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Time and Work (Man Days Concept (If M1 men can do W1 amount of work in D1…

- - - - 1) If A and B are opened simultaneously, find the time taken to fill the tank/the time after which the tank would overflow.
        
        Soln:
        (a) {1/A + 1/B} = 1/(Total time taken)
        (b) a + b = Total time taken
        Note: Questions of the same type will be given with slight alteration like asking B given A and total time taken.
      - 2) If C is outlet pipe located at the bottom of the tank, and all 3 pipes are operated simultaneously, then find the time taken for the tank to reach its full capacity.
        
        Soln:
        (a) {1/A + 1/B - 1/C} = 1/(Total time taken)
        (b) a + b - c = Total time taken
        Note: Questions of the same type will be given with slight alteration like asking C given A,B and total time taken.
      - 3) If C is outlet pipe located at the bottom of the tank, and all 3 pipes are operated in a series one after the other, then find the time taken for the tank to reach its full capacity.
        
        Soln:
        If the pipes are operated in a series for the same amount of time say 1 min each, then
        (1) Calculate the rate at which the tank is filled in 3 mins, i.e., 1st min A operates, in the 2nd min B operates and in the 3rd min C operates.
        (2) With this rate, calculate the time taken to fill the tank
        Note: In these kind of questions, order in which pipes operate plays a pivotal role. If the outlet pipe C operates for every 3 minutes, then we should not consider it in the last leap of 3rd min interval as the tank would be already full by the end of 2nd min. So always check the order of operation of the pipes and do small adjustments to final answer.
      - 4) Pipe B is opened 'n' minutes after pipe A is opened and the tank will overflow after 'x' minutes. Find n given x?
        
        Soln:
        (1) A works for x minutes
        (2) B works for x-n minutes
        => {x/A + (x-n)/B} = 1/x
        From this Eqn, given A, B and x, find n.
      - 5) Pipes A and B are opened simultaneously and Pipe B is closed 'n' minutes before the tank will overflow. Find the time taken for the tank to fill completely?
        
        Soln:
        (1) Assume B operates for x minutes
        => A operates for x+n minutes
        => {(x+n)/A + x/B} = 1/(x+n)
        From this Eqn, given A, B and n, find x.
- - - - Sol:
        A men can do a piece of work in n days
        => 1 man can do it in nA days
        Similarly, if B women can do a piece of work in m days
        => 1 woman can do it in mB days
        Therefore, time taken to complete the work if a man and a woman work together => (1/nA) + (1/mB) = 1
    - - Sol: Rate of A men = 1/n
        Rate of B women = 1/m
        Therefore, time taken to complete the work of all men and women work together
        =>Inverse of [(1/n) + (1/m) = 1]
    - - Sol:
        Let rate of work done by a boy = 1/g and a man = 1/h
        => a(1/g) + b(1/h) = 1/x ------- (1)
        => c(1/g) + d(1/h) = 1/y ------- (2)
        Solve (1) and (2) for g and h
        Then calculate 1/g + 1/h
- - - - Sol:
        (1/x) + (1/y) + (1/z) = 1
        Time taken will be the inverse of the above equation
    - - Sol:
        This is similar to the above question, but the sequence is important here.
        Assume x = 10, y = 20, z = 40
        => 1st day ---- 1/10
        => 2nd day ---- 1/20
        => 3rd day ---- 1/40
        Therefore, in the first 3 days, A, B and C can complete [(1/10) + (1/20) + (1/40)] th of the work
        => In 3 days, 7/40 of the work will be completed.
        => In 3x5 = 15 days, 35/40 part of the work will be completed.
        Left out is 5/40 part of the work, A can complete 4/40 in 1 day
        => Left out is 1/40 of the work. B can complete 2/40th work in 1 day
        => B takes half day to complete the remaining 1/40th work
        => Therefore, total time taken = 15 + 1 + 0.5 = 16.5 days
        Note: In this kind of questions, sequence of men is very important. That determines the answer.
    - - Sol:
        Let 'm' be the number of days taken for the work to complete after B left.
        => A worked for n + m days and B worked for n days
        => (n+m)/x + (n/y) = 1
        Calculate m, answer will be n + m
    - - Sol:
        Let 'm' be the total number of days taken for the work to complete.
        => A worked for m days and B worked for m - n days
        => (m)/x + (m - n/y) = 1
        Calculate m, answer will be m
    - - Sol:
        1st day ----- 25x1 = 25 sq.m.
        2nd day ----- (25+1)x1 = 26 sq.m.
        3rd day ----- (25+1+1)x1 = 27 sq.m.
        4th day ----- (25+1+1+1)x1 = 28 sq.m.
        ......so on
        => Time taken = 25 + 26 + 27 + ...... = 330 sq.m
        The above sequence is an AP with first term = 25 and CD = 1
        => Sum to n terms in an AP => (n/2)[2(25) + (n-1)1] = 330
        => n^2 + 49n - 660 = 0
        Solving for n, we get n = 11
    - - Sol: Ratio of their wages earned = ratio of their work done
        = (Rate of work)(Time taken)
        => Ratio of rates of work done by A, B and C = (1/x) : (1/y) : (1/z)
        => A and B worked for m days and C worked for m - n days
        => Therefore, ratio of work done = [(1/x)(m)] : [(1/y)(m)] : [(1/z)(m-n)]
        From the above ratio, calculate individual wages
        
        Question 1
        40 boys can construct a wall in 50 days and are paid 3000 in total. 80 boys and a few girls can construct the same wall in 20 days. Find the difference between the total amount earned by all the 80 boys and girls put together.
        
        Sol: 40 boys can construct a wall in 50 days => 80 boys can do it in 25 days
        => In 20 days, 80 boys can do 4/5th of work
        => Remaining 1/5th of work is done by some girls in 20 days
        => Efficiency of boy : Efficiency of girl = (4/5) : (1/5) = 4:1
        => Difference in amount = (3/5)(3000) = 1800