Aptitude Made Easy: Tips and Tricks on Work&Time Problems

It is difficult to imagine any aptitude exam without the questions on work and time. This topic is very important especially for campus placements, CSAT, Bank exams, CAT, MAT, XAT and other competitive exams.

Work_time_aptitude

The difficulty level of the questions on work and time can vary from very easy to very difficult. But there are some basic formulae and concepts which help you in solving any type of question, provided you have solved enough problems and have a good grip on the formulae and understand the concept well.

We are providing you the most important shortcuts and formulae which will help you to tackle most of these questions.

Ground Rules(Basic Funda – Must remember):

Rule1: If a man can finish a piece of work in ‘N’ days then he will complete only 1/N of the work on 1 day. Similarly if a man completes 1/N of the work assigned to him in 1 day then he will take ‘N’ days to finish the work.

Rule2: Efficieny matters: If A works ‘x’ times faster(efficient) than B then A will take 1/x of the time taken by B to finish the work. For example if A is twice as efficient as B then A will take only half of the time taken by B to finish the work.

Rule3: Most Important Formula (Forgetting this is not less than a crime 😉 )

If M1 men can finish a piece of work in D1 days(Team A) and M2 men can finish the same work in D2 days(Team B) then we can say, M1xD1 = M2xD2

Slight variations to this can be formed like this: If the men in the above case work for T1 and T2 time respectively then,            M1xD1xT1 = M2xD2xT2

If the persons in team A and B above have efficiency E1 and E2 respectively, then we can say that,                                             M1xD1xT1xE1 = M2xD2xT2xE2

Rule4: (Frequently used) If A can do a piece of work in D1 days and B can do the same piece of work in D2 days then A and B together can finish the work in D1xD2/(D1+D2) days.

Rule5: (To make your life easy) : There are three persons in the question now. If A and B finish some work in x days, B and C finish he same work in y days and C & A finish the work in Z days then the same work can be finished by,

A alone in : 2xyz/(xy+yz-zx) days

B alone in : 2xyz/(-xy+yz+zx) days

C alone in : 2xyz/(xy-yz+zx) days

A,B,C together in : 2xyz/(xy+yz+zx) days

Rule 6: If A can finish a piece of work in D1 days and B can do the same work in D2 days and C can complete the same work in D3 days then, they together can finish the work in D1*D2*D3/(D1+D2+D3) days

Rule 7: If A and B together can do a piece of work in D1 days and A alone can do the work in D2 days, then B completes the same work in D1*D2/(D2-D1) days

Important Note: The formulae and shortcuts discussed above appear directly as questions so practice them well.

We will come up with solved questions on this topic from previous year papers soon.

Happy Reading!

Team CareerShapers!!

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