Time and Work

Time and work aptitude questions are asked in every competitive exam. Placement papers for TCS, Infosys, Wipro, CTS, HCL, IBM or Bank exam or MBA exams like CAT, XAT, MAT, or other exams like GRE, GMAT tests always contain one or more aptitude questions from this section of quantitative aptitude. Problems on time and work which appear in CAT exams are quite advanced and complicated - But they can be solved easily if you know the basic formulas, shortcuts and tricks.

This section will provide shortcuts, tips and tricks to solve quantitative aptitude questions on time and work. These are similar to time and distance shortcuts or ratio and proportion shortcuts. So, all you have to do is be thorough with the basics and practice as many questions as you can.

Basic Concepts of Time and Work

Most of the aptitude questions on time and work can be solved if you know the basic correlation between time, work and man-hours which you have learnt in your high school class.

• Analogy between problems on time and work to time, distance and speed:

Speed is equivalent to rate at which work is done
Distance travelled is equivalent to work done.
Time to travel distance is equivalent to time to do work.

• Man - Work - Hour Formula:

• More men can do more work.
  More work means more time required to do work.
  More men can do more work in less time.
  MM men can do a piece of work in TT hours, then Total effort or work =MT man hoursTotal effort or work =MT man hours.
  Rate of work * Time = Work DoneRate of work * Time = Work Done
  1. If AA can do a piece of work in DD days, then AA's 1 day's work = 1/D 1D.
     Part of work done by AA for tt days = t/D tD.
  If AA's 1 day's work = 1/D 1D, then AA can finish the work in DD days.
  2. MDHW=Constant${\displaystyle \frac{MDH}{W}=\text{Constant}}$
     Where,
     

     M = Number of men
     D = Number of days
     H = Number of hours per day
     W = Amount of work
  3. If M1M1 men can do W1W1 work in D1D1 days working H1H1 hours per day and M2M2 men can do W2W2 work in D2D2 days working H2H2 hours per day, then
     
     M1D1H1W1=M2D2H2W2${\displaystyle \frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}}$
  If AA is xx times as good a workman as BB, then:
  Ratio of work done by AA and BB = x:1x:1
  Ratio of times taken by AA and BB to finish a work = 1:x1:x ie; AA will take (1/x)th(1x)th of the time taken by BB to do the same work.

Shortcuts for frequently asked time and work problems

• AA and BB can do a piece of work in 'a'′a′ days and 'b'′b′ days respectively, then working together:
  They will complete the work in
  1. aba+b${\displaystyle \frac{ab}{a+b}}$ days
  In one day, they will finish
  2. (a+bab)th${\displaystyle {\left(\frac{a+b}{ab}\right)}^{th}}$ part of work.
• If AA can do a piece of work in aa days, BB can do in bb days and CC can do in cc days then,
  
  A, B and C together can finish the same work inabcab+bc+ca days
  
• 
  
• 
  
• 
  
• 
  
• A${\displaystyle A}$ can do a work in x${\displaystyle x}$ days and A${\displaystyle A}$ and B${\displaystyle B}$ together can do the same work in y${\displaystyle y}$ days then,
  Number of days required to complete the work if B works alone=xyx−ydays${\displaystyle }$
• 
  
• If AA and BB together can do a piece of work in xx days, BB and CC together can do it in yy days and CC and AA together can do it in zz days, then number of days required to do the same work:
  
  1. If A, B, and C working together = 2xyzxy+yz+zx${\displaystyle \frac{2xyz}{xy+yz+zx}}$
  2. If A working alone = 2xyzxy+yz−zx${\displaystyle \frac{2xyz}{xy+yz-zx}}$
  3. If B working alone = 2xyz−xy+yz+zx${\displaystyle \frac{2xyz}{-xy+yz+zx}}$
  4. If C working alone = 2xyzxy−yz+zx${\displaystyle \frac{2xyz}{xy-yz+zx}}$
• If AA and BB can together complete a job in xx days.
  If AA alone does the work and takes aa days more than AA and BB working together.
  If BB alone does the work and takes bb days more than AA and BB working together.
  
  Then,x=ab−−√ days${\displaystyle x=\sqrt{ab}\text{ days}}$
If m1m1 men or b1b1 boys can complete a work in DD days, then m2m2 men and b2b2 boys can complete the same work in
•  Dm1b1m2b1+m1b2${\displaystyle \frac{D{m}_1{b}_1}{m_2{b}_1+m_1{b}_2}}$ days.
If mm men or ww women or bb boys can do work in DD days, then 1 man, 1 woman and 1 boy together can together do the same work in
•  Dmwbmw+wb+bm${\displaystyle \frac{Dmwb}{mw+wb+bm}}$ days

If the number of men to do a job is changed in the ratio a:ba:b, then the time required to do the work will be changed in the inverse ratio. ie; b:ab:a
If people work for same number of days, ratio in which the total money earned has to be shared is the ratio of work done per day by each one of them.
AA, BB, CC can do a piece of work in xx, yy, zz days respectively. The ratio in which the amount earned should be shared is
•  1x:1y:1z=yz:zx:xy${\displaystyle \frac{1}{x}:\frac{1}{y}:\frac{1}{z}=yz:zx:xy}$
If people work for different number of days, ratio in which the total money earned has to be shared is the ratio of work done by each one of them.

