>> Aptitude >> Pipes and Cisterns

PrimeFaces.cw("DataGrid","widget_frm_base_j_idt51",{id:"frm_base:j_idt51"});

medianet_width = "300";
medianet_height = "250";
medianet_crid = "565530822";
medianet_versionId = "3111299";

Contents:

- Aptitude
- Approximation
- Average
- Boat and Stream
- Compound interest
- Discount
- Linear Equations
- Mensuration
- Mixture and Allegation
- Number series
- Number System
- Partnership
- Percentage
- Permutation and combination
- Pipes and Cisterns
- Probability
- Problem on ages
- Profit and Loss
- Ratio and Proportions
- Simple and compound interest
- Time and Distance
- Time and work
- Trains
- Unitary Method
- Word problems
- Work and Wages

1). Two pipes A and B can fill a tank in 18 and 6 h, respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
Part filled by A in 1 h = \( \Large 3 \times \frac{1}{18} \) | ||||

2). There are two tanks A and B to fill up a water tank. The tank can be filled in 40 min, if both taps are on. The same tank can be filled in 60 min, if tap A alone is on. How much time will tap alone take, to fill up the same tank?
Part filled by tap A in 1 min = \( \Large \frac{1}{60} \) | ||||

3). A cistern can be filled up in 4 h by an inlet A. An outlet B can empty the cistern in 8 h. If both A and B are opened simultaneously, then after how much time the cistern get filled?
Part filled by A in 1 h = \( \Large \frac{1}{4} \) Part emptied by B in 1 h = \( \Large \frac{1}{8} \) Part filled by(A + B) in 1 h = \( \Large \frac{1}{4}+ \left(-\frac{1}{8}\right)=\frac{1}{4}-\frac{1}{8}=\frac{2-1}{8}=\frac{1}{8} \) Therefore, Required time to fill the cistern = 8 h Note \( \Large \frac{-1}{8} \) has been taken because it empties the tank. | ||||

4). A pipe can fill a tank in 20 h. Due to a leak in the bottom, it is filled in 40 h. If the tank is full, how much time will the leak take to empty it?
Let the leak empties the full tank in x h then | ||||

5). A pipe can fill a tank in 10 h, while an another pipe can empty it in 6 h. Find the time taken to empty the tank, when both the pipes are opened up simultaneously.
Part filled by 1st pipe in 1 h = \( \Large \frac{1}{10} \) Part emptied by 2nd pipe in 1 h = \( \Large \frac{1}{6} \) Part emptied when both the pipes are opened up = \( \Large \frac{1}{6} \) - \( \Large \frac{1}{10} \) = \( \Large \frac{5-3}{30} \)=\( \Large \frac{2}{30} \)=\( \Large \frac{1}{15} \) Hence, time taken to empty the full tank in 15 h. | ||||

6). Three taps are fitted in a cistern. The empty cistern is filled by the first and the second taps in 3 and 4 h, respectively. The full cistern is emptied by the third tap in 5 h. If all three taps are opened simultaneously, the empty cistern will be filled up in
Part of tank filled by first tap in 1 h = \( \Large \frac{1}{3} \) | ||||

7). Pipe A can fill a tank in 30 min, while pipe B can fill the same tank in 10 min and pipe C can empty the full tank in 40 min. If all the pipes are opened together, how much time will be needed to make the tank full?
Part filled by A in 1 min = \( \Large \frac{1}{30} \) Part filled by B in 1 min = \( \Large \frac{1}{10} \) Part emptied by C in 1 min = \( \Large \frac{1}{40} \) Net part filled in 1 h by (A + B + C) = \( \Large \left(\frac{1}{30}+\frac{1}{10}-\frac{1}{40}\right) \) = \( \Large \frac{4+12-3}{120} = \frac{13}{120} \) Therefore, Required time to fill the tank = \( \Large \frac{120}{13} \)h = \( \Large 9\frac{3}{13} \) h | ||||

8). Pipes A and B can fill a tank in 5 and 6 h, respectively. Pipe C can fill it in 30 h. If all the three pipes are opened together, then in how much time the tank will be filled up?
Part filled by A in 1 h = \( \Large \frac{1}{5} \) | ||||

9). Through an inlet, a tank takes 8 h to get filled up. Due to a leak in the bottom, it takes 2 h more to get it filled completely. If the tank is full, how much time will the leak take to empty it?
Let the leak takes x h to empty the tank. Now. part filled by inlet in 1 h = \( \Large \frac{1}{8} \) Part filled in 1 h when both tap and leak works together = \( \Large \frac{1}{8+2} \) = \( \Large \frac{1}{10} \) According to the question, = \( \Large \frac{1}{x} \) = \( \Large \frac{1}{8} \)-\( \Large \frac{1}{10} \)=\( \Large \frac{5-4}{40} \) = \( \Large \frac{1}{40} \) Therefore, x = 40 h | ||||

10). A tap can fill an empty tank in 12 h and a leakage can empty the tank in 20 h. If tap and leakage both work together, then how long will it take to fill the tank?
Part filled by tap in 1 h =\( \Large \frac{1}{12} \) Part emptied by leak in 1 h = \( \Large \frac{1}{20} \) Net part filled in 1 h when both (tap and leakage) work = \( \Large \frac{1}{12} \)-\( \Large \frac{1}{20} \)=\( \Large \frac{5-3}{60} \)=\( \Large \frac{2}{60} \)=\( \Large \frac{1}{30} \) Therefore, Required time to fill the tank = 30 h |