There were some persimmons in 3 baskets, S, T and U. 25% of the number of persimmons in Basket S was equal to 10% of the number of persimmons in Basket T. The number of persimmons in Basket U was 60% of the total number of persimmons. After Paul removed 20% of the persimmons in Basket U, there were 68 more persimmons in Basket U than in Basket T. In the end, how many persimmons should be transferred from Basket T to Basket U so that the number of persimmons in Basket S would be the same as Basket T?

Basket S	Basket T	Basket U
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket S	Basket T	Basket U
Before	4 u	10 u	21 u
Change			- 4.2 u
After	4 u	10 u	16.8 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Basket S = ¹₁₀ Basket T

Basket S : Basket T
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of persimmons in Basket S and Basket T is the combined repeated identity.  Make the total number of persimmons in Box S and Basket T the same.  LCM of 7 and 2 = 14

Number of persimmons removed from Basket U
= ²⁰₁₀₀ x 21 u
= 4.2 u

Number of persimmons left in Basket U
= 21 u - 4.2 u
= 16.8 u

Number of more persimmons in Basket U than Basket T
= 16.8 u - 10 u
= 6.8 u

6.8 u = 68

1 u = 68 ÷ 6.8 = 10

	Basket S	Basket T	Basket U
Before	4 u	10 u	16.8 u
Change		- 6 u	+ 6 u
After	4 u	4 u	22.8 u

Number of persimmons to be transferred from Basket T to Basket U so that the number of persimmons in Basket S would be the same as Basket T.
= 10 u - 4 u
= 6 u
= 6 x 10
= 60

Answer(s): 60