Shortcuts For Management Entrance Exam

Aptitude questions relating to time and work are asked in every competitive exams be it placement drives for reputed companies, bank exam, entrance exams like CAT, XAT, MAT,GRE, GMAT etc always contain a section of aptitude questions to test the quantitative ability. Problems relating to time and work in CAT aptitude exams are quite advanced and complicated and eat up much of the time and energy, unless you know the basic formulas, shortcuts and tricks to handle such scenarios. Hence it’s advisable to note down these tips and practice them over and over again to memorize them.

The Basic Concept of Word Problems

Relation between variables of problem on Time, Work, Distance and Speed 

• Speed is equivalent to rate at which work is done
• Distance travelled is equivalent to work done
• Time taken to cover distance is equivalent to time taken to complete the work
• Relation between variables of problem on Men, Work and Hour
• More men can do more work
• More work with will take more time to finish itself.
• More men can do more work in less time.
• If M men can do a piece of work in T hours, then total work = M*T
• Work done = rate of work * time taken
• If A can do a piece of work in D days, then A’s 1 day’s work = 1/D

            So, the amount of work done in T days =T*1/D

• If A’s 1 day’s work = 1/D, then A can finish the work in =1/(1/D)=D days
• If ,M= Number of men
      D = Number of days
      H = Number of hours per day
      W= Amount of work

Then, MDH/W=Constant

• If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day, then

(M1D1H1)/W1=(M2D2H2)/W2

• If A is x times as good a worker as B is, then following deduction can be made:

1. Ratio of work done by A and B = x:1
2. Ratio of times taken by A and B to finish a work = 1:x which suggests ,A will take (1/x)th of the time taken by B to do the same work.

Shortcuts for time and work problems

• If A and B can do a piece of work in ‘a’ and ‘b’ days respectively, then together:

1. They will complete the work in ab/(a+b) days
2. In one day, they will do [(a+b)/ab]th part of work.

• If A can do a piece of work in ‘a’ days, B can do it in ‘b’ days and C can do it in ‘c’ days then, A, B and C together can finish the same work in abc/(ab+bc+ca)days
• If A can do a work in x days and A and B together can do the same work in y days then, Number of days required to complete the work if B works alone=xy/(x-y)days
• 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, then number of days required to do the same work:

If A, B, and C working together = 2xyz/(xy+yz+zx) days
If A working alone = 2xyz/(xy+yz-zx) days
If B working alone = 2xyz/(-xy+yz+zx) days
If C working alone = 2xyz/(xy-yz+zx) days

• If A and B can together complete a job in x days.
  If A alone does the work and takes a days more than A and B working together.
  If B alone does the work and takes b days more than A and B working together.

            Then,x=(ab)^1/2 days

• If m1 men or b1 boys can complete a work in D days, then m2 men and b2 boys can complete the same work in (Dm1b1)/(m2b1+m1b2) days.
• If m men or w women or b boys can do work in D days, then 1 man, 1 woman and 1 boy together can together do the same work in (Dmwb)/(mw+wb+bm) days
• 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 changed in the inverse ratio. ie; b: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.
A, B, C can do a piece of work in x, y, z days respectively. The ratio in which the amount earned should be shared is (1/x):(1/y):(1/z)=yz:zx:xy
·     If people work for different number of days,the 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 e.g.: A and B work together for few days, then C joins them, after few days B leaves the job. To solve such problems, following procedure can be adopted.
1. Let the entire job be completed in D days.
2. Let sum of parts of the work completed by each person = 1.
3. Find out part of work done by each person with respect to D. This can be easily found out if you calculate how many days each person worked with respect to D.
4. 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 D days. If there were m more men, the work can be done in d days less. How many men were there initially?

Let the initial number of men be M
Number of man days to complete work = MD
If there are M+m men, days taken = D-d
So, man days = (M+m)(D-d)
ie; MD=(M+m)(D-d)
M(D–(D-d))=m(D-d)
M=m(D-d)/d

• A certain no of men can do the work in D days. If there were m less men, the work can be done in d days more. How many men were initially?

Let the initial number of men be M
Number of man days to complete work = MD
If there are M-m men, days taken = D+d
So, man days = (M-m)(D+d)
ie; MD=(M-m)(D+d)
M(D+d–D)=m(D+d)
M=m(D+d)/d

• Given A takes a days to do work. B takes b days to do the same work. Now A and B started the work together and n days before the completion of work A leaves the job. Find the total number of days taken to complete work?

Let D be the total number of days to complete work.
A and B work together for D-n days.
So, (D-n)(1/a+1/b)+n(1/b)=1
D(1/a+1/b)–n/a-n/b+n/b=1
D(1/a+1/b)=(n+a)/a
D=b(n+a)/(a+b) days.

The above tricks and shortcuts may be difficult to remember and may go against the systematic procedure that one is habitual of but with practice and logic it comes more naturally than the tiring conventional practices that eats up time and energy during examination.

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