61). Two pipes A and B can fill a tank in 24 and 32 min, respectively. If both the pipes are opened together, after how much time pipe B should be closed so that the tank is full in 9 min?
A). 40 min 
B). 30 min 
C). 10 min 
D). 20 min 
Correct Answer: 20 min
Part filled by A in 1 min = \( \Large \frac{1}{24} \)
Part filled by B in 1 min = \( \Large \frac{1}{32} \)
Let B is closed after x min. Then, [Part filled by (A + B) in x mm] + [Part filled by A in (9x) min] = 1
Therefore,
\( \Large x \left(\frac{1}{24}+\frac{1}{32}\right)+ \left(9x\right) \times \frac{1}{24}=1 \)
=> \( \Large x \left(\frac{4+3}{96}\right)+\frac{ \left(9x\right) }{24}=1 \)
=> \( \Large \frac{7x}{96}+\frac{ \left(9x\right) }{24}=1 \)
=> \( \Large \frac{7x+4 \left(9x\right) }{96}=1 \)
=> \( \Large 7x+4 \left(9x\right)=96 \)
=> 7x+364x=96
=> 7x  4x = 96  36
= 3x = 60 => x = \( \Large \frac{60}{3} \) = 20
Hence, B must be closed after 20 min.

62). Two pipes A and B can fill a cistern in 15 and 20 min, respectively. Both the pipes are opened together, but after 2 min, pipe A is turned off. What is the total time required to fill the tank?
A). \( \Large \frac{46}{3} \) min 
B). \( \Large \frac{52}{3} \) min 
C). \( \Large \frac{43}{3} \) min 
D). \( \Large \frac{41}{3} \) min 
Correct Answer: \( \Large \frac{52}{3} \) min
Part filled by both in 2 min = \( \Large 2x \left(\frac{1}{15}+\frac{1}{20}\right)=2x \left(\frac{4+3}{60}\right) \)
=\( \Large 2 \times \frac{7}{60}= \frac{7}{3} \)
Part unfilled = \( \Large 1  \frac{7}{30}=\frac{307}{30}=\frac{23}{30} \)
Now, B fills \( \Large \frac{1}{20} \) part in 1 min.
Therefore, \( \Large \frac{23}{30} \) part will be filled by B In
= \( \Large \left(20 \times \frac{23}{30}\right)min\ or\ i\ \frac{46}{3}min \)
Therefore, Required time taken to fill the tank
=\( \Large \left(2+\frac{46}{3}\right)=\frac{52}{3}min. \)

63). A tank can be filled by a tap in 20 min and by another tap in 60 min. Both the tape are kept open for 5 min and then the 1st tap is shut off. After this. how much time the tank will be completely filled?
A). 20 min 
B). 30 min 
C). 45 min 
D). 40 min 
Correct Answer: 40 min
Part of the tank filled by both taps in 5 min
=\( \Large 5 \times \left(\frac{1}{20}+\frac{1}{60}\right)
= \frac{5 \times \left(6+2\right) }{120}=\frac{8}{24}=\frac{1}{3} \)
Therefore, Remaining part = \( \Large \left(1\frac{1}{3}\right)=\frac{2}{3} \)
Therefore, \( \Large \frac{1}{60} \) part is now filled in 1 min.
Therefore, \( \Large \frac{2}{3} \) part is now filled in \( \Large 60 \times \frac{2}{3} \) = 40 mm

64). If two pipes function together, the tank will be filled in 12 h. One pipe fills the tank in 10 h faster than the other. How many hours does the faster pipe take to fill up the tank?
A). 20 h 
B). 60 h 
C). 15 h 
D). 25 h 
Correct Answer: 20 h
Let one pipe takes m h to fill the tank.
Then, the other pipe takes \( \Large \left(m10\right)h \)
According to the question,
Therefore, \( \Large \frac{1}{m}+\frac{1}{ \left(m10\right) }=\frac{1}{12} \)
=> \( \Large \frac{m10+m}{m \left(m10\right) }=\frac{1}{12} \)
=> \( \Large 12 \left(m10+m\right)=m \left(m10\right) \)
=> \( \Large m^{2}34m+120 = 0 \)
=> \( \Large m^{2}30m4m+120 = 0 \)
=> \( \Large \left(m30\right) \left(m4\right)=0 \)
m = 30 or 4
Faster pipe will take \( \Large \left(3010\right)h \) = 20 h to fill the tank.

