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

Bag W	Bag X	Bag Y
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Bag W	Bag X	Bag Y
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 W = ¹₁₀ Bag X

Bag W : Bag X
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of dragonfruits in Bag W and Bag X is the combined repeated identity.  Make the total number of dragonfruits in Box W and Bag X the same.  LCM of 3 and 2 = 6

Number of dragonfruits removed from Bag Y
= ²⁵₁₀₀ x 9 u
= 2.25 u

Number of dragonfruits left in Bag Y
= 9 u - 2.25 u
= 6.75 u

Number of more dragonfruits in Bag Y than Bag X
= 6.75 u - 4 u
= 2.75 u

2.75 u = 55

1 u = 55 ÷ 2.75 = 20

	Bag W	Bag X	Bag Y
Before	2 u	4 u	6.75 u
Change		- 2 u	+ 2 u
After	2 u	2 u	8.75 u

Number of dragonfruits to be transferred from Bag X to Bag Y so that the number of dragonfruits in Bag W would be the same as Bag X.
= 4 u - 2 u
= 2 u
= 2 x 20
= 40

Answer(s): 40