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 (9-x) min] = 1
Therefore,
\( \Large x \left(\frac{1}{24}+\frac{1}{32}\right)+ \left(9-x\right) \times \frac{1}{24}=1 \)
=> \( \Large x \left(\frac{4+3}{96}\right)+\frac{ \left(9-x\right) }{24}=1 \)
=> \( \Large \frac{7x}{96}+\frac{ \left(9-x\right) }{24}=1 \)
=> \( \Large \frac{7x+4 \left(9-x\right) }{96}=1 \)
=> \( \Large 7x+4 \left(9-x\right)=96 \)
=> 7x+36-4x=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 = \( \Large 1 - \frac{7}{30}=\frac{30-7}{30}=\frac{23}{30} \)
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 \( \Large \left(m-10\right)h \)
According to the question,
Therefore, \( \Large \frac{1}{m}+\frac{1}{ \left(m-10\right) }=\frac{1}{12} \)
=> \( \Large \frac{m-10+m}{m \left(m-10\right) }=\frac{1}{12} \)
=> \( \Large 12 \left(m-10+m\right)=m \left(m-10\right) \)
=> \( \Large m^{2}-34m+120 = 0 \)
=> \( \Large m^{2}-30m-4m+120 = 0 \)
=> \( \Large \left(m-30\right) \left(m-4\right)=0 \)
m = 30 or 4
Faster pipe will take \( \Large \left(30-10\right)h \) = 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.
= \( \Large \frac{1}{6}- \left(\frac{1}{12}+\frac{1}{15}\right)=\frac{1}{6}- \left(\frac{5+4}{60}\right) \)
=\( \Large \left(\frac{1}{6}-\frac{3}{2}\right)=\frac{1}{60} \)
One-sixtieth part is emptied in 1 min.
Therefore, Three-fourth 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 = \( \Large \left(7-x\right) \)
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,
\( \Large \frac{7-x}{10}-\frac{x}{15}=\frac{11}{30} \)
=> \( \Large \frac{3 \left(7-x\right)-2x}{30}=\frac{11}{30} \)
=> 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 = \( \Large \left(42+56-48\right) \) = 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. |