There were some chikoos in 3 bags, P, Q and R. 25% of the number of chikoos in Bag P was equal to 10% of the number of chikoos in Bag Q. The number of chikoos in Bag R was 60% of the total number of chikoos. After Archie removed 10% of the chikoos in Bag R, there were 356 more chikoos in Bag R than in Bag Q. In the end, how many chikoos should be transferred from Bag Q to Bag R so that the number of chikoos in Bag P would be the same as Bag Q?

Bag P	Bag Q	Bag R
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Bag P	Bag Q	Bag R
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 Bag P = ¹₁₀ Bag Q

Bag P : Bag Q
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

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

Number of chikoos removed from Bag R
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of chikoos left in Bag R
= 21 u - 2.1 u
= 18.9 u

Number of more chikoos in Bag R than Bag Q
= 18.9 u - 10 u
= 8.9 u

8.9 u = 356

1 u = 356 ÷ 8.9 = 40

	Bag P	Bag Q	Bag R
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 Bag Q to Bag R so that the number of chikoos in Bag P would be the same as Bag Q.
= 10 u - 4 u
= 6 u
= 6 x 40
= 240

Answer(s): 240