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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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