There were some starfruits in 3 bags, H, J and K. 25% of the number of starfruits in Bag H was equal to 10% of the number of starfruits in Bag J. The number of starfruits in Bag K was 70% of the total number of starfruits. After Justin removed 10% of the starfruits in Bag K, there were 582 more starfruits in Bag K than in Bag J. In the end, how many starfruits should be transferred from Bag J to Bag K so that the number of starfruits in Bag H would be the same as Bag J?

Bag H	Bag J	Bag K
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Bag H	Bag J	Bag K
Before	6 u	15 u	49 u
Change			- 4.9 u
After	6 u	15 u	44.1 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag H = ¹₁₀ Bag J

Bag H : Bag J
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

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

Number of starfruits removed from Bag K
= ¹⁰₁₀₀ x 49 u
= 4.9 u

Number of starfruits left in Bag K
= 49 u - 4.9 u
= 44.1 u

Number of more starfruits in Bag K than Bag J
= 44.1 u - 15 u
= 29.1 u

29.1 u = 582

1 u = 582 ÷ 29.1 = 20

	Bag H	Bag J	Bag K
Before	6 u	15 u	44.1 u
Change		- 9 u	+ 9 u
After	6 u	6 u	53.1 u

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

Answer(s): 180