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Pipes A and B can fill a tank in 5 and 3 hrs respectively. Pipe C can empty empty it in 15 h. The tank is half full. All the three pipes are in operation simultaneously. After how much time the tank will be full ?
  • a)
    1(7/15)hrs
  • b)
    2(1/11)hrs
  • c)
    1(1/14)hrs
  • d)
    2(3/11)hrs
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Pipes A and B can fill a tank in 5 and 3 hrs respectively. Pipe C can ...
In 1 hr = 1/5+1/3 – 1/15 = 3+5-1/15 = 7/15
½ tank filled by 3 pipes = 15/7*1/2 = 15/14 =1(1/14)
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Most Upvoted Answer
Pipes A and B can fill a tank in 5 and 3 hrs respectively. Pipe C can ...
Given:
Pipe A can fill the tank in 5 hrs
Pipe B can fill the tank in 3 hrs
Pipe C can empty the tank in 15 hrs
The tank is half full.

Approach:
Let the capacity of the tank be C.
As the tank is half full, the remaining capacity of the tank is C/2.
Let's assume that the tank is filled in 't' hours.
In 't' hours, pipe A fills t/5th of the tank.
In 't' hours, pipe B fills t/3rd of the tank.
In 't' hours, pipe C empties t/15th of the tank.
As all pipes are operated simultaneously, the net amount of water filled in the tank in 't' hours is given by the sum of the amount of water filled by pipes A and B minus the amount of water emptied by pipe C.

Calculation:
The equation for the net amount of water filled in the tank is given as:
(t/5 + t/3) - (t/15) = C/2
On solving the above equation, we get:
t = 17.5 hours

Therefore, the tank will be full after 1(1/14) hours.

Answer: Option (c)
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Community Answer
Pipes A and B can fill a tank in 5 and 3 hrs respectively. Pipe C can ...
1/5+1/3-1/15=7/15, so 1/2 empty tank can be filled in 15/7*2=15/14 hrs =1(1/14) hrs. Option C
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