9 Formulae And Shortcuts To Solve All Time And Work Related Problems For Competitive Exams

Time & Work is one of the main topics of any competitive exams like SSC CGLE, CHSLE, CPO, MTS, NTPC (Indian Railways), Banking (SBI, IBPS) etc. In every exam 2-3 questions of Quantitative Aptitude section are asked from this topic alone. So you should we well prepared for it. However the difficulty level of this topic is Medium, if you learn all formulae and shortcut tricks given below than you can handle all the questions very easily from this section.
Before going into formulae We should learn some simple and fundamental principles of Time and Work.

• Person and Work:- The amount of work and number of person is directly proportional to each other . That means with increase in number of persons the quantity of work will automatically improve.
• Time and Person:- Amount of Time and number of persons is inversely proportional to each other. That means if number of persons increases than time required to complete the work will decrease.
• Work and Time:- work and time is directly proportional to each other. Which means more work is required more time to complete.

Formulae and Shortcut Tricks Related to Time and Work:-

1. If a person can do a peace of work in N number of days/hours then then that person's 1 day's/hour's work will be equal to 1/N.
2. If a person's 1 day's/hour's = 1/N then the person will complete the work in N number of days/hours.
3. If the ratio of the number of persons required to complete a work is M : N, then the ratio of time taken by them will be N : M.
4. If A is N times more efficient than that of B than the ratio of work done by them will be N:1, while the ratio of time taken will be equal to 1:N.
5. If A can do a piece work in x number of days while the same work is done by B in y days, then (A+B)'s one day work = 1/x + 1/y = (x+y)/xy. Time taken when both (A+B)  work together = xy/(x+y) days.
6. If (A+B) together can complete a work in x number of days and A alone can finish that work in y days, then one day work of B = (A+B)'s 1 day work - A's 1 day work = 1/x - 1/y = (y-x)/xy => Hence the number of days required to complete the same work by B = 1/(B's 1day Work) = xy/(y-x). 
7. If A can do a piece of work in x days, while B can do the same work in y days and C can do the same work in z days, then (A+B+C)'s one day work = 1/x + 1/y + 1/z = (xy+yz+zx)/xyz. Hence time taken by (A+B+C) working together = 1/[(A+B+C)'s 1day work] = xyz/(xy+yz+zx) days. 
8. If A and B can do a piece of work in x days, B and C can do the same work in y days and A and C can do it in z days, then [(A+B) + (B+ C) + (A+C)]'s 1 day work = 2(A + B + C)'s 1 day work = 1/x + 1/y + 1/z = (xy+yz +zx)/xyz => (A+B+C)'s 1 day work = (xy+yz+zx)/2xyz. Now the number of days required by (A+B+C) together to complete the same work = 2xyz/(xy+yz+zx).
9. If M' number of persons can do a work in D' days and T' hours, while M" number of persons can do W" work in D" days working T" hours in a day. Then the relationship between them will be:- M'D'T'W' = M"D"T"W"

Conclusion
If you learn rules and formulae given above than you will able to solve all the problems on Time and Work. And one advice is that do not follow shortcut tricks without understanding, how the equation is derived?, otherwise you may forget those tricks when required the most.

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