There were some guavas in 3 bags, U, V and W. 25% of the number of guavas in Bag U was equal to 10% of the number of guavas in Bag V. The number of guavas in Bag W was 70% of the total number of guavas. After Luke removed 20% of the guavas in Bag W, there were 968 more guavas in Bag W than in Bag V. In the end, how many guavas should be transferred from Bag V to Bag W so that the number of guavas in Bag U would be the same as Bag V?

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

	Bag U	Bag V	Bag W
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 U = ¹₁₀ Bag V

Bag U : Bag V
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

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

Number of guavas removed from Bag W
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of guavas left in Bag W
= 49 u - 9.8 u
= 39.2 u

Number of more guavas in Bag W than Bag V
= 39.2 u - 15 u
= 24.2 u

24.2 u = 968

1 u = 968 ÷ 24.2 = 40

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

Number of guavas to be transferred from Bag V to Bag W so that the number of guavas in Bag U would be the same as Bag V.
= 15 u - 6 u
= 9 u
= 9 x 40
= 360

Answer(s): 360