$10$ men and $15$ women together can complete a work in $6$ days. It takes $100$ days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?

$P$ 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?

$A$ can do a piece of work in $30 \text{ days}$ while $B$ can do it in $40 \text{ days}$. $A$ and $B$ working together can do it in (in $\text{days}$)

$A, B$ and $C$ can do a job in $6$ days,$12$ days and $15$ days,respectively. After $\frac{1}{8}$ of the work is completed,$C$ leaves the job. The rest of the work is done by $A$ and $B$ together. The total time taken to finish the work is (in days):

Twenty-four men can complete a work in sixteen $days$. Thirty-two women can complete the same work in twenty-four $days$. Sixteen men and sixteen women started working and worked for twelve $days$. How many more men are to be added to complete the remaining work in $2$ $days$?

If $10$ men or $18$ boys can do a work in $15$ days,then the number of days required by $15$ men and $33$ boys to do twice the work is