of chikoos in Box N would be the same as Box P?

Box N	Box P	Box Q
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Box N	Box P	Box Q
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 Box N = ¹₁₀ Box P

Box N : Box P
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of chikoos in Box N and Box P is the combined repeated identity.  Make the total number of chikoos in Box N and Box P the same.  LCM of 7 and 2 = 14

Number of chikoos removed from Box Q
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of chikoos left in Box Q
= 21 u - 2.1 u
= 18.9 u

Number of more chikoos in Box Q than Box P
= 18.9 u - 10 u
= 8.9 u

8.9 u = 267

1 u = 267 ÷ 8.9 = 30

	Box N	Box P	Box Q
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of chikoos to be transferred from Box P to Box Q so that the number of chikoos in Box N would be the same as Box P.
= 10 u - 4 u
= 6 u
= 6 x 30
= 180

Answer(s): 180