There were some pears in 3 bags, W, X and Y. 20% of the number of pears in Bag W was equal to 10% of the number of pears in Bag X. The number of pears in Bag Y was 60% of the total number of pears. After David removed 20% of the pears in Bag Y, there were 96 more pears in Bag Y than in Bag X. In the end, how many pears should be transferred from Bag X to Bag Y so that the number of pears 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			- 1.8 u
After	2 u	4 u	7.2 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 pears in Bag W and Bag X is the combined repeated identity.  Make the total number of pears in Box W and Bag X the same.  LCM of 3 and 2 = 6

Number of pears removed from Bag Y
= ²⁰₁₀₀ x 9 u
= 1.8 u

Number of pears left in Bag Y
= 9 u - 1.8 u
= 7.2 u

Number of more pears in Bag Y than Bag X
= 7.2 u - 4 u
= 3.2 u

3.2 u = 96

1 u = 96 ÷ 3.2 = 30

	Bag W	Bag X	Bag Y
Before	2 u	4 u	7.2 u
Change		- 2 u	+ 2 u
After	2 u	2 u	9.2 u

Number of pears to be transferred from Bag X to Bag Y so that the number of pears in Bag W would be the same as Bag X.
= 4 u - 2 u
= 2 u
= 2 x 30
= 60

Answer(s): 60