The acid and water in two vessels $A$ and $B$ are in the ratio $4:3$ and $2:3$. In what ratio should the liquids in both the vessels be mixed to obtain a new mixture in vessel $C$ containing half acid and half water?

$A$ mixture contains $80 \%$ acid and the rest water. The part of the mixture that should be removed and replaced by the same amount of water to make the ratio of acid and water $4:3$ is

$A$ vessel of $160 \text{ litres}$ is filled with milk and water. $70 \%$ of milk and $30 \%$ of water is taken out of the vessel. After this task,it is found that the vessel is now filled by $55 \%$ of its original total quantity. What is the original quantity of milk and water in the vessel respectively?

$A$ and $B$ are two alloys of gold and copper prepared by mixing metals in the ratio $7:2$ and $7:11$ respectively. If equal quantities of the alloys are melted to form a third alloy $C$,the ratio of gold and copper in $C$ will be

$A$ drum contains $80$ $litres$ of ethanol. $20$ $litres$ of this liquid is removed and replaced with water. $20$ $litres$ of this mixture is again removed and replaced with water. How much water (in $litres$) is present in this drum now?

How much water should be added to $14$ litres of milk worth ₹ $5.40$ a litre so that the value of the mixture may be ₹ $4.20$ a litre? (in litres)