There were some passion fruits in 3 bags, K, L and M. 20% of the number of passion fruits in Bag K was equal to 10% of the number of passion fruits in Bag L. The number of passion fruits in Bag M was 60% of the total number of passion fruits. After Luke removed 10% of the passion fruits in Bag M, there were 164 more passion fruits in Bag M than in Bag L. In the end, how many passion fruits should be transferred from Bag L to Bag M so that the number of passion fruits in Bag K would be the same as Bag L?

Bag K	Bag L	Bag M
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Bag K	Bag L	Bag M
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 K = ¹₁₀ Bag L

Bag K : Bag L
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of passion fruits in Bag K and Bag L is the combined repeated identity.  Make the total number of passion fruits in Box K and Bag L the same.  LCM of 3 and 2 = 6

Number of passion fruits removed from Bag M
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of passion fruits left in Bag M
= 9 u - 0.9 u
= 8.1 u

Number of more passion fruits in Bag M than Bag L
= 8.1 u - 4 u
= 4.1 u

4.1 u = 164

1 u = 164 ÷ 4.1 = 40

	Bag K	Bag L	Bag M
Before	2 u	4 u	8.1 u
Change		- 2 u	+ 2 u
After	2 u	2 u	10.1 u

Number of passion fruits to be transferred from Bag L to Bag M so that the number of passion fruits in Bag K would be the same as Bag L.
= 4 u - 2 u
= 2 u
= 2 x 40
= 80

Answer(s): 80