Saturday, June 2, 2012

Time and Work Solved Problems

Difficulty Level - Medium
      
      1.     A can work on 1km railway track in 1 day. In how many days, will he able to complte the work on 12km railway track?

Soln: no. of days = total work / work done in 1 day
Therefore, no. of days taken = 12/1 = 12 days


      2.     A can complete the work in 15 days. What fraction of work will be completed in 1 day?

Soln.:   Let the total work is 1 unit.
Work in 1day = total work/no. of days to complete
                                    = 1/15th of work

      3.     A can do a piece of work in 3 days and B can do a piece of work in 5 days. In how many days will the work be completed if both A and B work together?

Soln.: Using formula:
                  Work done by A in 1 day = 1/3
                  Work done by B in 1 day = 1/5
                  Total work done by A and B in 1 day = 1/3 + 1/5 = 8/15
                  Therefore, no. of days to complete work by A and B together = 1/(Total work) = 1/(8/15) = 15/8 days which is less than 3 and 5

    Using shortcut/analysis/assumption
                  Let us consider the total work be 15 units (LCM of 3 and 5)
                  So work done by A in 1 day = 15/3 = 5 units
                  Similarly work done by B in 1 day = 15/5 = 3 units
                  So total work done by A and B in 1 day = 5 + 3 = 8 units
                  Therefore, no. of days to complete total work i.e. 15 units = total work/work done in 1 day = 15/8 days
       
      Note:
                a.     Work done by A and B in 1 day will always be greater than that of A and B individually
          b. No. of days taken by A and B together will always be less than that of A and B individually

      4.     A can do a piece of work in 6 days, B can do a piece of work in 4 days and C can do a piece of work in 12 days. Find the no. of days to complete the work if A, B and C work together?

Soln.: Using formula:
                  Work done by A in 1 day = 1/6
                  Work done by B in 1 day = ¼
                  Work done by C in 1 day = 1/12
                  Total work done by A, B and C in 1 day = 1/6 + ¼ + 1/12 = 12/24 = 1/2
                  Therefore, no. of days to complete work by A, B and C together = 1/(Total work) = 1/(1/2) = 2 days which is less than 4, 6, 12

      Using shortcut/analysis/assumption
                  Let us consider the total work be 24 units (LCM of 4, 6, 12)
                  So work done by A in 1 day = 24/4 = 6 units
                  work done by B in 1 day = 24/6 = 4 units
                  work done by C in 1 day = 24/12 = 2 units
                  So total work done by A, B and C in 1 day = 6 + 4 + 2 = 12 units
                  Therefore, no. of days to complete total work i.e. 24 units = total work/work done in 1 day = 24/12 = 2 days

The above Note is valid here as well.

      5.     A can do a piece of work in 6 days and B can do a piece of work in 12. Find the no. of days to complete the work if A and B work alternatively?

Soln.: Using formula:
                  Work done by A in 1 day = 1/6
                  Work done by B in 1 day = 1/12
                  Total work done by A and B working 1 day each = 1/6 + 1/12 = 3/12 = ¼
                  Therefore, 1/4th of work is done in 2days.
                  No. of days to complete total work if A and B work alternatively = 1/((1/4)/2) = 8 days

      Using shortcut/analysis/assumption
                  Let us consider the total work as 12 units (LCM of 6, 12)
                  So work done by A in 1 day = 12/6 = 2 units
                  work done by B in 1 day = 12/12 = 1 unit
                  Total work done by A and B working 1 day each = 2 + 1 = 3 units in 2 days
                  Therefore, work done in 1 day = work/no. of days = 3/2 units
                        No. of days to complete work = total work/work in 1 day = 12/(3/2) = 8 days

      6.     30 men can complete a job in 40 days. Then 25 men can complete the same job in how many days?

Soln.: As per M1D1 = M2D2
               30 * 40 = 25 * x  => x = 30 * 40/25 = 48 days

      7.   30 men can complete 1500 units in 24 days working 6hrs a day. In how many days can 18 men can complete 1800 units working 8 hrs a day?



Soln.: As per the formula  (from my earlier blog), M1D1h1/W1 = M2D2h2/W2
          => 30*24*6/1500 = 18*x*8/1800
          => x = 36 days


      8.     A and B can do a work in 10 and 15 days respectively. Then combinedly A & B, in how many days the work will be completed?

      Soln.: As per the formula  (from my earlier blog), x*y/(x + y)
               A and B together can complete the work in 10 * 15/(10 + 15) = 6 days

      9.   A can do a work in 10 and, A and B together can do a work in 6 days. In how many days B can complete the same work?


      Soln.: As per the formula  (from my earlier blog), x*y/(x - y)
               B alone can complete the work in 10 * 6/(10 - 6) = 15 days


      10.  A is twice faster than B and B can complete in 12 days alone. Find the number of days to complete if A and B together work?

            Soln.: Given B works in 12 days
      A is twice faster than B => A takes 2 times less time than B
      Therefore, A completes work in 12/2 = 6 days
            A and B together can complete in 12 * 6/(12 + 6) = 4 days


5 comments:

  1. hi i am pretty confusd abut why and where we used m1d1 forumula ..plz explain ...

    ReplyDelete
  2. If m1 can do a work w1 in d1 days and m2 can complete work w2 in d2 days then

    (m1*d1)/w1=(m2*d2)/w2

    If the both works w1 and w2 are same then

    m1*d1=m2*d2

    ReplyDelete
  3. Excuse me ! Please Explain M1D1h1/W1 = M2D2h2/W2 and x*y/(x + y) formulaes.

    ReplyDelete
  4. thanks man well informative! :)

    ReplyDelete