Work done by the waste pipe in 1 minute = [-ve sign means emptying]
Volume of part = 3 gallons
Volume of whole = (3 x 40) gallons = 120 gallons.
1/10 + 1/15 = 1/6 x 3 = 1/2
1 - 1/2 = 1/2
1/10 + 1/15 - 1/x = 1/2
x = 8.57 min
Let the filling capacity of the pump be x /min.
Then, emptying capacity of the pump=(x+10) /min.
so,
Clearly,pipe B is faster than pipe A and so,the tank will be emptied.
part to be emptied = 2/5
part emptied by (A+B) in 1 minute=
so, the tank will be emptied in 6 min
1/8 - 1/x = 1/10
=> x = 40 hrs
The work to be done = Capacity of reservoir = 30 litres.
1st Minute ---> inlet pipe opened ---> 5 lit filled
2nd minute ---> inlet pipe closed; outlet pipe opened ---> 4 lit emptied
In 2 minutes (5 litres - 4 litres = 1lit) is filled into the reservoir.
It takes 2 minutes to fill 1lit ---> it takes 50 minutes to fill 25 litres into the reservoir.
In the 51st minute inlet pipe is opened and the reservoir is filled.
Volume of water collected in the tank in 1 hour
? (0.3 × 0.2 × 20km × 1000mts) = 1200 m cubic
If after t hours, the water is at height of 12m,
1200t=200×150×12
? t = 300 Hours.
Capacity of the tank =(12 x 13.5) liters =162 liters.
Capacity of each bucket =9 liters
Number of buckets needed = 162/9 =18.
Let the slower pipe alone fill the tank in x minutes.
Then,faster pipe will fill it in x/3 minutes.
Given A alone can fill the tank of capacity 240 lit in 16 hrs.
=> A can fill in 1 hr = 240/16 = 15 lit
=> B alone can fill the tank of capacity 240 lit in 12 hrs.
=> B can fill in 1 hr = 240/12 = 20 lit
Now, (A + B) in 1 hr = 15 + 20 = 35 lit
But they are opened for 2 hrs
=> 2 x 35 = 70 lit rae filled
Remaining water to be filled in tank of 240 lit = 240 - 70 = 170 lit.
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