There were some chikoos in 3 bags, E, F and G. 25% of the number of chikoos in Bag E was equal to 10% of the number of chikoos in Bag F. The number of chikoos in Bag G was 60% of the total number of chikoos. After Albert removed 20% of the chikoos in Bag G, there were 272 more chikoos in Bag G than in Bag F. In the end, how many chikoos should be transferred from Bag F to Bag G so that the number of chikoos in Bag E would be the same as Bag F?

Bag E	Bag F	Bag G
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Bag E	Bag F	Bag G
Before	4 u	10 u	21 u
Change			- 4.2 u
After	4 u	10 u	16.8 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag E = ¹₁₀ Bag F

Bag E : Bag F
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

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

Number of chikoos removed from Bag G
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of chikoos left in Bag G
= 21 u - 4.2 u
= 16.8 u

Number of more chikoos in Bag G than Bag F
= 16.8 u - 10 u
= 6.8 u

6.8 u = 272

1 u = 272 ÷ 6.8 = 40

	Bag E	Bag F	Bag G
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of chikoos to be transferred from Bag F to Bag G so that the number of chikoos in Bag E would be the same as Bag F.
= 10 u - 4 u
= 6 u
= 6 x 40
= 240

Answer(s): 240