Some staff promised to do a job in $18 \, days,$ but $6$ of them went on leave. So,the remaining men took $20 \, days$ to complete the job. How many men were there originally?

$A$ does $\frac{2}{5}$ of a work in $9$ days. Then $B$ joined him and they together completed the remaining work in $6$ days. $B$ alone can finish the whole work in (in days):

$18$ women can complete a work in $12$ days and $12$ men can complete the same work in $9$ days. In how many days will $8$ men and $8$ women complete that work? (in days)

$A$ man is twice as fast as a woman and a woman is twice as fast as a boy in doing a work. If all of them, a man, a woman and a boy, can finish the work in $7$ days, in how many days will a boy do it alone?

$A$ can do a work in $5 \, \text{days}$ less than the time taken by $B$ to do it. If both of them together take $11 \frac{1}{9} \, \text{days}$, then the time taken by $B$ alone to do the same work (in $\text{days}$) is

$A$ and $B,$ working separately,can complete a piece of work in $9$ and $12$ days,respectively. If they work for a day alternately,$A$ beginning,in how many days will the work be completed?