There were some starfruits in 3 bags, Q, R and S. 25% of the number of starfruits in Bag Q was equal to 10% of the number of starfruits in Bag R. The number of starfruits in Bag S was 70% of the total number of starfruits. After Vaidev removed 20% of the starfruits in Bag S, there were 484 more starfruits in Bag S than in Bag R. In the end, how many starfruits should be transferred from Bag R to Bag S so that the number of starfruits in Bag Q would be the same as Bag R?

Bag Q	Bag R	Bag S
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Bag Q	Bag R	Bag S
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 Q = ¹₁₀ Bag R

Bag Q : Bag R
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of starfruits in Bag Q and Bag R is the combined repeated identity.  Make the total number of starfruits in Box Q and Bag R the same.  LCM of 7 and 3 = 21

Number of starfruits removed from Bag S
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of starfruits left in Bag S
= 49 u - 9.8 u
= 39.2 u

Number of more starfruits in Bag S than Bag R
= 39.2 u - 15 u
= 24.2 u

24.2 u = 484

1 u = 484 ÷ 24.2 = 20

	Bag Q	Bag R	Bag S
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of starfruits to be transferred from Bag R to Bag S so that the number of starfruits in Bag Q would be the same as Bag R.
= 15 u - 6 u
= 9 u
= 9 x 20
= 180

Answer(s): 180