There were some mangoes in 3 bags, X, Y and Z. 20% of the number of mangoes in Bag X was equal to 10% of the number of mangoes in Bag Y. The number of mangoes in Bag Z was 60% of the total number of mangoes. After John removed 25% of the mangoes in Bag Z, there were 110 more mangoes in Bag Z than in Bag Y. In the end, how many mangoes should be transferred from Bag Y to Bag Z so that the number of mangoes in Bag X would be the same as Bag Y?

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

	Bag X	Bag Y	Bag Z
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 X = ¹₁₀ Bag Y

Bag X : Bag Y
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

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

Number of mangoes removed from Bag Z
= ²⁵₁₀₀ x 9 u
= 2.25 u

Number of mangoes left in Bag Z
= 9 u - 2.25 u
= 6.75 u

Number of more mangoes in Bag Z than Bag Y
= 6.75 u - 4 u
= 2.75 u

2.75 u = 110

1 u = 110 ÷ 2.75 = 40

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

Number of mangoes to be transferred from Bag Y to Bag Z so that the number of mangoes in Bag X would be the same as Bag Y.
= 4 u - 2 u
= 2 u
= 2 x 40
= 80

Answer(s): 80