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    Pipe Q can fill the tank in 60 hours while pipe R may fill in 45 hours. Q and R pipes are opened together for 6 hours after which pipe W is also opene
    Question

    Pipe Q can fill the tank in 60 hours while pipe R may fill in 45 hours. Q and R pipes are opened together for 6 hours after which pipe W is also opened to empty the tank. All three pipes are opened simultaneously for 24 hours to reach the half-level mark. How much time (in hours) will pipe W alone take to empty the entire tank?

    A.

    48

    B.

    42

    C.

    36

    D.

    30

    Correct option is C

    Given:
    Pipe Q fills the tank in 60 hours.
    Pipe R fills the tank in 45 hours.
    Pipes Q and R are opened together for 6 hours.
    After 6 hours, pipe W is also opened to empty the tank, and all three pipes are then opened for 24 hours.
    After 24 hours of combined operation, the tank reaches half capacity.    
    Formula Used:  
    Total work = Efficiency ×\times Total time 
    Solution: 
    LCM of 60, 45 = 180 
    Let total work is 180 unit 
    then, 

    Pipe Q efficiency = 18060=3 unit/h\frac{180}{60} = 3 \ unit/h  

    Pipe R efficiency = 18045=4 unit/h\frac{180}{45} = 4 \ unit/h 

    Pipe P and Q 6 hours work = 6×(3+4)=42 unit6 \times(3+4) = 42\ unit 
    Now, pipe W  (empty) open and all three opened for 24 hours half the tank is filled i.e = 1802=90 unit\frac{180}{2} =90\ unit 
    42 unit tank already filled by Pipe Q and R So, 
    Remaining = 90 - 42 = 48 unit, is filled by P, R, and W in 24 hours 
    So, Combined efficiency of P+ R - W = 4824=2 unit/hour\frac{48}{24} = 2 \ unit/hour 
    Now, Efficiency of W = 7 - 2 = 5 unit/hour 
    Time taken to empty the full tank by W = 1805=36 hours\bf \frac{180}{5} = 36 \ hours

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