| 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 \[\frac{XY}{Y-X}\] 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 \[\frac{XY}{Y-X}\] 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,
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| 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 a1 men and b1 boys can complete a work in x days, while a2 men and b2 boys can complete the same work in y days, then \[\frac{\text{One day work of 1 man}}{\text{One day work of 1 boy}}=\frac{\left(yb_2-xb_1\right)}{\left(xa_1-ya_2\right)}\] |
| 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 \[T=\frac{(b-x)a}{a+b}\] |
| 15 | If ‘M1’ persons can do ‘W1’ works in ‘D1’ days and ‘M2’persons can do ‘W2’ works in ‘D2’ days then M1D1W2 = M2D2W1 If T1 and T2 are the working hours for the two groups then M1D1W2T1 = M2D2W1T2 Similarly, M1D1W2T1E1 = M2D2W1T2E2, where E1 and E2 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. |