Abi had some starfruits and persimmons in 2 baskets. In Basket E, the number of starfruits to the number of persimmons was in the ratio of 3 : 10. In Basket F, there were twice as many starfruits as persimmons. After Abi transferred ⁴₅ of the persimmons from Basket E to Basket F, the number of fruits left in Basket E was 140 and the ratio of the number of starfruits to the number of persimmons in Basket B became 6 : 7. How many fruits were in Basket F in the end?

	Basket E		Basket F
	Starfruits	Persimmons	Starfruits	Persimmons
Before	3 u	10 u	2x3 =  6 p	1x3 = 3 p
Change		- 8 u		+ 8 u (+ 4 p)
After	3 u	2 u	6x1 = 6 p	7x1 = 7 p

Number of persimmons that Abi transferred from Basket E to Basket F
= ⁴₅ x 10 u
= 8 u

Total number of fruits in Basket E in the end
= 3 u + 2 u
= 5 u

5 u = 140
1 u = 140 ÷ 5 = 28

The number of starfruits in Basket F is the unchanged quantity.
LCM of 2 and 6 is 6.

Number of persimmons that Abi transferred from Basket E to Basket F
= 8 u
= 8 x 28
= 224

Increase in the number of persimmons that due to the transfer from Basket E to Basket F
= 7 p - 3 p
= 4 p

4 p = 8 u 
4 p = 224

1 p = 224 ÷ 4 = 56

Number of fruits in Basket F in the end
= 6 p + 7 p
= 13 p
= 13 x 56
= 728

Answer(s): 728