SBI Clerk 2018 Coaching Day 2 - Time and Work (Shortcuts & Practice Tests)

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Hello friends, as some people are asking me to take gap for Compound Interest, I have decided to postpone the lesson on Compound Interest for another day. Today I am sharing a simple and important topic 'Time and Work' with shortcuts. The main advantage of Time and Work problems is, you can easily solve them without even knowing the basic formulas or shortcuts. But the shortcut methods help us solving the problems in less time. Let's start today's lesson.

Time and Work Shortcuts

Time and Work Formulas :

1st of all you need to remember the basic formula,

i.e., Work = Strength x Time

in Simple words, the Work done by you is depends upon the amount of Strength you put on that work and the amount of Time you spent to do that work.

Simple, isn't it ?

Now let's derive some shortcut formulas from this main formula.

Strength x Time = Work

If you send the work to right hand side, then it will become,

(Strength x Time) / Work = 1

This condition will be true for every case, No matter whatever the numbers are given.

 So, (SxT)/W will be same for all the cases.  Keep this point in your mind.

Now lets see some more important conditions :

If they give you Days :

• If A can do some work in n days, then he can do 1/n work in One Day.

If they give you Work :

• If A can do 1/n work in One Day, he can finish it in n days.

If A is TWICE as good a workman as B :

• This means, A can do 2 times work than B. SO,
• The ratio of work done by A and B is 2 : 1
• The ratio of time taken by A and B to finish the work is 1:2

Now let's work with some models so that you will understand better. As usual I am explaining one problem per model and giving 3 problems as homework for your practice. Solve them and share your answers at the comments section below.

Time & Work : Examples and Practice Problems

Model 1 :

1. 36 men can do a piece of work in 25 days. In how many days can 30 men do it ?
   1. 28
   2. 30
   3. 32
   4. 36
   5. None of these

Solution :

In this problem the work is same for both groups of men.

So, let's equate it                   

                    (S X T) / W = (s X t) / w
 
                  = >       (S X T) / W = (s X t) / w    [ Cancel W for both sides]

               Now, substitute the values...

                   = >  36 X 25  = 30 X t  

                     = >  (36 X 25) / 30  =  t  => t = 900/3  =  30

Practice Problems on Model 1 

1. 63 Men can do a piece of work in 52 days. In how many days can 91 men do it ?
   1. 28
   2. 30
   3. 32
   4. 36
   5. None of these
2. 48 Men can do a piece of work in 36 days. In how many days can 54 men do it ?
   1. 28
   2. 30
   3. 32
   4. 33
   5. None of these
3. 36 Men can do a piece of work in 35 days. In how many days can 45 men do it ?
   1. 28
   2. 30
   3. 32
   4. 33
   5. None of these

Model 2 :

1. 32 men can do a piece of work in 15 days working for 6 hours a day. In how many days will 40 men can finish it if they work for 8 hours a day?
   1. 8
   2. 9
   3. 10
   4. 12
   5. None of these

Solution :

                  32 x 15 x 6  = 40 x d x 8 = 9

Practice Problems on Model 2

1. 63 men can do a piece of work in 36 days working for 6 hours a day. In how many days will 27 men finish it if they work for 7 hours a day ?
   1. 25
   2. 66
   3. 72
   4. 92
   5. None of these
2. 36 Men can do a piece of work in 25 days working for 7 hours a day. In how many days will 42 men finish it if they work for 6 hours per day ?
   1. 25
   2. 66
   3. 72
   4. 92
   5. None of these
3. 55 Men can do a piece of work in 45 days working for 8 hours a day. In how many days will 40 men finish it if they work for 7.5 hours a day ?
   1. 25
   2. 66
   3. 72
   4. 92
   5. None of these 

Model 3 :

1. If 16 men can build a wall of 52 m long in 25 days working for 8 hours a day, in how many days can 64 men build a similar wall of 260m long working for 10hrs a day ?
   1. 12
   2. 20
   3. 25
   4. 28
   5. None of these

Solution :

We already know that (Strength X Time) / Work = 1

We can write =>

          (16 * 200) / 52  =  (64 * X * 10) / 260 

                                =>  X = 25

Practice Problems on Model 3 :

1. If 60 men can build a wall of 52 m long in 42 days working for 8 hours a day, in how many days can 35 men build a similar wall of 26 m long working for 9 hours a day ?
   1. 32
   2. 20
   3. 25
   4. 28
   5. None of these
2. If 36 men can build a wall of 51 m long in 45 days working for 8 hours a day, in how many days can 120 men build a similar wall of 85 m long working for 7.5 hours a day ?
   1. 12
   2. 20
   3. 25
   4. 24
   5. None of these
3. If 96 men can build a wall of 76 m long in 44 days working for 6 hours a day, in how many days can 55 men build a similar wall of 260 m long working for 8 hours a day ?
   1. 50
   2. 60
   3. 72
   4. 90
   5. None of these

