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Aptitude Pipes and Cisterns Practice QA

2804.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. The time required by the first pipe is:
6 hours
10 hours
15 hours
30 hours
Explanation:

Suppose, first pipe alone takes $ x $ hours to fill the tank .

Then, second and third pipes will take$\left ( x -5\right)$ and $\left( x - 9\right)$ hours respectively to fill the tank.

$\therefore \dfrac{1}{x} $+$ \dfrac{1}{(x - 5)} $=$ \dfrac{1}{(x - 9)} $

$\Rightarrow \dfrac{x - 5 + x}{x(x - 5)} $=$ \dfrac{1}{(x - 9)} $

$\Rightarrow$ $\left(2 x - 5\right)$$\left( x - 9\right)$ = $ x $$\left( x - 5\right)$

$\Rightarrow x $2 - 18$ x $ + 45 = 0

$\left( x - 15\right)$$\left( x - 3\right)$ = 0

$\Rightarrow x $ = 15.    [neglecting $ x $ = 3]

2808.Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
10
12
14
16
Explanation:

Part filled in 2 hours =$ \dfrac{2}{6} $=$ \dfrac{1}{3} $

Remaining part =$ \left(1 -\dfrac{1}{3} \right) $=$ \dfrac{2}{3} $.

$\therefore$ $\left(A + B\right)$s 7 hours work =$ \dfrac{2}{3} $

$\left(A + B\right)$s 1 hours work =$ \dfrac{2}{21} $

$\therefore$ Cs 1 hours work = { (A + B + C)s 1 hours work } - { (A + B)s 1 hours work }

   =$ \left(\dfrac{1}{6} -\dfrac{2}{21} \right) $=$ \dfrac{1}{14} $

$\therefore$ C alone can fill the tank in 14 hours.

2810.Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
10 min. 20 sec.
11 min. 45 sec.
12 min. 30 sec.
14 min. 40 sec.
Explanation:

Part filled in 4 minutes = 4$ \left(\dfrac{1}{15} +\dfrac{1}{20} \right) $=$ \dfrac{7}{15} $.

Remaining part =$ \left(1 -\dfrac{7}{15} \right) $=$ \dfrac{8}{15} $.

Part filled by B in 1 minute =$ \dfrac{1}{20} $

$\therefore \dfrac{1}{20} $:$ \dfrac{8}{15} $:: 1 : $ x $

$ x $ =$ \left(\dfrac{8}{15} \times 1 \times 20\right) $= 10$ \dfrac{2}{3} $min = 10 min. 40 sec.

$\therefore$ The tank will be full in $\left(4 min. + 10 min. + 40 sec.\right)$ = 14 min. 40 sec.

2812.Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
6 hours
6$ \dfrac{2}{3} $hours
7 hours
7$ \dfrac{1}{2} $hours
Explanation:

$\left(A + B\right)$s 1 hours work =$ \left(\dfrac{1}{12} +\dfrac{1}{15} \right) $=$ \dfrac{9}{60} $=$ \dfrac{3}{20} $.

$\left(A + C\right)$s hours work =$ \left(\dfrac{1}{12} +\dfrac{1}{20} \right) $=$ \dfrac{8}{60} $=$ \dfrac{2}{15} $.

Part filled in 2 hrs =$ \left(\dfrac{3}{20} +\dfrac{2}{15} \right) $=$ \dfrac{17}{60} $.

Part filled in 6 hrs =$ \left(3 \times\dfrac{17}{60} \right) $=$ \dfrac{17}{20} $.

Remaining part =$ \left(1 -\dfrac{17}{20} \right) $=$ \dfrac{3}{20} $.

Now, it is the turn of A and B and $ \dfrac{3}{20} $ part is filled by A and B in 1 hour.

$\therefore$ Total time taken to fill the tank = $\left(6 + 1\right)$ hrs = 7 hrs.

2815.Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
$ \dfrac{5}{11} $
$ \dfrac{6}{11} $
$ \dfrac{7}{11} $
$ \dfrac{8}{11} $
Explanation:

Part filled by $\left(A + B + C\right)$ in 3 minutes = 3$\left(\dfrac{1}{30} +\dfrac{1}{20} +\dfrac{1}{10} \right) $=$ \left(3 \times\dfrac{11}{60} \right) $=$ \dfrac{11}{20} $.

