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?
A). \( \Large 4\frac{1}2{} \) h 
B). 7 h 
C). 6 h 
D). 10 h 
Correct Answer: \( \Large 4\frac{1}2{} \) h Part filled by A in 1 h = \( \Large 3 \times \frac{1}{18} \)
Part filled by B in 1 h = \( \Large \frac{1}{6} \)
Part filled by (A + B) =
\( \Large \frac{1}{18} \)+\( \Large \frac{1}{6} \) = \( \Large \frac{1+3}{18} \) = \( \Large \frac{4}{18} \) = \( \Large \frac{2}{9} \)
Here, both the pipes together will fill the tank in \( \Large \frac{9}{2} \) h or \( \Large 4\frac{1}{2} \)

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?
A). 64 min 
B). 80 min 
C). 96 min 
D). 120 min 
Correct Answer: 120 min Part filled by tap A in 1 min = \( \Large \frac{1}{60} \)
Let tap B fills the tank in x min
Then, part filled by tap, B in 1 min = \( \Large \frac{1}{x} \)
According to the question,
= \( \Large \frac{1}{60} \) + \( \Large \frac{1}{x} \) = \( \Large \frac{1}{40} \)
= \( \Large \frac{1}{x} \) = \( \Large \frac{1}{40} \)  \( \Large \frac{1}{60} \)
=> \( \Large \frac{1}{x} \) = \( \Large \frac{32}{120} \) => \( \Large \frac{1}{x} \) = \( \Large \frac{1}{120} \)
Therefore, Tap B can fill the tank in 120 min.

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?
A). 5 h 
B). 7 h 
C). 8 h 
D). 6 h 
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?
A). 40 h 
B). 30 h 
C). 50 h 
D). 30 h 
Correct Answer: 40 h Let the leak empties the full tank in x h then
Part emptied in 1 h by leak = \( \Large \frac{1}{x} \)
Part filled by inlet in 1 h = \( \Large \frac{1}{20} \)
According to the question.
\( \Large \frac{1}{20} \)+\( \Large \frac{1}{x} \)=\( \Large \frac{1}{40} \)
Therefore, \( \Large \frac{1}{x} \) = \( \Large \frac{1}{40} \)\( \Large \frac{1}{20} \)
= \( \Large \frac{12}{40} \)= \( \Large  \frac{1}{40} \)
[Negative sign indicates emptying]
Clearly, leak will empty the full tank in 40 h

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.
A). 11 h 
B). 15 h 
C). 18 h 
D). 16 h 
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
A). \( \Large 1\frac{14}{23} \) 
B). \( \Large 2\frac{14}{23} \) 
C). 2 h 40 min 
D). 1 h 56 min 
Correct Answer: \( \Large 2\frac{14}{23} \) Part of tank filled by first tap in 1 h = \( \Large \frac{1}{3} \)
Part of tank filled by second tap in 1 h = \( \Large \frac{1}{4} \)
Part of tank emptied by third tap in 1 h = \( \Large \frac{1}{5} \)
Part of the tank filled by all pipes opened simultaneously in 1 h
= \( \Large \frac{1}{3} \) + \( \Large \frac{1}{4} \)  \( \Large \frac{1}{5} \)
= \( \Large \frac{20+1512}{60} \) = \( \Large \frac{23}{60}\)
Time taken by all the taps to fill the tank when it is empty
= \( \Large \frac{60}{23} \)h = \( \Large 2\frac{14}{23} \) h

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?
A). \( \Large 9\frac{3}{13} \) h 
B). \( \Large 9\frac{4}{13} \) h 
C). \( \Large 9\frac{7}{13} \) h 
D). \( \Large 9\frac{9}{13} \) h 
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?
A). \( \Large 3\frac{3}{14} \) h 
B). \( \Large 2\frac{1}{2} \) h 
C). \( \Large 3\frac{9}{14} \) h 
D). \( \Large 2\frac{1}{14} \) h 
Correct Answer: \( \Large 2\frac{1}{2} \) h Part filled by A in 1 h = \( \Large \frac{1}{5} \)
Part filled by B in 1 h = \( \Large \frac{1}{6} \)
Part filled by C in 1 h = \( \Large \frac{1}{30} \)
Net part filled by (A + B + C) in 1 h
= \( \Large \left(\frac{1}{5}+\frac{1}{6}+\frac{1}{30}\right) \)
= \( \Large \frac{6+5+1}{30} = \frac{12}{30} \) = \( \Large \frac{2}{5} \)
Therefore, Required time to fill the tank = \( \Large \frac{5}{2} \)h = \( \Large 2\frac{1}{2} \) h

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?
A). 16 h 
B). 20 h 
C). 32 h 
D). 40 h 
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?
A). 25 h 
B). 40 h 
C). 30 h 
D). 35 h 
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 