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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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)
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