There were some dragonfruits in 3 bags, H, J and K. 20% of the number of dragonfruits in Bag H was equal to 10% of the number of dragonfruits in Bag J. The number of dragonfruits in Bag K was 60% of the total number of dragonfruits. After Elijah removed 10% of the dragonfruits in Bag K, there were 82 more dragonfruits in Bag K than in Bag J. In the end, how many dragonfruits should be transferred from Bag J to Bag K so that the number of dragonfruits in Bag H would be the same as Bag J?

Bag H	Bag J	Bag K
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Bag H	Bag J	Bag K
Before	2 u	4 u	9 u
Change			- 0.9 u
After	2 u	4 u	8.1 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Bag H = ¹₁₀ Bag J

Bag H : Bag J
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of dragonfruits in Bag H and Bag J is the combined repeated identity.  Make the total number of dragonfruits in Box H and Bag J the same.  LCM of 3 and 2 = 6

Number of dragonfruits removed from Bag K
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of dragonfruits left in Bag K
= 9 u - 0.9 u
= 8.1 u

Number of more dragonfruits in Bag K than Bag J
= 8.1 u - 4 u
= 4.1 u

4.1 u = 82

1 u = 82 ÷ 4.1 = 20

	Bag H	Bag J	Bag K
Before	2 u	4 u	8.1 u
Change		- 2 u	+ 2 u
After	2 u	2 u	10.1 u

Number of dragonfruits to be transferred from Bag J to Bag K so that the number of dragonfruits in Bag H would be the same as Bag J.
= 4 u - 2 u
= 2 u
= 2 x 20
= 40

Answer(s): 40