51). Two pipes A and B can fill a tank in 1 h and 75 min, respectively. There is also an outlet C. If all the three pipes are opened together. The tank is full in 50 min. How much time will be taken by C to empty the full tank?
A). 100 min |
B). 150 min |
C). 200 min |
D). 125 min |
Correct Answer: 100 min
Work done by C in 1 min
= \( \Large \left(\frac{1}{60}+\frac{1}{75}-\frac{1}{50}\right) \)
= \( \Large \frac{5+4-6}{300}=\frac{3}{300}=\frac{1}{100} \)
Hence, C can empty the full tank in 100 min. |

52). A tank has a leak which would empty it in 8 h. A tap is turned on which admits 3 L a min into the tank and it is now emptied in 12 h. How many litres does the tank hold?
A). 4320 L |
B). 4000 L |
C). 2250 L |
D). 4120 L |
Correct Answer: 4320 L
Work done by the inlet in 1 h
= \( \Large \left(\frac{1}{8}\right)-\frac{1}{12} = \frac{1}{24} \)
Work done by the inlet in 1 min 1 min
= \( \Large \frac{1}{24} \times \frac{1}{60}=\frac{1}{1440} \)
Therefore, Volume of \( \Large \frac{1}{1440} \) part = 3 L
Therefore, Volume of the whole
= \( \Large 3 \times 1440 \) = 4320 L |

53). A, B and C are three pipes connected to a tank. A and B together fill the tank in 6 h, B and C together fill the tank in 10 h and A and C together fill the tank in 12 h. In how much time A, B and C fill up the tank together?
A). 9 h |
B). \( \Large 5\frac{3}{7} \) h |
C). \( \Large 5\frac{2}{7} \) h |
D). \( \Large 5\frac{5}{7} \) h |
Correct Answer: \( \Large 5\frac{5}{7} \) h
Part filled bv (A + B) in 1 h = \( \Large \frac{1}{6} \)
Part filled by (B + C') in 1 h = \( \Large \frac{1}{10} \)
Part filled by (A + C) in 1 h = \( \Large \frac{1}{12} \)
Part filled by 2 (A + B + C) in 1 h
= \( \Large \frac{1}{6}+\frac{1}{10}+\frac{1}{12} = \frac{10+6+5}{60}=\frac{21}{60}=\frac{7}{20} \)
Therefore, Part filled by (A + B + C) in 1 h
= \( \Large \frac{7}{2 \times 20}=\frac{7}{40} \)
Therefore, Required time = \( \Large \frac{40}{7}=\frac{5}{7} \) |

54). Two pipes A and B are opened together to fill a tank. Both pipes fill the tank in a certain time. If A separately takes 16 min more than the time taken by (A + B) and B takes 9 min more than the time taken by (A + B). Find the time taken by A and B to fill the tank when both the pipes are opened together.
A). 10 min |
B). 12 min |
C). 15 min |
D). 8 min |
Correct Answer: 12 min
Here, a = 16 and b = 9
Required time = \( \Large \sqrt{ab} \)
= \( \Large \sqrt{16 \times 9} = 4 \times 3 \) =12 min |

55). Two pipes can fill a tank in 20 and 24 min, respectively and a waste pipe can empty 6 gallon per min. All the three pipes working together can fill the tank in 15 min. Find the capacity of the tank.
A). 210 gallon |
B). 50 gallon |
C). 150 gallon |
D). 240 gallon |
Correct Answer: 240 gallon
Part filled by 1st pipe = \( \Large \frac{1}{20} \)
Part filled by 2nd pipe in 1 min = \( \Large \frac{1}{24} \)
Part filled by all the pipes in 1 min = \( \Large \frac{1}{15} \)
Work done by the waste pipe in 1 min
= \( \Large \frac{1}{15}- \left(\frac{1}{20}+\frac{1}{24}\right)=\frac{1}{15}- \left(\frac{6+5}{120}\right) \)
= \( \Large \frac{1}{15}-\frac{11}{20} \)
= \( \Large \frac{8-11}{120}= \left(-\frac{3}{120}\right)= \left(-\frac{1}{40}\right) \) [-ve sign indicates emptying]
Now, volume of \( \frac{1}{40} \) part = 6 gallon
Therefore, Volume of whole tank = \( \Large 40 \times 6 \) = 240 gallon |

56). Inlet A is four times faster than inlet B to fill a tank. If A alone can fill it in 15 min, how long will it take if both the pipes are opened together?
A). 10 min |
B). 12 min |
C). 15 min |
D). 14 min |
Correct Answer: 12 min
Time taken by A to fill the tank, m = 15 min
Therefore, Time taken by B to fill the tank, n = \( \Large 15 \times 4 \) = 60 min
Therefore, Required time taken \( \Large \frac{m \times n}{m + n} \)
= \( \Large \frac{15 \times 60}{15 + 60}=\frac{15 \times 6}{75} \) = 12 min. |

