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

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

	Bag J	Bag K	Bag L
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 J = ¹₁₀ Bag K

Bag J : Bag K
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

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

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

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

Number of more starfruits in Bag L than Bag K
= 39.2 u - 15 u
= 24.2 u

24.2 u = 484

1 u = 484 ÷ 24.2 = 20

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

Answer(s): 180