# Simple tricks for Time and Work problems | Time and Work Shortcuts

## Time and Distance Shortcuts

### METHOD 1

M1 persons do a work in D1 days and M2 persons do the same work in D2 days then
we have the equation
M1*d1= M2* D2

### METHOD 2

M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days then
we have (M1 D1)/W1 = (M2 D2)/W2

### METHOD 3

if A can do a work in x days, B can do a work in y days then
A and B together (A+B) can do the same work in ( xy )/(x+y) days

### METHOD 4

A alone can do a work in x days , A+B can do a work in y days then .
B alone can do the work in (xy)/(x-y)

### METHOD 5

here A+B means A and B working together
if A working alone takes x days more than (A+B) and B working alone takes y days more than (A+B) then
the number of days taken by A and B together is given by
root of (xy)

### METHOD 6

here A+B+C means A B C working together
A can do a work in x days , B can do the same work in y days , C can also do the same work in z days then
A+B+C can do the work in (xyz)/(xy+yz+zx) days

### METHOD 7

here A+B+C means A B C working together
A+B can do a work in x days , B+C can do the work in y days, C+A can do the work in z days then
A+B+C can do the work in (2xyz )/xy+yz+zx days

### METHOD 8

if x1 men or y1 women can reap the field in D days ,then
x2 men and y2 women take to reap it in D(x1y1)/(x2 y1+x1 y2) days

### METHOD 9

a certain men can do a work in D days if there are x men less it could be finished in d days more,then
the no of men orginaly are x(D+d)/d

### METHOD 10

a certain men can do a work in D days if there are x men more it could be finished in d days more,then
the no of men orginaly are x(D-d)/d

### METHOD 11

if x1 men or x2 women or x3 boys can do a work in D days then
1 men + 1 women + 1 boy can do the same work in d(x1x2x3)/(x1x2+x2x3+x3x1)