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.
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
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.
Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24
Suppose the tank is filled in x minutes.
Then, x/2(1/24 + 1/40) = 1
(x/2) * (1/15) = 1 => x = 30 min.
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