57). There are two inlets A and B connected to a tank. A and B can fill the tank in 16 h and 10 h, respectively. If both the pipes are opened alternately for 1 h, starting from A, then how much time will the tank take to be filled?
A). \( \Large 13\frac{1}{4} \) |
B). \( \Large 11\frac{6}{8} \) |
C). \( \Large 12\frac{2}{5} \) |
D). \( \Large 12\frac{1}{4} \) |
Correct Answer: \( \Large 12\frac{2}{5} \)
Part filled by A in 1 h = \( \Large \frac{1}{16} \)
Part filled by B in 1 h = \( \Large \frac{1}{10} \)
Part filled by (A+B) in 2 h
= \( \Large \frac{1}{16}+\frac{1}{10}=\frac{13}{80} \)
Part filled by (A+ B) in 12 h = \( \Large \frac{6 \times 13}{80} = \frac{78}{80} \)
Therefore, Remaining part = \( \Large 1 - \frac{78}{80}=\frac{2}{80} = \frac{1}{40} \)
Total time taken by A to fill \( \frac{1}{40} \) part of the tank
= \( \Large \frac{1}{40} \times 16 = \frac{2}{5}h \)
Therefore, Total time taken = \( \Large \left(12+\frac{2}{5}\right) = 12\frac{2}{5}h \) |

58). Two pipes X and Y can fill a cistern in 6 and 7 min, respectively. Starting with pipe X, both the pipes are opened alternately, each for 1 min. In what time will they fill the cistern?
A). \( \Large 6\frac{2}{7} \) |
B). \( \Large 6\frac{3}{7} \) |
C). \( \Large 6\frac{5}{7} \) |
D). \( \Large 6\frac{1}{7} \) |
Correct Answer: \( \Large 6\frac{3}{7} \)
Part filled by X in 1st min and Y in the 2nd min
= \( \Large \left(\frac{1}{6}+\frac{1}{7}\right) = \frac{13}{42} \)
Part filled by (X+Y) working alternately in 6 min
= \( \Large \frac{1}{2} \times \frac{13}{42} \times 6 = \frac{13}{14} \)
Therefore, Remaining part = \( \Large 1 - \frac{13}{14}=\frac{1}{14} \)
Now, it is the turn of x, one-sixth part is filled in 1 min. One-fourteenth part is filled in \( \Large \left(6 \times \frac{1}{14}\right) \) min = \( \Large \frac{3}{7} \)min
Therefore, Required time = \( \Large \left(6 + \frac{3}{7}\right) = 6\frac{3}{7} \) min |

59). A tap having diameter 'd' an empty a tank in 40 min. How long another tap having diameter '2d' take to empty the same tank?
A). 5 min |
B). 29 min |
C). 10 min |
D). 40 min |
Correct Answer: 10 min
Area of tap ... Work done by pipe. When diameter is doubled, area will be four times. So, it will work four times faster.
Hence, required time taken to empty the tank = \( \Large 40 \times \frac{1}{4} \) = 10 min. |

60). Two pipes can fill a cistern in 14 and 16 h, respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom, it took 92 min more to fill the cistern. When the cistern is full, in what time will the leak empty it?
A). \( \Large 43\frac{19}{23} \) h |
B). \( \Large 43\frac{17}{23} \) h |
C). \( \Large 43\frac{13}{23} \) h |
D). \( \Large 43\frac{15}{23} \) h |
Correct Answer: \( \Large 43\frac{19}{23} \) h
Part filled by 2nd pipe in 1 h = \( \Large \frac{1}{14} \)
Part filled by 2nd pipe in 1 h = \( \Large \frac{1}{16} \)
Part filled by the two pipes in 1 h = \( \Large \left(\frac{1}{14}+\frac{1}{16}\right)=\frac{8+7}{112}=\frac{15}{112} \)
Therefore, Time taken by these two pipes cistern
= \( \Large \frac{112}{15} \) = 7 h 28 min
Due to leakage, the time taken = 7 h 28 min + 92 min = 9 h
Therefore, Work done by (two pipes + leak) in 1 h = \( \Large \frac{1}{9} \)
Work done by the leak in 1 h = \( \Large \frac{1}{9}-\frac{15}{112}=\frac{112-135}{1008}
= - \frac{23}{1008} \)
Therefore, Time taken by leak to empty the full cistern
= \( \Large \frac{1008}{23} = 43\frac{19}{23} \) |