$A$ dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains $25 \%$. The percentage of water in the mixture is: (in $\%$)

Vessels $A$ and $B$ contain acid and water in the ratios $4:3$ and $5:3$ respectively. In what ratio must these mixtures be mixed to obtain a new mixture in vessel $C$ containing acid and water in the ratio $3:2$?

$100$ $litres$ of a mixture contains $20 \%$ water and the rest is milk. The amount of milk that must be added so that the resulting mixture contains $87.5 \%$ milk is (in $litres$):

In a mixture of $25\, L$,the ratio of acid to water is $4: 1$. Another $3\, L$ of water is added to the mixture. The ratio of acid to water in the new mixture is

There are two vessels $A$ and $B.$ Vessel $A$ contains $40$ $litres$ of pure milk and vessel $B$ contains $22$ $litres$ of pure water. From vessel $A,$ $8$ $litres$ of milk is taken out and poured into vessel $B.$ Then,$6$ $litres$ of the mixture (milk and water) is taken out from vessel $B$ and poured into vessel $A.$ What is the ratio of the quantity of pure milk in vessel $A$ to the quantity of pure water in vessel $B\,?$

$A$ merchant mixes three varieties of rice costing $Rs. 20$ per $kg$,$Rs. 24$ per $kg$,and $Rs. 30$ per $kg$ and sells the mixture at a profit of $20\%$ at $Rs. 30$ per $kg$. How many $kgs$ of the second variety will be in the mixture if $2$ $kgs$ of the third variety is there in the mixture?