Pipes and Cisterns level - 1
Q1–10 of 50A pipe can fill a tank in $x$ hours and another pipe can empty it in $y$ ($y > x$) hours. If both the pipes are open, in how many hours will the tank be filled?
∴ The tank will be filled in $\left(\frac{xy}{y-x}\right)$ hours.
Answer: (d)
A tap can completely fill a water tank in 8 hours. The water tank has a hole in it through which the water leaks out. The leakage will cause the full water tank to get empty in 12 hours. How much time will it take for the tap to fill the tank completely with the hole?
Net part filled in 1 hour $= \left(\frac{1}{8} - \frac{1}{12}\right) = \frac{1}{24}$.
∴ The tank will be filled in 24 hours.
Answer: (c)
A tap can fill a tank in 48 minutes whereas another tap can empty it in 2 hours. If both the taps are opened at 11:40 A.M, then the tank will be filled at
Net part filled in 1 hour $= \left(\frac{1}{48} - \frac{1}{120}\right) = \frac{3}{240} = \frac{1}{80}$.
∴ The tank will be filled in 80 mins i.e. 1 hour 20 min after 11:40 A.M., i.e. at 1 P.M.
Answer: (b)
A tank with capacity $T$ litres is empty. If water flows into the tank from pipe X at the rate of $x$ litres per minute and water is pumped out by Y at the rate of $y$ litres per minute and $x > y$, then in how many minutes will the tank be filled?
Net volume filled in 1 minute $= (x - y)$ litres.
∴ Time taken to fill the tank $= \frac{T}{x-y}$ minutes.
Answer: (c)
Pipes A and B can fill a tank in 20 hours and 30 hours respectively and pipe C can empty the full tank in 40 hours. If all the pipes are opened together, how much time will be needed to make the tank full?
Net part filled in 1 hour $= \left(\frac{1}{20} + \frac{1}{30} - \frac{1}{40}\right) = \frac{7}{120}$.
∴ The tank will be full in $\frac{120}{7}$ i.e. $17\frac{1}{7}$ hours.
Answer: (c)
A pipe can fill a tank in 3 hours. There are two outlet pipes from the tank which can empty it in 7 and 10 hours respectively. If all the three pipes are opened simultaneously, then the tank will be filled in
Net part filled in 1 hour $= \frac{1}{3} - \left(\frac{1}{7} + \frac{1}{10}\right) = \frac{1}{3} - \frac{17}{70} = \frac{19}{210}$.
∴ The tank will be filled in $\frac{210}{19}$ hrs i.e. $11\frac{1}{19}$ hrs $\approx 11$ hrs.
Answer: (d)
In what time would a cistern be filled by three pipes whose diameters are 1 cm, $1\frac{1}{3}$ cm and 2 cm running together, when the largest alone will fill it in 61 minutes, the amount of water flowing in by each pipe, being proportional to the square of its diameter?
Let $t_1, t_2, t_3$ be the times taken by pipes with diameters 1 cm, $\frac{4}{3}$ cm and 2 cm respectively.
Since time taken to fill the tank is inversely proportional to the amount of water flowing through it, $t \propto \frac{1}{r^2}$, i.e. $t_3 = \frac{k}{r^2} \Rightarrow 61 = \frac{k}{4} \Rightarrow k = 244$.
Thus, $t_1 = \frac{k}{(1)^2} = 244$; $t_2 = \frac{k}{(4/3)^2} = \frac{9}{16} \times 244 = \frac{549}{4}$.
Net part filled in 1 min $= \frac{1}{244} + \frac{1}{61} + \frac{4}{549} = \frac{1}{61}\left(\frac{1}{4} + 1 + \frac{4}{9}\right) = \frac{1}{61} \times \frac{61}{36} = \frac{1}{36}$.
Hence, all three pipes together would fill the tank in 36 minutes.
Answer: (c)
A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
Time taken by one tap to fill half the tank = 3 hrs.
Part filled by the four taps in one hour $= \left(4 \times \frac{1}{6}\right) = \frac{2}{3}$.
Remaining part $= \frac{1}{2}$.
$\frac{2}{3} : \frac{1}{2} :: 1 : x$, or $x = \left(\frac{1}{2} \times 1 \times \frac{3}{2}\right) = \frac{3}{4}$ hrs, i.e. 45 min.
So, total time taken = 3 hrs 45 min.
Answer: (b)
A cistern has two pipes. One can fill it with water in 8 hours and other can empty it in 5 hours. In how many hours will the cistern be emptied if both the pipes are opened together when $\frac{3}{4}$ of the cistern is already full of water?
Net part emptied in one hour $= \left(\frac{1}{5} - \frac{1}{8}\right) = \frac{3}{40}$.
$\frac{3}{40} : \frac{3}{4} :: 1 : x$, or $x = \left(\frac{3}{4} \times 1 \times \frac{40}{3}\right) = 10$ hrs.
So, the cistern will be emptied in 10 hrs.
Answer: (c)
A vessel has three pipes connected to it, two to supply liquid and one to draw liquid. The first alone can fill the vessel in $4\frac{1}{2}$ hours, the second in 3 hours and the third can empty it in $1\frac{1}{2}$ hours. If all the pipes are opened simultaneously when the vessel is half full, how soon will it be emptied?
Net part emptied in 1 hour $= \frac{2}{3} - \left(\frac{2}{9} + \frac{1}{3}\right) = \left(\frac{2}{3} - \frac{5}{9}\right) = \frac{1}{9}$.
$\frac{1}{9} : \frac{1}{2} :: 1 : x$, or $x = \left(\frac{1}{2} \times 9\right) = 4\frac{1}{2}$ hours.
So, the tank will be emptied in $4\frac{1}{2}$ hours.
Answer: (a)