If $28$ men complete $\frac{7}{8}$ of a piece of work in a week,then the number of men,who must be engaged to get the remaining work completed in another week is

$40$ men can complete a work in $40$ days. They started the work together. But at the end of each $10^{th}$ day,$5$ men left the job. The work would have been completed in (in days):

$A$ certain number of persons can complete a piece of work in $55 \text{ days}$. If there were $6$ persons more,the work could be finished in $11 \text{ days}$ less. How many persons were originally there?

Three men $A, B, C$ working together can do a job in $6$ hours less time than $A$ alone,in $1$ hour less time than $B$ alone,and in one-half the time needed by $C$ when working alone. Then $A$ and $B$ together can do the job in (in hours):

$B$ would have taken $10 \, \text{hours}$ more than what $A$ would have taken to complete a task if each of them worked alone. Working together they can complete the task in $12 \, \text{hours}$. How many $\text{hours}$ would $B$ take to do $50 \%$ of the task?

$A$ work can be completed by $P$ and $Q$ in $12$ days,$Q$ and $R$ in $15$ days,and $R$ and $P$ in $20$ days. In how many days can $P$ alone finish the work?