Man days are an equivalent unit of work. For example, if we say that 20 men can do a work in 30 days, this means that total work is \(20 \times 30\) man days = 600 man days, which means that the same work can be done by 10 men in 60 days  or 60 men can do in 10 days only etc. So for a constant work number of man days is constant. Following examples are based on this concept.

Example 13: 10 men or 20 women or 30 children can do a work in 15 days. If 10 men, 12 women and 18 children work on the same work, find the time in which work can be completed.

Solution: We see that 10 men are equivalent to 20 women that is equivalent to 30 children

Hence 1 man \( \equiv \)2 women \( \equiv \)3 children

Total work is \(10 \times15\) = 150 man days

10 men, 12 women and 18 children are equivalent to 10 + (12/2) + (18/3) = 22 men.

Hence time taken to complete the work

= 150/22 = 75/11 = \(6\frac{9}{{11}}\) days.

Example 14: 30 man can do a work in 10 days working together. If this work is started by 1 man next day one more man joins and so on. Find on which day work is complete.

Solution:  Total work = \(30 \times 10{\rm{ }} = {\rm{ }}300\) man days

Work done on the first day = 1 man day

Work done on the second day = 2 man days

Work done on the third day = 3 man days and so on.

After \(n\) days, the total work done = \(\frac{{n(n + 1)}}{2}\) 

\( = {\rm{ }}300 \Rightarrow n\left( {n + {\rm{ }}1} \right){\rm{ }} = {\rm{ }}600\) or \(n = 24\).

Hence on 24th day work will be completed  

Example 15: 4 men and 6 women finish a work in 8 days, while 3 men and 7 women finish it in 10 days. In how many days will 10 women finish it?

Solution:  Suppose one day’s work of a man and a woman are \(x\) and \(y\) respectively. From the given information,

\(4x + 6y = \frac{1}{8}\) and \(3x + 7y = \frac{1}{{10}}\)

Solving these equations we get, \(x = \frac{{11}}{{400}}\) and 

\(y =\frac{1}{{400}}\)

Work done by the 10 women in one day 

= \(\frac{{10}}{{400}}\)=\(\frac{1}{{40}}\)

Hence total time required = 40days.

Example 16: If 5 men and 10 women can do a work in 6 days, same work can be done by 2 men and 9 women in 10 days. In how many days work will be completed by 3 men and 4 women.

Solution:  \(5m + 10w\) can do a work in 6 days thus \(6(5m + 10w)\) can do the work in 1 day. 

\(2m + 9w\) can do the work in 10 days

Thus, \(10(2m + 9w)\) can do the same work in 1 day

So, \(6(5m + 10w) = 10(2m + 9w)\)

\(30m + {\rm{ }}60w = {\rm{ }}20m + {\rm{ }}90w \Rightarrow \;m = {\rm{ }}3w\)

Thus, efficiency of 1 man is thrice of the efficiency of a woman.

Now, \(5m + {\rm{ }}10w = {\rm{ }}5{\rm{ }} \times {\rm{ }}3w + {\rm{ }}10w\) and efficiency of 3 men + 4 women is equal to \(3 \times 3\) women + 4 women = 13 woman \((1m = 3w)\)

when, \({m_1} = {\rm{ }}25,{d_1} = {\rm{ }}6,{m_2} = {\rm{ }}13,{d_2} = {\rm{ }}?\)

\(25{\rm{ }} \times {\rm{ }}6{\rm{ }} \times {\rm{ }}1{\rm{ }} = {\rm{ }}13{\rm{ }} \times {d_2} \times {\rm{ }}1\) (men days concept)

\({d_2}\) = \(\frac{{150}}{{13}}\) i.e. 11\(\frac{7}{{13}}\) days