Since, 20% of 5 litre = 1 litre or 1000 ml of chemical was emptied. so 4 litres or 4000 ml is left.
And 60% of 4000 ml of antifreeze = 2400 ml
So percent of water in the chemical = 4000 - 2400 = 1600 ml
Now, Amount of antifreeze in final mixture is 50%
So, Amount of water in final mixture is also 50%
Now, let the X amount of pure water must be added.
So, on taking ratio of antifreeze to water we get
2400/(1600+X) = 50/50
=> 2400/(1600+X) = 1
=> 2400 = 1600 +X
=> X = 800 ml or 4/5 litres.
Another method,
Since,
Now 2400 is 60% of 4000 ml solution.
And antifreeze is 50% of Y ml solution, where Y is new total of antifreeze and water, after water being added to the solution
So,
Y * 50% = 2400
=> Y = 2400 * 100 / 50 = 4800
Therefore, the amount of water added to the solution = final solution - initial solution = 4800 - 4000 = 800ml or 4/5 litres
Hope this is what you want