In this chapter, there are many parameters like number of men, working hours, efficiency, amount of work etc. 

Assuming that all employees work with same efficiency, we can conclude that if the work done is constant then the number of days is inversely proportional to the number of employees working. 

\(M \propto \frac{1}{D}\), where \(M\) is the number of employees.

\( \Rightarrow MD\) = constant 

Thus we obtain a relationship 

\({M_1}{D_1} = {M_2}{D_2}\)

If work is not constant, then it is directly proportional to both number of employees and number of days 

\(\frac{{{M_1}{D_1}}}{{{W_1}}} = \frac{{{M_2}{D_2}}}{{{W_2}}}\) = constant

Important points:

1. If \(A\) can do a piece of work in \(X\) days, then A’s one day’s work = \(\frac{1}{X}th\) part of whole work.
2. If A’s one days’ work = \(\frac{1}{X}th\) part of whole work, then \(A\) can finish the work in \(X\) days.
3. If A can do a piece of work in \(X\) days and \(B\) can do it in \(Y\) days then \(A\) and \(B\) working together will do the same work in\(\frac{{XY}}{{X + Y}}\) days.
4. If \(A, B\) and \(C\) can do a work in \(X,Y\) and \(Z\) days respectively then all of them working together can finish the work in \(\frac{{XYZ}}{{XY + YZ + XZ}}\) days.

Example 1: \(A\) can do a piece of work in 20 days and \(B\) can do it in 12 days. How long will they take, if both work together? 

Solution: There are different methods to solve this problem. 

(i) Unitary Method: A can do \(\frac{1}{{20}}th\) of the work in one day, and B can do \(\frac{1}{{12}}th\) of the work in one day. Hence one day‘s work of \(A\) and \(B\) together will be \(\frac{1}{{20}} + \frac{1}{{12}} = \frac{{3 + 5}}{{60}} = \frac{8}{{60}}\) 
Hence total number of days required =\(\frac{{60}}{8} = 7\frac{1}{2}\,\,days\)

(ii) LCM Method: \(A\) can do the work in 20 days while \(B\) can do the same work in 12 days. 

Let the total work be \(LCM (20, 12) = 60\) units

Contribution of \(A\) per day = \(\frac{{60}}{{20}} = 3\) unit

Contribution of \(B\) per day = \(\frac{{60}}{{12}} = 5\) unit 

Thus, contribution of both is 3 + 5 = 8 units in a day. Thus, total days required is = \(\frac{{60}}{8} = 7\frac{1}{2}\;days\)

Example 2: If \(X\) can do a work in 10 day, \(Y\) can do the same work in 20 days, \(Z\) can do double of the work in 30 days. If all 3 started working together, find the total time required to complete the work. 

Solution:

(i) Unitary Method: One day’s work of \(X\) = \(\frac{1}{{10}}\)

One day’s work of \(Y\) and \(Z\) will be \(\frac{1}{{20}}\) and \(\frac{1}{{15}}\)

Because \(Z\) completes double of the work in 30 days, 

Combined work done in one day 

\(\frac{1}{{10}} + \frac{1}{{20}} + \frac{1}{{15}} = \frac{{6 + 3 + 4}}{{60}} = \frac{{13}}{{60}}\)

Hence total time taken =  \(\frac{{60}}{{13}}\,\, = \,\,4\frac{8}{{13}} = 4.61\) days

(ii) LCM Method: \(X\) can do the work in 10 days while \(Y\) can do the same work in 20 days and \(Z\) can do it in 15 days.

Let the total work be \(LCM (10, 20, 15) = 60\) units

Contribution of \(X\) per day = \(\frac{{60}}{{10}} = 6\) units

Contribution of \(Y\) per day = \(\frac{{60}}{{20}} = 3\) units

Contribution of \(Z\) per day = \(\frac{{60}}{{15}} = 4\) units

Thus, contribution of all of them is \(6 + 3 + 4 = 13\) units in a day. 

Thus, total days required is = \(\frac{{60}}{{13}} = 4\frac{8}{{13}}\;days\)

Example 3: Working alone, Anil can do a job in 12 days and Vinay can do the same job in 15 days. Anil starts working and works for 6 days and then Vinay joins him. In how many days will the job be completed? 

Solution: Working alone, Anil can do the job in 12 days, Therefore, in 6 days he will complete half the job. 

Vinay and Anil together can complete \(\frac{1}{{12}} + \frac{1}{{15}}\) 

\( = \frac{{5 + 4}}{{60}} = \frac{3}{{20}}th\)of the job a day. Therefore, it will take \(\frac{{20}}{3}\) days to complete the whole work and they will take only \(\frac{{10}}{3}\) days to complete half of the work. Hence, the total time taken 

= 6 + \(\frac{{10}}{3}days\,\, = \,\,\frac{{28}}{3}days\) \( = \,9\,\frac{1}{3}days\)

Alternate method: Let total work be 60 units (LCM of 12 days and 15 days). That means, Anil can do 5 units/day and Vinay can do 4 units/day. Anil worked 6 days alone so he does 30 units in these days remaining 30 units will be done by Anil and Vinay together. Together they can do 

5 + 4 = 9 units/day so they will finish the remaining work in \(\frac{{30}}{9}\) days.  Hence, the total time taken = \(6 +\frac{{10}}{3}days\,\, = \,\,\frac{{28}}{3}days\)\( = \,9\,\frac{1}{3}days\).