Model 4 :

1. A man engaged 10 laborers to make 320 toys in 5 days. After 3 days he found that only 120 toys were made. How many additional men should he engage to finish the work in time ?
   1. 15
   2. 20
   3. 25
   4. 30
   5. None of these

Solution : 

Here, equate the complete work with the remaining work

                       (10 x 3) / 120 = (10+x) x 2 / 200 

                          = >   10+x = 25   =>   x = 15

Practice Problems on Model 4 :

1. A man engaged 15 laborers to make 320 toys in 10 days. After 7 days he found that only 175 toys were made. How many additional men should he engage to finish the work in time ?
   1. 4
   2. 14
   3. 29
   4. 30
   5. None of these
2. A man engaged 45 laborers to make 350 toys in 21 days. After 15 days he found that only 210 toys were made. How many additional men should he engage to finish the work in time ?
   1. 4
   2. 14
   3. 29
   4. 30
   5. None of these
3. A man engaged 20 laborers to make 360 toys in 20 days. After 12 days he found that only 200 toys were made. How many additional men should he engage to finish the work in time ?
   1. 4
   2. 14
   3. 29
   4. 30
   5. None of these

Model 5 :

1. A can do a task in 20 days and B can do it in 30 days. In how many days can they finish it if they work together ?
   1. 10
   2. 12
   3. 15
   4. 16
   5. None of these

Solution :

A can do a task in 20 days. That means A's one day's work = 1/20

Similarly, B's one day's work = 1/30

         => Their one day's work = (1/20)+(1/30)  =  (3+2)/60  =  5/60  =  1/12

               This is their one day's work TOGETHER.

                               So, obviously the number of days will be = 12

One More Short Cut : calculate Product/Sum   =  (20 X 30) / 50 = 12 That's it ;)

Practice Problems on Model 5 :

1. A can do a job in 60 days and B in 40 days. In how many days can they finish it if they work together ?
   1. 10
   2. 12
   3. 15
   4. 24
   5. None of these
2. A can do a job in 24 days and B in 40 days. In how many days can they finish it if they work together ?
   1. 10
   2. 12
   3. 15
   4. 24
   5. None of these
3. A can do a job in 120 days and B in 180 days. In how many days can they finish it if they work together ?
   1. 60
   2. 72
   3. 90
   4. 96
   5. None of these

Model 6 :

1. A, B and C can do a job in 20 days, 30 days and 60 days respectively. If they work together, in how many days will the work be finished ?
   1. 9
   2. 10
   3. 12
   4. 15
   5. None of these

Solution :

          (1/20)+(1/30)+(1/60) = (3+2+1)/60  =  6/60  = 1/10

               So, the number of days is = 10

Practice Problems on Model 6 :

1. A, B and C can do a job in 40 days, 30 days and 60 days respectively. If they work together, in how many days will three such jobs be finished ?
   1. 30
   2. 40
   3. 60
   4. 72
   5. None of these
2. A, B and C can do a job in 20 days, 30 days and 36 days respectively. If they work together, in how many days will the work will be finished ?
   1. 9
   2. 10
   3. 12
   4. 15
   5. None of these
3. A, B and C can do a job in 20 days, 30 days and 24 days respectively. If they work together, in how many days will the work be finished ?
   1. 9
   2. 10
   3. 12
   4. 15
   5. None of these

Model 7 :

1. Two taps A and B can fill a tank in 10 hours and 15 hours respectively. a third tap C can empty the full tank in 12 hours. How many hours will be required if all of them are opened simultaneously to fill in an empty tank completely ?
   1. 9
   2. 10
   3. 12
   4. 15
   5. None of these

Solution :

        Here, first two are Inlets (which can fill the tank) and the last one is Outlet (which can empty the tank),

If there is Inlet use '+', if there is Outlet use '-'

          So,      (1/10)+(1/15)-(1/12) =  (6+4-5)/60  =  5/60  = 1/12

               So, the answer is 12

Practice Problems on Model 7 :

