Time and Work

Time and Work


In this post, we shall discuss about the Time and Work. Lets start with some basic formulas.

Work = Strength X Time
 =>   Strength X Time = Work
=>   (Strength X Time ) / Work = 1   :)

So, For every Case
(Strenth X Time) / Work = 1      [ No matter the values]

So, (SxT)/W will be same for all the cases.

 I mean   st/w  =  ST/W

Keep this point in your mind.

Now lets see some more important formulas

If Days are Given :

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

If Work is Given :

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

If A is TWICE as good a work man as B, then 

• The ratio of work done by A and B = 2:1
• The ratio of time taken by A and B to finish the work = 1:2 (Please Dont be confused)

Now lets do some problems on Time and Work

Time and Work - Problems


1.  36 men can do a piece of work in 25 days. In how many days can 30 men do it?

    Here , the work is same.. So,
                   
                    (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

2.  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?

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

3. 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?
     As we know that (Strength X Time) / Work = 1 (Read details Here)

We can write =>

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

                                =>  X = 25

4. 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?

            here, equate the complete work with the remaining work

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

                          = >   10+X = 25   =>   X = 15

5.  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?


        A's one day's work = 1/20

         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

     Short Cut : calculate Product/Sum   =  (20 X 30) / 50 = 12 Thats it

6.  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?

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

               So, the number of days is = 10

7.  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?

        Here, first two are Inlets and the last one is Outlet,

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

               So, our answer is 12

8. 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?

       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 :)

9.  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???

         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

  10. 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?

    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