Rule  Description 

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 day’s 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 \[\frac{XYZ}{XY+YZ+ZX}\] days. 
5  If (A + B) can do a piece of work in X days, (B + C) can do a place of work in Y days and (C + A) can do a piece of work in Z days. Then, (A + B + C) can do a piece of work in \[\frac{2XYZ}{XY+YZ+ZX}\] days. 
6  If A and B together can do a piece of work in X days and A alone can do it in Y days, then B alone can do the work in \[\frac{XY}{YX}\] days. 
7  If (A + B + C) can do a piece of work in X days and (B + C) can do a piece of work in Y days, then A can do a piece of work in \[\frac{XY}{YX}\] days. 
8  A and B can do a work in ‘X’ and ‘Y’ days respectively. They started the work together but A left ‘a’ days before completion of the work. Then, time taken to finish the work is \[\frac{Y(X+a)}{X+Y}\] 
9  If ‘A’ is ‘a’ times efficient than B and A can finish a work in X days, then working together, they can finish the work in \[\frac{aX}{a+1}\] days 
10  If A is ‘a’ times efficient than B and working together they finish awork in Z days then,

11  If A working alone takes ‘x’ days more than A and B together, and B working along takes ‘y’ days more than A and B together, then number of days taken by A and B working together is given by \[\sqrt{xy}\] days. 
12  If a_{1} men and b_{1} boys can complete a work in x days, while a_{2} men and b_{2} boys can complete the same work in y days, then \[\frac{\text{One day work of 1 man}}{\text{One day work of 1 boy}}=\frac{\left(yb_2xb_1\right)}{\left(xa_1ya_2\right)}\] 
13  If n men or m women can do a piece of work in X days, then N men and M women together can finish the work in \[\frac{nmX}{nM+mN}\] days. 
14  A and B do a piece of work in a and b days, respectively. Both begin together but after some days, A leaves off and the remaining work is completed by B in x days. Then, the time after which A left, is given by \[T=\frac{(bx)a}{a+b}\] 
15  If ‘M_{1}’ persons can do ‘W_{1}’ works in ‘D_{1}’ days and ‘M_{2}’persons can do ‘W_{2}’ works in ‘D_{2}’ days then M_{1}D_{1}W_{2} = M_{2}D_{2}W_{1} If T_{1} and T_{2} are the working hours for the two groups then M_{1}D_{1}W_{2}T_{1} = M_{2}D_{2}W_{1}T_{2} Similarly, M_{1}D_{1}W_{2}T_{1}E_{1} = M_{2}D_{2}W_{1}T_{2}E_{2}, where E_{1} and E_{2} are the efficiencies of the two groups. 
16  If the number of men to do a job is changed in the ratio a : b, then the time required to do the work will be in the ratio b : a, assuming the amount of work done by each of them in the given time is the same, or they are identical. 
17  A is K times as good a worker as B and takes X days less than B to finish the work. Then the amount of time required by A and B working together is \[\frac{K \times X}{K^2  1}\] days. 
18  If A is n times as efficient as B, i.e. A has n times as much capacity to do work as B, then A will take \[\frac{1}{n}\] of the time taken by B to do the same amount of work. 