$A$ works thrice as good as $B$ and is,therefore,able to finish a piece of work in $60 \text{ days}$ less than $B$. Find the time in which they can complete it,working together (in $\text{days}$).

$A$ tyre has two punctures. The first puncture alone would have made the tyre flat in $9 \,minutes$ and the second alone would have done it in $6 \,minutes.$ If air leaks out at a constant rate,how long does it take both the punctures together to make it flat? (in $minutes$)

$A$ takes twice as much time as $B$ and $C$ takes thrice as much time as $B$ to finish a piece of work. Working together they can finish the work in $12$ days. The number of days needed for $A$ to do the work alone is:

$A$ and $B$ can do a job in $6$ and $12$ days,respectively. They began the work together,but $A$ leaves after $3$ days. Then,the total number of days needed for the completion of the work is:

$P$ can complete $\frac{1}{4}$ of a work in $10$ days,$Q$ can complete $40\%$ of the same work in $15$ days,$R$ can complete $\frac{1}{3}$ of the work in $13$ days,and $S$ can complete $\frac{1}{6}$ of the work in $7$ days. Who will be able to complete the work first?

$A$ and $B$ can complete a piece of work in $15$ days and $10$ days respectively. They contracted to complete the work for $Rs. 30000$. The share of $A$ in the contracted money will be (in $Rs.$)