Special cases of time and work problems

Given a number of people work together/alone for different time periods to complete a work, for eg: AA and BB work together for few days, then CC joins them, after few days BB leaves the job. To solve such problems, following procedure can be adopted.

Let the entire job be completed in DD days.
Let sum of parts of the work completed by each person = 1.
Find out part of work done by each person with respect to DD. This can be easily found out if you calculate how many days each person worked with respect to DD.
Substitute values found out in Step 3 in Step 2 and solve the equation to get unknowns.
A certain no of men can do the work in DD days. If there were mm more men, the work can be done in dd days less. How many men were there initially?
• Let the initial number of men be MM
  Number of man days to complete work = MDMD
  If there are M+mM+m men, days taken = D−dD-d
  So, man days = (M+m)(D−d)(M+m)(D-d)
  ie; MD=(M+m)(D−d)MD=(M+m)(D-d)
  M(D–(D−d))=m(D−d)M(D–(D-d))=m(D-d)
  

  M=m(D−d)d${\displaystyle M=\frac{m(D-d)}{d}}$
• A certain no of men can do the work in DD days. If there were mm less men, the work can be done in dd days more. How many men were initially?
  
  Let the initial number of men be MM
  Number of man days to complete work = MDMD
  If there are M−mM-m men, days taken = D+dD+d
  So, man days = (M−m)(D+d)(M-m)(D+d)
  ie; MD=(M−m)(D+d)MD=(M-m)(D+d)
  M(D+d–D)=m(D+d)M(D+d–D)=m(D+d)
  

  M=m(D+d)d${\displaystyle M=\frac{m(D+d)}{d}}$
• Given AA takes aa days to do work. BB takes bb days to do the same work. Now AA and BB started the work together and nn days before the completion of work AA leaves the job. Find the total number of days taken to complete work?
  Let DD be the total number of days to complete work.
  AA and BB work together for D−nD-n days.
  So, (D−n)(1a+1b)+n(1b)=1${\displaystyle (D-n)(\frac{1}{a}+\frac{1}{b})+n\left(\frac{1}{b}\right)=1}$
  D(1a+1b)–na−nb+nb=1${\displaystyle D(\frac{1}{a}+\frac{1}{b})–\frac{n}{a}-\frac{n}{b}+\frac{n}{b}=1}$
  D(1a+1b)=n+aa${\displaystyle D(\frac{1}{a}+\frac{1}{b})=\frac{n+a}{a}}$
  

  D=b(n+a)a+b${\displaystyle D=\frac{b(n+a)}{a+b}}$ days.

Frequently asked questions in quantitative aptitude test on time and work

After going through the questions given below, it will be good for you if you can take our practice placement test. At the end of the test, you can have a look at solutions provided for each question with answers.

Given A takes x days to do work. B takes y days to do the same work. If A and B work together, how many days will it take to complete the work.

If A and B together can do a piece of work in x days, B and C together can do it in y days and C and A together can do it in z days, find how many days it takes for each of them to complete the work if they worked individually. How many days will it take to complete the work if they worked together?

Give A is n times efficient than B. Also A takes n days less than B to complete the work. How many days will it take to complete the work if they worked together?

Given A takes x days to do work. B takes y days to do the same work. Now A & B together begins a work. After few days one of them leaves. Also given the other takes n more days to complete the work.

Find total number of days to complete the work.
How many days did they work together?

Given A takes x days to do work. B takes y days to do the same work. A started the work and B joined him after n days.

How long did it take to complete the work?
How many days did they work together? Or How long did B work?
Case 5 with 3 people joining work one after the other.

Given A takes x days to do work. B takes y days to do the same work. If A and B works on alternate days ie A alone works on first day, B alone works on next day and this cycle continues, in how many days will the work be finished

Given A alone can complete a job in x days and also B is b% efficient than A. How many days will it take to complete work if B works alone.
Problems where combinations of workers [men, women, girls and boys] take some days to do a work. 

These problems are solved using man days concept.

You have to calculate for another combination of them to complete the work.
How long will one set of people take to complete the entire work?

A certain combination starts the job and after few days leaves the work. Find the number of people from the category who are required to finish the remaining work.

Problems related to wages from work. How much each person earns from the work done.