There were some lemons in 3 bags, H, J and K. 25% of the number of lemons in Bag H was equal to 10% of the number of lemons in Bag J. The number of lemons in Bag K was 60% of the total number of lemons. After Bryan removed 10% of the lemons in Bag K, there were 267 more lemons in Bag K than in Bag J. In the end, how many lemons should be transferred from Bag J to Bag K so that the number of lemons in Bag H would be the same as Bag J?

Bag H	Bag J	Bag K
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Bag H	Bag J	Bag K
Before	4 u	10 u	21 u
Change			- 2.1 u
After	4 u	10 u	18.9 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag H = ¹₁₀ Bag J

Bag H : Bag J
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of lemons in Bag H and Bag J is the combined repeated identity.  Make the total number of lemons in Box H and Bag J the same.  LCM of 7 and 2 = 14

Number of lemons removed from Bag K
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of lemons left in Bag K
= 21 u - 2.1 u
= 18.9 u

Number of more lemons in Bag K than Bag J
= 18.9 u - 10 u
= 8.9 u

8.9 u = 267

1 u = 267 ÷ 8.9 = 30

	Bag H	Bag J	Bag K
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of lemons to be transferred from Bag J to Bag K so that the number of lemons in Bag H would be the same as Bag J.
= 10 u - 4 u
= 6 u
= 6 x 30
= 180

Answer(s): 180