There were some chikoos in 3 bags, N, P and Q. 20% of the number of chikoos in Bag N was equal to 10% of the number of chikoos in Bag P. The number of chikoos in Bag Q was 60% of the total number of chikoos. After Sam removed 25% of the chikoos in Bag Q, there were 55 more chikoos in Bag Q than in Bag P. In the end, how many chikoos should be transferred from Bag P to Bag Q so that the number of chikoos in Bag N would be the same as Bag P?

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

	Bag N	Bag P	Bag Q
Before	2 u	4 u	9 u
Change			- 2.25 u
After	2 u	4 u	6.75 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Bag N = ¹₁₀ Bag P

Bag N : Bag P
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

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

Number of chikoos removed from Bag Q
= ²⁵₁₀₀ x 9 u
= 2.25 u

Number of chikoos left in Bag Q
= 9 u - 2.25 u
= 6.75 u

Number of more chikoos in Bag Q than Bag P
= 6.75 u - 4 u
= 2.75 u

2.75 u = 55

1 u = 55 ÷ 2.75 = 20

	Bag N	Bag P	Bag Q
Before	2 u	4 u	6.75 u
Change		- 2 u	+ 2 u
After	2 u	2 u	8.75 u

Number of chikoos to be transferred from Bag P to Bag Q so that the number of chikoos in Bag N would be the same as Bag P.
= 4 u - 2 u
= 2 u
= 2 x 20
= 40

Answer(s): 40