Pipe A, B, C can empty a full tank in 12, 15, 20 hrs. When will they empty the full tank when pipe A is constantly opened and pipe B & pipe C opened alternatively one hour each? *
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Answer:
Given : A,B,C taps fill the tank in 12, 15,20.
To solve this fist we to find out total work that would be LCM of 12,15,20 = 60 (capacity of tank)
Now find out the individual works of A,B,C
Individual work of A=5units
Similarly B=4units , C = 3units/hr
A is open all the time and B and C are open alternatively
A+B = 5+4 =9units/hr
A+C=5+3=8units/hr
Total of 17units in 2hrs
Total capacity = 60/17 = 3.5 ×2hrs
Tank will be filled in 3.5×2hrs = 7hrs
Answer: 7hrs
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