1. Two taps A and B can fill a tank in 20 hours and 15 hours respectively. A third tap C can empty the full tank in 12 hours. How many hours will be required if all of them are opened simultaneously to fill an empty tank completely ?
   1. 9
   2. 10
   3. 30
   4. 15
   5. None of these
2. Two taps A and B can fill a tank in 25 min and 15 min respectively. A third tap C can empty the full tank in 10 min. How many hours will be required if all of them are opened simultaneously to fill in an empty tank completely ?
   1. 2
   2. 2.5
   3. 3
   4. 3.5
   5. None of these
3. Two taps A and B can fill a tank in 24 min and 36 min respectively. A third tap C can empty the full tank in 15 mins. How many minutes will be required if all of them are opened simultaneously to fill in an empty tank completely ?
   1. 6
   2. 8
   3. 12
   4. 15
   5. None of these

Model 8 :

1. A and B can do a job in 12 days. B and C in 15 days and C and A in 20 days. In how many days can they finish it if they work TOGETHER ?
   1. 9
   2. 10
   3. 12
   4. 15
   5. None of these

Solution : 

       A+B = 12

        B+C = 15

       C+A = 20

So here,

               A+B's One day's work = >   1/ (A+B) = 1/12

              B+C's one day's work  =>  1/ (B+C) =  1/15

               C+A's one day's work = > 1/(C+A) = 1/20

              Just Add them  =>  1/( 2A+2B+2C)  =  12/60  =  1/5

                                  =>  1/2(A+B+C)  =  1/5

                                     =>1/(A+B+C)  =  1/10   [this is their one day's work TOGETHER]
             

                      So, they can finish it in 10 days :)

Practice Problems on Model 8 :

1. A and B can do one third of a job in 20 days, B and C in 15 days and C and A in 30 days. In how many days can they finish full job if they work together ?
   1. 36
   2. 40
   3. 48
   4. 60
   5. None of these
2. A and B can do a job in 27 days, B and C in 45 days and C and A in 36 days. In how many days can they finish 47 such jobs if they work together ?
   1. 450
   2. 480
   3. 540
   4. 600
   5. None of these
3. A and B can do a job in 60 days, B and C in 90 days and C and A in 100 days. In how many days can they finish 17 such jobs if they work together ?
   1. 450
   2. 480
   3. 540
   4. 600
   5. None of these

Model 9 :

1. A and B can do a job in 12 days. B and C can do the same job in 15 days. C and A in 20 days. In how many days can A alone finish the whole task ? 

1. 24
2. 28
3. 30
4. 32
5. None of these

Solution :
     
         A+B = 1/12

         B+C = 1/15

         C+A = 1/20

Here we need A, so take a pair which is NOT HAVING A  and subtract it from the others,

               so,  A+B-(B+C)+C+A  =  A+B-B-C+C+A =  2A  
    

But according to our Problem, 2A =  (1/12)-(1/15)+(1/20)

                                    =>A = 1/30 (this is A's one day work but we need A's total work)
                       
                                           => A = 30 Days
Practice Problems on Model 9 :

1. A and B can do a job in 20 days. B and C in 15 days and C and A in 30 days. In how many days can B alone finish the whole work in the above problem ?
   1. 24
   2. 28
   3. 30
   4. 32
   5. None of these
2. A and B can do a job in 27 min. B and C in 45 min and C and A in 36 min. In how many hours can C alone finish 7 such jobs ?
   1. 24
   2. 28
   3. 30
   4. 18
   5. None of these
3. A and B can do a job in 60 days, B and C in 90 days and C and A in 100 days. In how many days can C alone finish the whole work ?
   1. 450
   2. 480
   3. 540
   4. 600
   5. None of these

Model 10 :

1. A and B can do a piece of work in 20 days. A alone can do it in 30 days. In how many days can B alone do it ?
   1. 40
   2. 45
   3. 50
   4. 60
   5. None of these

Solution :

    Per day work of A and B = 1/20

         Work done by A=  1/30

       So, B's one day work =  1/20  - 1/30  =  (3-2)/60  =  1/60 

                     =>  B's work is 60 Days

Short Cut :   Product/diff  =  600/10  =  60

Practice Problems on Model 10 :

1. A and B can do a piece of work in 20 days. A alone can do it in 24 days. In how many days can B alone do it ?
   1. 100
   2. 108
   3. 120
   4. 150
   5. None of these
2. A and B can do a piece of work in 20 days. A alone can do it in 25 days. In how many days can B alone do it ?
   1. 100
   2. 108
   3. 120
   4. 150
   5. None of these
3. A and B can do a piece of work in 40 days. A alone can do it in 30 days. In how many days can B alone do it ?
   1. 96
   2. 120
   3. 150
   4. 180
   5. None of these

Work with these models and solve the given practice problems. Share your solutions in the comments section below. Will post post more methods along with practice problems soon. Hit the REFRESH button after completing these 10 models. All the Best :)

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