If $m$ men can do a job in $p$ days, then $(m + r)$ men can do the job in how many days?(A)$(p + r)$days (B) $\dfrac{{mp}}{{m + r}}$days (C) $\dfrac{p}{{m + r}}$days (D) $\dfrac{{m + r}}{p}$days

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Hint: To solve this question in detail one should have a deep knowledge of the Unitary method. In this whole solution, we will apply the unitary method many times.

By applying unitary method we can get,
Work done by $m$men in $p$ days$= 1$ Statement (1)
$\therefore$ Work done by $m$men in $1$ day$= \dfrac{1}{p}$part Statement (2)
And Work done by $1$man in $1$ day$= \dfrac{1}{{mp}}$part Statement (3)
Now, we have to find the number of days required to complete the job by $(m + r)$men,
Therefore after statement (3) we can write that,
Work done by $(m + r)$men in $1$ day$= \dfrac{{(m + r)}}{{mp}}$ part
Now again by applying the unitary method in a different way to find the time taken by $(m + r)$men to do the job,
Time taken by $(m + r)$men to do $\dfrac{{(m + r)}}{{mp}}$part of work$= 1$ day
$\therefore$ Time taken by $(m + r)$men to do full work (it means$1$)$= \dfrac{{1 \times 1}}{{(\dfrac{{m + r}}{{mp}})}} = \dfrac{{mp}}{{m + r}}$ days.
Hence $(m + r)$ men can do the job in $\dfrac{{mp}}{{m + r}}$days.
If ${M_1}$men can do ${W_1}$ work in ${D_1}$ days working ${H_1}$hours per day and ${M_2}$men can do ${W_2}$ work in ${D_2}$ days working ${H_2}$hours per day, then
$\dfrac{{{M_1}{D_1}{H_1}}}{{{W_1}}} = \dfrac{{{M_2}{D_2}{H_2}}}{{{W_2}}}$
If you do not have one of these values then put $1$in place of that.