Pipes and Cisterns Problems and Solutions

• 8. Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in.

  1. \begin{aligned} 3\frac{9}{5} \end{aligned}
  2. \begin{aligned} 3\frac{9}{17} \end{aligned}
  3. \begin{aligned} 3\frac{7}{5} \end{aligned}
  4. \begin{aligned} 3\frac{7}{17} \end{aligned}

  Answer :

  Option B

  Explanation:

  Net part filled in 1 hour =
  \begin{aligned}
  \left(\frac{1}{5}+\frac{1}{6}-\frac{1}{12}\right) \\
  = \frac{17}{60} hrs \\
  = 3\frac{9}{17}
  \end{aligned}

• 9. A tank can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tank from empty state if B is used for the first half time and then A and B fill it together for the other half.

  1. 15 mins
  2. 20 mins
  3. 25 mins
  4. 30 mins

  Answer :

  Option D

  Explanation:

  Let the total time be x mins.
  Part filled in first half means in x/2 = 1/40
  
  Part filled in second half means in x/2 = \begin{aligned}
  \frac{1}{60}+\frac{1}{40} \\
  = \frac{1}{24} \\
  \text{ Total = } \\
  \frac{x}{2}*\frac{1}{40} + \frac{x}{2}*\frac{1}{24} = 1 \\
  
  => \frac{x}{2} \left(\frac{1}{40}+\frac{1}{24} \right) = 1 \\
  => \frac{x}{2}*\frac{1}{15} = 1 \\
  => x = 30 mins
  \end{aligned}

• 10. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank ?

  1. 10 hours
  2. 15 hours
  3. 17 hours
  4. 18 hours

  Answer :

  Option B

  Explanation:

  Suppose, first pipe alone takes x hours to fill the tank .
  
  Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.
  
  As per question, we get
  \begin{aligned}
  \frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9} \\
  => \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\\
  => (2x - 5)(x - 9) = x(x - 5)\\
  => x^2 - 18x + 45 = 0
  \end{aligned}
  
  After solving this euation, we get
  (x-15)(x+3) = 0,
  As value can not be negative, so x = 15
  

• 11. Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long it will take to fill the tank ?

  1. 10 mins
  2. 12 mins
  3. 15 mins
  4. 20 mins

  Answer :

  Option B

  Explanation:

  In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result, as
  
  Part filled by A in 1 min = 1/20
  Part filled by B in 1 min = 1/30
  
  Part filled by (A+B) in 1 min = 1/20 + 1/30
  = 1/12
  
  So both pipes can fill the tank in 12 mins.
  

• 12. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres) is

  1. 5780 litres
  2. 5770 litres
  3. 5760 litres
  4. 5750 litres

  Answer :

  Option C

  Explanation:

  Work done by the inlet in 1 hour =
  \begin{aligned}
  \frac{1}{6}-\frac{1}{8} = \frac{1}{24} \\
  \text{Work done by inlet in 1 min} \\
  = \frac{1}{24}*\frac{1}{60}\\
  = \frac{1}{1440} \\
  => \text{Volume of 1/1440 part = 4 liters} \\
  
  \end{aligned}
  
  Volume of whole = (1440 * 4) litres = 5760 litres.
  
  

• 13. A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time ?

  1. 10 mins
  2. 15 mins
  3. 20 mins
  4. 25 mins

  Answer :

  Option C

  Explanation:

  How we can solve this question ?
  First we will calculate the work done for 10 mins, then we will get the remaining work, then we will find answer with one tap work, As
  
  Part filled by Tap A in 1 min = 1/20
  Part filled by Tap B in 1 min = 1/60
  
  (A+B)'s 10 mins work =
  \begin{aligned}
  10*\left(\frac{1}{20}+\frac{1}{60}\right) \\
  = 10*\frac{4}{60} = \frac{2}{3} \\
  \text{Remaining work = } 1-\frac{2}{3} \\
  = \frac{1}{3} \\
  
  \text{METHOD 1} \\
  => \frac{1}{60}:\frac{1}{3}=1:X \\
  => X = 20 \\
  
  \text{METHOD 2} \\
  
  1/60 \text{ part filled by B in} = 1 min \\
  1/3 \text{ part will be filled in} \\
  = \frac{\frac{1}{3}}{\frac{1}{60}} \\
  = \frac{60}{3} = 20
  \end{aligned}

• 14. Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?

  1. 3 hours
  2. 5 hours
  3. 7 hours
  4. 10 hours

  Answer :

  Option B

  Explanation:

  (A+B)'s 2 hour's work when opened =
  \begin{aligned}
  \frac{1}{6}+\frac{1}{4} = \frac{5}{12} \\
  
  (A+B)'s \text{ 4 hour's work} = \frac{5}{12}*2 \\
  = \frac{5}{6}
  
  \text{Remaining work = } 1-\frac{5}{6} \\
  = \frac{1}{6} \\
  
  \text{Now, its A turn in 5 th hour} \\
  \frac{1}{6} \text{ work will be done by A in 1 hour}\\
  \text{Total time = }4+1 = 5 hours
  \end{aligned}