Question 1

If two pipes function together, the tank will be filled in 12 h . One pipe fills the tank in 10 h faster than the other. How many hours does the faster pipe take to fill up the tank?

Accepted Answer

Let one pipe takes m h to fill the tank. Then, the other pipe takes (m - 10) h. According to the question. ∴ 1/m + 1/(m - 10) = 1/12 ⇒ (m - 10 + m)/{m(m - 10)} = 1/12 ⇒ 12m( m - 10 + m ) = ( m - 10 ) ⇒ m2 - 34m + 120 = 0 ⇒ m2 - 30m - 4m + 120 = 0 ⇒ ( m - 30 )( m - 4 ) = 0 ∴ m = 30 or 4 ∴ Faster pipe will take ( 30 - 10 ) h = 20 h to fill the tank.

Question 2

A pipe P can fill a tank in 12 min and another pipe R can fill it in 15 min. But, the 3rd pipe M can empty it in 6 min. The 1st two pipes P and R are kept open for double the 2.5 min in beginning and then the 3rd pipe is also opened. In what time is…

Accepted Answer

According to the question, Double the 2.5 min = 5 min Now, part filled in 5 min = 5 x ( 1/12 + 1/15 ) = [5 x ( 5 + 4)]/60 = (5 x 9)/60 = 3/4 Part emptied in 1 min when P, R and M, all are opened. = 1/6 - ( 1/12 + 1/5 ) = 1/6 - (5 + 4)/60 = 1/60 One - sixtieth part is emptied in 1 min. ∴ Three - fourth part will be emptied in 60 x (3/4) = 15 x 3 = 45 min.

Question 3

Three pipes A, B and C can fill a tank in 30 min, 20 min and 10 min, respectively. When the tank is empty, all the three pipes are opened. If A, B and C discharge chemical solution P, Q and R respectively, then the part of solution R in the liquid i…

Accepted Answer

Total quantity of solution P, Q and R from A, B and C receptively, after 3min = 3/30 + 3/20 + 3/10 = 3 x (2 + 3 + 6)/60 = 11/20 Quantity of solution R in liquid in 3 min = 3/10 ∴ part of solution R = [3/10] / [11/20 ] = (3 x 20)/(10 x 11) = 6/11

Question 4

There are 7 pipes attached with a tank out of which some are inlets and some are outlets. Every inlet can fill the tank in 10 h and every outlet can empty the tank in 15 h. When all the pipes are opened simultaneously, the tank is filled up in 2 8/ …

Accepted Answer

Let number of outlets be x ∴ Number of inlets = ( 7 - x ) Time taken to fill the tank when all the pipes are opened = 30/11h Part of tank filled in 1 h when all the pipes are opened = 11/30 h According to the question, (7 - x)/10 - x/15 = 11/30 ⇒ {3(7 - x) - 2x}/30 = 11/30 ⇒ 21 - 3x - 2x = 11 ⇒ 5x = 10 ∴ x = 2 Hence, number of outlets = 2 and number of inlets = 7 - 2 = 5

Question 5

Three taps A, B and C fill a tank in 20 min, 15 min and 12 min, respectively. If all the taps are opened simultaneously, how long will they take to fill 40% of the tank?

Accepted Answer

Part of the tank filled in 1 min by A, B and C = 1/20 + 1/15 +1/12 = (3 + 4 + 5)/60 = 12/60 = 1/5 ∴ Time taken by A, B and C to fill the tank = 5 min ∴ Time taken by A, B and C to fill 40% of the tank = 40% of 5 = (40/100) x 5 = 2 min.

Question 6

Capacity of tap B is 80% more than that of A. If both the taps are opened simultaneously, they take 45 h to fill the tank. How long will B take to fill the tank alone?

Accepted Answer

Let time taken by B to fill the tank = T h. ∴ Time taken by A to fill the tank = T + (T x 80)/100 = 9T/5 h According to the formula, Time taken by both the taps to fill the tank = ab /(a + b) ⇒ 45 = (T x 9T/5)/(T + 9T/5) ⇒ 45 x 14T/5 = 9T2/5 ∴ T = 45 x 14/9 = 70 h

Question 7

Taps A, B and C are attached with a tank and velocity of water coming through them are 42 L/h, 56 L/h and 48 L/h, respectively. A and B are inlets and C is outlet. If all the taps are opened simultaneously, tank is filled in 16 h. What is the capaci…

Accepted Answer

Quantity of water admitted by tap A in 1 h = 42 L Quantity of water admitted by tap B in 1h = 56 L Quantity of water removed by tap C in 1 h = 48 L So, quantity of water filled in the tank in 1 h = ( 42 + 56 - 48 ) L = 50 L ∴ Quantity for water filled in 16 h = 16 x 50 = 800 L Hence, capacity of tank = 800 L

Question 8

Two taps A and B can fill a tank in 25 min and 20 min, respectively. But taps are not opened properly so the taps A and B allow 5/6th and 2/3rd part of water, respectively. How long will they take to fill the tank?

