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?
View Answer Correct Answer: \( \Large 4\frac{1}2{} \) h 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?
View Answer Correct Answer: 120 min 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?
View Answer Correct Answer: 8 h 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 = \( \Large \frac{1}{4}+ \left(\frac{1}{8}\right)=\frac{1}{4}\frac{1}{8}=\frac{21}{8}=\frac{1}{8} \) 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?
View Answer Correct Answer: 40 h 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.
View Answer Correct Answer: 15 h 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 = \( \Large \frac{1}{6} \)  \( \Large \frac{1}{10} \) = \( \Large \frac{53}{30} \)=\( \Large \frac{2}{30} \)=\( \Large \frac{1}{15} \) 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
View Answer Correct Answer: \( \Large 2\frac{14}{23} \) 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?
View Answer Correct Answer: \( \Large 9\frac{3}{13} \) h 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) = \( \Large \left(\frac{1}{30}+\frac{1}{10}\frac{1}{40}\right) \) = \( \Large \frac{4+123}{120} = \frac{13}{120} \) 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?
View Answer Correct Answer: \( \Large 2\frac{1}{2} \) h 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?
View Answer Correct Answer: 40 h 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, = \( \Large \frac{1}{x} \) = \( \Large \frac{1}{8} \)\( \Large \frac{1}{10} \)=\( \Large \frac{54}{40} \) = \( \Large \frac{1}{40} \) 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?
View Answer Correct Answer: 30 h 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 = \( \Large \frac{1}{12} \)\( \Large \frac{1}{20} \)=\( \Large \frac{53}{60} \)=\( \Large \frac{2}{60} \)=\( \Large \frac{1}{30} \) Therefore, Required time to fill the tank = 30 h

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