Gold is $19$ times as heavy as water and copper is $9$ times. In what ratio should these metals be mixed so that the mixture may be $15$ times as heavy as water?

There are two containers of equal capacity. The ratio of milk to water in the first container is $3:1$,and in the second container is $5:2$. If they are mixed together,then the ratio of milk to water in the mixture will be:

In three vessels,the ratio of water and milk is $6:7$,$5:9$,and $8:7$,respectively. If the mixtures of the three vessels are combined,what will be the final ratio of water to milk?

$A$ 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

$A$ $20\, \text{litre}$ mixture of milk and water contains milk and water in the ratio $3:2$. $10\, \text{litres}$ of the mixture is removed and replaced with pure milk,and the operation is repeated once more. At the end of the two removals and replacements,what is the ratio of milk and water in the resultant mixture?

$A$ mixture of milk and water measures $60 \, ltr$. It contains $20 \%$ water. How many litres of water should be added to it so that water may be $25 \% $ (in $ltr$)?