There were some pomegranates in 3 baskets, K, L and M. 25% of the number of pomegranates in Basket K was equal to 10% of the number of pomegranates in Basket L. The number of pomegranates in Basket M was 60% of the total number of pomegranates. After Xavier removed 10% of the pomegranates in Basket M, there were 89 more pomegranates in Basket M than in Basket L. In the end, how many pomegranates should be transferred from Basket L to Basket M so that the number of pomegranates in Basket K would be the same as Basket L?

Basket K	Basket L	Basket M
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket K	Basket L	Basket M
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 Basket K = ¹₁₀ Basket L

Basket K : Basket L
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of pomegranates in Basket K and Basket L is the combined repeated identity.  Make the total number of pomegranates in Box K and Basket L the same.  LCM of 7 and 2 = 14

Number of pomegranates removed from Basket M
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of pomegranates left in Basket M
= 21 u - 2.1 u
= 18.9 u

Number of more pomegranates in Basket M than Basket L
= 18.9 u - 10 u
= 8.9 u

8.9 u = 89

1 u = 89 ÷ 8.9 = 10

	Basket K	Basket L	Basket M
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of pomegranates to be transferred from Basket L to Basket M so that the number of pomegranates in Basket K would be the same as Basket L.
= 10 u - 4 u
= 6 u
= 6 x 10
= 60

Answer(s): 60