There were some grapefruits in 3 bags, V, W and X. 25% of the number of grapefruits in Bag V was equal to 10% of the number of grapefruits in Bag W. The number of grapefruits in Bag X was 70% of the total number of grapefruits. After Andy removed 20% of the grapefruits in Bag X, there were 726 more grapefruits in Bag X than in Bag W. In the end, how many grapefruits should be transferred from Bag W to Bag X so that the number of grapefruits in Bag V would be the same as Bag W?

Bag V	Bag W	Bag X
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Bag V	Bag W	Bag X
Before	6 u	15 u	49 u
Change			- 9.8 u
After	6 u	15 u	39.2 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag V = ¹₁₀ Bag W

Bag V : Bag W
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of grapefruits in Bag V and Bag W is the combined repeated identity.  Make the total number of grapefruits in Box V and Bag W the same.  LCM of 7 and 3 = 21

Number of grapefruits removed from Bag X
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of grapefruits left in Bag X
= 49 u - 9.8 u
= 39.2 u

Number of more grapefruits in Bag X than Bag W
= 39.2 u - 15 u
= 24.2 u

24.2 u = 726

1 u = 726 ÷ 24.2 = 30

	Bag V	Bag W	Bag X
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of grapefruits to be transferred from Bag W to Bag X so that the number of grapefruits in Bag V would be the same as Bag W.
= 15 u - 6 u
= 9 u
= 9 x 30
= 270

Answer(s): 270