Vessel contains 3 parts of soda and 5 parts of rum.
∴ Total solution in the vessel = (3+5) = 8 parts
Assuming,
initially in the vessel, there are 8 litres of solution in which 3 litres are soda and 5 litres is rum.
Now, some solution (assuming S litre) has been taken out and replaced with soda (taken as same amount as S litre) so that the solution contains an equal amount of soda and rum (1:1).
As, S litre of solution has been taken out-
remaining amount of soda in the (8-S) litre solution = 3-3/8S {As, S litre solution also has 3 parts of soda and 5 parts of rum }
remaining amount of rum in the (8-S) litre solution = 5- 5/8S
Now, S litre soda also been added in the vessel.
Total amount of soda in the vessel = 3-3/8S+S
Now, as per the criteria, (3-3/8S) & $ (5- 5/8s) needs to be same.
∴ 3-(3/8)S+S = 5- (5/8)S
Or, (24-3S+8S)/8 = (40-5S)/8
Or, 24+5S = 40-5S
Or, 10S = 40-24
Or, S = 16/10 = 8/5
∴ 8/5 Lt. solution has been removed initially, which is equal to1/5 of 8 Lt. OR 1/5 of initial solution has been taken out and replaced with soda so that the solution contains equal amount of soda and rum.