Pipes and Cistern
- A tank has a leak which would empty it in 8 h. A tap is turned on which admits 3 L a min into the tank and it is now emptied in 12 h. How many litres does the tank hold?
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Work done by the inlet in 1 h = (1/8 - 1/24) = 1/24
Work done by the inlet in 1 min = 1/24 x 1/60 = 1/1440
∴ Volume of 1/1440 part = 3 LCorrect Option: A
Work done by the inlet in 1 h = (1/8 - 1/24) = 1/24
Work done by the inlet in 1 min = 1/24 x 1/60 = 1/1440
∴ Volume of 1/1440 part = 3 L
∴ Volume of the whole = 3 x 1440 = 4320 L
- Inlet A is four times faster than inlet B to fill a tank. If A alone can fill it in 15 min, how long will it take if both the pipes are opened together?
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Time taken by A to fill the tank, m = 15 min
∴ Time taken by B to fill the tank, n = 15 x 4 = 60
∴ Required time taken m x n/(m + n)Correct Option: B
Time taken by A to fill the tank, m = 15 min
∴ Time taken by B to fill the tank, n = 15 x 4 = 60
∴ Required time taken m x n/(m + n)
= (15 x 60)/(15 + 60) = (15 x 60)/75 = 12 min
- There are two inlets A and B connected to a tank. A and B can fill the tank in 16 h and 10 h, respectively. If both the pipes are opened alternately for 1 h, starting from A, then how much time will the tank take to filled?
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Part filled by A in 1 h = 1/16
Part fill by B in 1 h = 1/10
Part filled by ( A + B ) in 2 h = 1/16 + 1/10 = 13/80
∴ Part filled by ( A + B ) in 12 h = 6 x 13/80 = 78/80
∴ Remaining part = 1 - 78/80 = 2/80 = 1/40Correct Option: C
Part filled by A in 1 h = 1/16
Part fill by B in 1 h = 1/10
Part filled by ( A + B ) in 2 h = 1/16 + 1/10 = 13/80
∴ Part filled by ( A + B ) in 12 h = 6 x 13/80 = 78/80
∴ Remaining part = 1 - 78/80 = 2/80 = 1/40
Now, it is the turn of A
Time taken by A to fill 1/40 part of the tank = (1/40) x 16 = 2/5 h
∴ Total time taken ( 12 + 2/5) h = 122/5 h
- Two pipes X and Y can fill a cistern in 6 and 7 min, respectively. Starting with pipe X, both the pipes are opened alternately, each for 1 min. In what time will they fill the cistern?
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Part filled by X in 1st min and Y in the 2nd min = ( 1/6 + 1/7) = 13/42
Part filled by ( X + Y ) working alternately in 6 min = 1/2 x 13/42 x 6 =13/14
∴ Remaining Part = ( 1- 13/14 ) = 1/14Correct Option: B
Part filled by X in 1st min and Y in the 2nd min = ( 1/6 + 1/7) = 13/42
Part filled by ( X + Y ) working alternately in 6 min = 1/2 x 13/42 x 6 =13/14
∴ Remaining Part = ( 1- 13/14 ) = 1/14
Now, it is the turn of X, one-sixth part is filled in 1 min.
One-fourteenth part is filled in (6 x 1/14) min = 3/7
∴ Required time = ( 6 + 3/7 ) = 63/7 min
- A tap having diameter 'd' empty a tank in 40 min. How long another tap having diameter '2d' take to empty the same tank?
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Area of tap ∝ Work done by pipe.
When diameter is doubled, area will be four times. so, it will work four times faster.Correct Option: B
Area of tap ∝ Work done by pipe.
When diameter is doubled, area will be four times. so, it will work four times faster.
Hence, required time taken to empty the tank = 40 x 1/4 = 10 min.