Accepted Answer

Part of the tank filled with A and B in 1 min = 1/25 x 5/6 + 1/20 x 2/3 = 1/30 + 1/30 = 2/30 = 1/15 Hence, Time taken to fill the tank = 15 min

Question 9

Two taps A and B can fill a tank in 20 min and 30 min, respectively. An outlet pipe C can empty the full tank in 15 min. A, B and C are opened alternatively, each for 1 min. How long will the tank take to be filled?

Accepted Answer

Part of the tank filled by the taps A, B and C in 3 min = 1/20 + 1/30 - 1/15 = (3 + 2 - 4)/60 = 1/60 ∴ Time taken to fill [ 1 - ( 1/20 + 1/30 )] or 55/60th part of the tank = 3 x 55 = 165 min Remaining part of the tank = 1 - 55/60 = 5/60 = 1/12 Tap A fill 1/0 part in 1 min, then Remaining part = 1/12 - 1/20 = (5 -3)/60 = 2/60 = 1/30 i.e, 1/30th part is filled by B in 1 min Hence, required time to fill the whole tank = (165 + 1 +1 ) min = 167 min

Question 10

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?

Accepted Answer

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

Question 11

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?

Accepted Answer

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

Question 12

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?

Accepted Answer

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

Question 13

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?

Accepted Answer

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

Question 14

A tap having diameter 'd' empty a tank in 40 min. How long another tap having diameter '2d' take to empty the same tank?

Accepted Answer

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.

Question 15

Two pipes A and B can fill a tank in 24 and 32 min, respectively. If both the pipes are opened together, after how much time pipe B should be closed so that the tank is full in 9 min?

Accepted Answer

Here, A = 24 min, B = 32, T = 9 min ∴ Required time = B(1 - T/A ) = 32 ( 1 - 9/24 ) = 32 x 15/24 = 20 min.

Question 16

A pipe can fill a cistern in 12 min and another pipe can fill it in 15 min, but a third pipe can empty it in 6 min. The first two pipes are kept open 5 min in the beginning and then the third pipe is also opened.Time taken to empty the cistern is?

Accepted Answer

Let the number of minutes taken to empty the cistern be T min. According to the questions, T/6 - (T + 5)/12 - (T + 5)/15 = 0 ⇒ T/6 - T/12 - 5/12 - T/12 - T/12 - T/15 = 0 ⇒ T/6 - T/12 - T/15 = 5/12 + 5/15 ⇒ (10T - 5T - 4T)/ 60 = (25 + 20)/60 ⇒ T/60 = 45/60 ⇒ T = 45 min.

Question 17

A tank be filled by one tap in 20 min. and by another in 25 min. Both the taps are kept open for 5 min. and then the second is turned off. In how many minutes more is the tank completely filled?

Accepted Answer

Work done by both the taps in 5 min = 5 (1/20 +1/25) = (5 x 9)/100 = 9/20 Remaining part = (1 - 9/20) = 11/20 Now, 1/20 part is filled in 1 min. So, 11/20 part will be filled in 11 min. Hence, the tank will be full in 11 more min.

Question 18

A cistern has two taps which fill it in 12 min. and 15 min. respectively. There is also a waste pipe in the cistern. When all the pipes are opened the empty cistern is full in 20 min. how long will the waste pipe take to empty a full cistern?

Accepted Answer

Work done by waste pipe in 1 min. = (1/12 + 1/15) - 1/20 = (3/20 - 1/20) = 1/10 ∴ Waste pipe can empty the cistern in 10 min.

Question 19

Two taps can separately fill a cistern in 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in?

Accepted Answer

Work done by waste pipe in 1 min. = (1/10) + (1/15) - (1/18) = (1/6) - (1/18) = 1/9 ∴ Waste pipe can empty the cistern in 9 min.

Question 20

A tank is filled by a pipe A in 32 minutes and pipe B in 36 minutes. When full, it in can be emptied by a pipe C in 20 minutes. If all the three pipes are opened simultaneously, half of the tank will be filled in?

Accepted Answer

Net filling in 1 min. = (1/32 + 1/36 - 1/20) = 13/1440 ∴ Time taken to fill the tank = (1440/13) min. Time taken to fill half of the tank = (1440/13) x 2 min. = (720/13) min. = 555/13 min.

Pipes and Cistern Questions