65). A pipe P can fill a tank in 12 min and another pipe R can fill it in 15 min. But, the 3rd pipe M can empty it in 6 min. The 1st two pipes P and R are kept open for double the 2.5 min in the beginning and then the 3rd pipe is also opened. In what time is the tank emptied?
A). 30 min 
B). 25 min 
C). 45 min 
D). 35 min 
Correct Answer: 45 min
According to the question,
Double the 2.5 min = 5 min
Now, part filled in 5 min
= \( \Large 5 \times \left(\frac{1}{12}+\frac{1}{15}\right) = 5 \times \left(\frac{5+4}{60}\right) \)
= \( \Large 5 \times \frac{9}{60} = \frac{3}{4} \)
Part emptied in 1 min when R R and all are opened.
= \( \Large \frac{1}{6} \left(\frac{1}{12}+\frac{1}{15}\right)=\frac{1}{6} \left(\frac{5+4}{60}\right) \)
=\( \Large \left(\frac{1}{6}\frac{3}{2}\right)=\frac{1}{60} \)
Onesixtieth part is emptied in 1 min.
Therefore, Threefourth part will be emptied in\( \Large 60 \times \frac{3}{4}=15 \times 3 \) = 45 min.

66). There are 7 pipes attached with a tank out of which some are inlets and some are outlets. Every inlet can fill the tank in 10 h and every outlet can empty the tank in 15 h. When all the pipes are opened simultaneously, the tank is filled up in \( \Large \Large 2\frac{8}{11} \) h. Find the numbers of inlets and outlets.
A). 5.2 
B). 6.1 
C). 4.3 
D). 3.4 
Correct Answer: 4.3
Let numbr of outlets be x.
Therefore, Number of inlets = \( \Large \left(7x\right) \)
Timer taken to fill the tank when all the pipes are opened = \( \Large \frac{30}{11}h \)
Part of tank filled in 1 h when all the pipes are opened = \( \Large \frac{11}{30}h \)
According to the question,
\( \Large \frac{7x}{10}\frac{x}{15}=\frac{11}{30} \)
=> \( \Large \frac{3 \left(7x\right)2x}{30}=\frac{11}{30} \)
=> 21  3x  2x = 11
=> 5x = 10
Therefore, x = 2
Hence, number of outlets = 2
and number of inlets = 7  2 = 5

67). Capacity of tap B is 80% more than that of A. If both the taps are opened simultaneously, they take 45 h to fill the tank. How long will B take to fill the tank alone?
A). 72 h 
B). 48 h 
C). 66 h 
D). 70 h 
Correct Answer: 70 h
Let time taken by B to fill the tank, a = xh.
Therefore, Time taken by A to fill the tank
b = \( \Large x+\frac{x \times 80}{100}=\frac{9x}{5}h \)
According to the formula,
Time taken by both the taps to fill the tank
\( \Large t = \frac{ab}{a+b} \)
=> \( \Large 45 = \frac{x \times \frac{9x}{5}}{x+\frac{9x}{5}} \)
=> \( \Large 45 \times \frac{14x}{5}=\frac{9x^{2}}{5} \)
Therefore, x = \( \Large \frac{45 \times 14}{9} \)

68). Three taps A, B and C fill a tank in 20 min, 15 min and 12 min, respectively. If all the taps are opened simultaneously, how long will they take to fill 40% of the tank?
A). 1 min 
B). 2 min 
C). 3 min 
D). 4 min 
Correct Answer: 2 min
Part of the tank filled in 1 min by A, B and C.
= \( \Large \frac{1}{20}+\frac{1}{15}+\frac{1}{12} \)
= \( \Large \frac{3+4+5}{60} = \frac{12}{60} = \frac{1}{5} \)
Therefore, Time taken by A, B and C to fill the tank = 5 min.
Therefore, Time taken by A, B and C to fill 40% of the tank
= 40% of 5 = \( \Large \frac{40}{100} \times 5 \)

69). Taps A, B and C are attached with a tank and velocity of water coming through them are 42 L/h, 56 L/h and 48 L/h, respectively. A and B are inlets and C is outlet. If all the taps are opened simultaneously, tank is filled in 16 h. What is the capacity of the tank?
A). 2346 L 
B). 1600 L 
C). 800 L 
D). 960 L 
Correct Answer: 800 L
Quantity of water admitted by tap 1 in 1 h = 42 L
Quantity of water admitted by tap 2 in 1 h = 56 L
Quantity of water removed by tap 3 in 1 h = 48 L
So, quantity of water filled in the tank in 1 h = \( \Large \left(42+5648\right) \) = 50 L
Therefore, Quantity of water filled in 16 h = \( \Large 16 \times 50 = 800\ L \)
Hence, capacity of tank = 800 L

70). Two taps A and B can fill a tank in 25 min and 20 min, respectively. But taps are not opened properly, so the taps A and B allow \( \Large \Large \frac{5}{6} \)th and \( \Large \Large \frac{2}{3} \)rd part of water respectively. How long will they take to fill the tank?
A). 12 min 
B). 13 min 
C). 14 mm 
D). 15 min 
Correct Answer: 15 min
Part of the tank filled with A and B in 1 min
= \( \Large \frac{1}{25} \times \frac{5}{6}+\frac{1}{20} \times \frac{2}{3}=\frac{1}{30}+\frac{1}{30} \)
=\( \Large \frac{2}{30} = \frac{1}{15} \)
Hence, time taken to fill the tank = 15 min.
