There were some persimmons in 3 boxes, J, K and L. 20% of the number of persimmons in Box J was equal to 10% of the number of persimmons in Box K. The number of persimmons in Box L was 60% of the total number of persimmons. After Luis removed 10% of the persimmons in Box L, there were 205 more persimmons in Box L than in Box K. In the end, how many persimmons should be transferred from Box K to Box L so that the number of persimmons in Box J would be the same as Box K?

Box J	Box K	Box L
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Box J	Box K	Box L
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 Box J = ¹₁₀ Box K

Box J : Box K
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

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

Number of persimmons removed from Box L
= ¹⁰₁₀₀ x 9 u
= 0.9 u

Number of persimmons left in Box L
= 9 u - 0.9 u
= 8.1 u

Number of more persimmons in Box L than Box K
= 8.1 u - 4 u
= 4.1 u

4.1 u = 205

1 u = 205 ÷ 4.1 = 50

	Box J	Box K	Box L
Before	2 u	4 u	8.1 u
Change		- 2 u	+ 2 u
After	2 u	2 u	10.1 u

Number of persimmons to be transferred from Box K to Box L so that the number of persimmons in Box J would be the same as Box K.
= 4 u - 2 u
= 2 u
= 2 x 50
= 100

Answer(s): 100