Part filled by C in 3 minutes =$ \dfrac{3}{10} $.

$\therefore$ Required ratio =$ \left(\dfrac{3}{10} \times\dfrac{20}{11} \right) $=$ \dfrac{6}{11} $.

2819.Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 40 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?
28 hr
16 hr
22 hr
32 hr
Explanation:

Part filled by pipe A in 1 hour $=\dfrac{1}{5}$

Part filled by pipe B in 1 hour =$\dfrac{1}{20}$

Part filled by pipe A and B in 1 hour $=\dfrac{1}{5}+\dfrac{1}{20}$=$\dfrac{1}{4}$

i.e., A and B together can fill the tank in 4 hours.

Due to leakage, it took 40 minutes more to fill the tank.

i.e., due to the leakage, the tank got filled in

4$\dfrac{40}{60}$ hour =4$\dfrac{2}{3}$ hour =$\dfrac{14}{3}$ hour.

Net part filled by pipe A, pipe B and the leak in 1 hour=$\dfrac{3}{14}$

Therefore, part emptied by the leak in 1 hour=$\dfrac{1}{4}-\dfrac{3}{14}=\dfrac{1}{28}$

i.e., the leak can empty the tank in 28 hours.

2820.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?
3 hrs 15 min
3 hrs 45 min
4 hrs
4 hrs 15 min
Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =$ \left(4 \times\dfrac{1}{6} \right) $=$ \dfrac{2}{3} $.

Remaining part =$ \left(1 -\dfrac{1}{2} \right) $=$ \dfrac{1}{2} $.

$\therefore \dfrac{2}{3} $:$ \dfrac{1}{2} $:: 1 : $ x $

$\Rightarrow x $ =$ \left(\dfrac{1}{2} \times 1 \times\dfrac{3}{2} \right) $=$ \dfrac{3}{4} $hours i.e., 45 mins.

So, total time taken = 3 hrs. 45 mins.

2822.A water tank is two-fifth full. Pipe A can fill a tank in 12 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?
2.8 min
4.2 min
4.8 min
5.6 min
Explanation:

Since pipe B is faster than pipe A, the tank will be emptied.

Part filled by pipe A in 1 minute $=\dfrac{1}{12}$

Part emptied by pipe B in 1 minute $=\dfrac{1}{6}$

Net part emptied by pipe A and B in 1 minute

$=\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}$

Time taken to empty $\dfrac{2}{5}$ of the tank

$=\dfrac{\left(\dfrac{2}{5}\right)}{\left(\dfrac{1}{12}\right)}$=$\dfrac{2}{5}$×12=4.8 min

2823.Two pipes A and B can fill a cistern in 37$ \dfrac{1}{2} $ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
5 min.
9 min.
10 min.
15 min.
Explanation:

Let B be turned off after $ x $ minutes. Then,

Part filled by $\left(A + B\right)$ in $ x $ min. + Part filled by A in $\left(30 - x \right)$ min. = 1.

$\therefore x \left(\dfrac{2}{75} +\dfrac{1}{45} \right) $+ $\left(30 - x \right)$.$ \dfrac{2}{75} $= 1

$\Rightarrow \dfrac{11x}{225} $+$ \dfrac{(60 -2x)}{75} $= 1

$\Rightarrow$ 11$ x $ + 180 - 6$ x $ = 225.

$\Rightarrow x $ = 9.

2824.Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
1 hour
2 hours
6 hours
8 hours
Explanation:

Let the cistern be filled by pipe A alone in $ x $ hours.

Then, pipe B will fill it in $\left( x + 6\right)$ hours.

$\therefore \dfrac{1}{x} $+$ \dfrac{1}{(x + 6)} $=$ \dfrac{1}{4} $

$\Rightarrow \dfrac{x + 6 + x}{x(x + 6)} $=$ \dfrac{1}{4} $

$\Rightarrow x $2 - 2$ x $ - 24 = 0

$\Rightarrow$ $\left( x -6\right)$$\left( x + 4\right)$ = 0

$\Rightarrow x $ = 6.     [neglecting the negative value of $ x $]

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