11). Three taps A, B and C together can fill an empty cistern in 10 min. The tap A alone can fill it in 30 min and the tap B alone can fill it in 40 min. How long will the tap C alone take to fill it?
A). 16 min |
B). 24 min |
C). 32 min |
D). 40 min |
Correct Answer: 24 min
Part filled by (A+B+C) in 1 min = \( \Large \frac{1}{10} \)
Part filled by A in 1 min = \( \Large \frac{1}{30} \)
Part filled by B in 1 min = \( \Large \frac{1}{40} \)
Part filled by (A + B) in 1 min = \( \Large \frac{1}{30} \) + \( \Large \frac{1}{40} \)
= \( \Large \frac{4+3}{120} \) = \( \Large \frac{7}{120} \)
Therefore, Part filled by C in 1 min
= \( \Large \frac{1}{10} \) - \( \Large \frac{7}{120} \)=\( \Large \frac{12-7}{120} \)=\( \Large \frac{5}{120} \)=\( \Large \frac{1}{24} \)
Therefore, Tap C will fill the cistern in 24 min. |

12). Two pipes P and Q can fill a cistern in 12 and 15 min, respectively. If both are opened together and at the end of 3 min, the first is closed. How much longer will the cistern take to fill?
A). \( \Large 8\frac{1}{4} \) |
B). \( \Large 8\frac{3}{4} \) |
C). 5 min |
D). \( \Large 8\frac{1}{2} \) |
Correct Answer: \( \Large 8\frac{1}{4} \)
Part filled by pipe P in 1 min = \( \Large \frac{1}{12} \)
Part filled by pipe Q in 1 min = \( \Large \frac{1}{15} \)
Part filled by both pipes in 1 min
= \( \Large \frac{1}{12} \)+ \( \Large \frac{1}{15} \)= \( \Large \frac{5+4}{60} \)= \( \Large \frac{9}{60} \)
Now, part filled by both pipes in 3 min
= \( \Large \frac{3 \times 9}{60}=\frac{27}{60}=\frac{9}{20} \)
Remaining part = \( \Large 1 - \frac{9}{20} = \frac{11}{20} \)
Let the remaining part is filled by pipe Q in x min.
Then, \( \Large x \times \frac{1}{15} = \frac{11}{20} \)
\( \Large x = \frac{15 \times 11}{20} = \frac{3 \times 11}{4}
= \frac{33}{4} = 8\frac{1}{4} min \) |

13). There are three pipes connected with a tank. The first pipe can fill \( \Large \frac{1}{2} \) part of the tank in 1 h, second pipe can fill \( \Large \frac{1}{3} \) part of the tank in 1 h. Third pipe is connected to empty the tank. After opening all the three pipes, \( \Large 7\frac{1}{2} \) part of the tank can be filled in 1 h, then how long will third pipe take to empty the full tank?
A). 3 h |
B). 4 h |
C). 5 h |
D). 6 h |
Correct Answer: 4 h
1st pipe takes 1 h to fill \( \Large \frac{1}{2} \) part of the tank.
So, time taken to fill the whole tank (m) = 2 h
2nd pipe takes 1 h to \( \Large \frac{1}{3} \) part of the tank
So. time taken to fill the whole tank (n) = 3 h
Let 3rd pipe takes P h to empty the tank = x
Therefore, \( \Large \frac{1}{m} \)+ \( \Large \frac{1}{n} \)- \( \Large \frac{1}{x} \)= \( \Large \frac{7}{12} \)
= \( \Large \frac{1}{2} \)+ \( \Large \frac{1}{3} \)- \( \Large \frac{1}{x} \) = \( \Large \frac{7}{12} \)
= \( \Large \frac{1}{x}=\frac{6+4-7}{12}=\frac{3}{12}=\frac{1}{4} \)
Therefore, x = 4 h |

14). A pipe can fill a cistern in 12 min and another pipe can fill it in 15 min, but a third pipe can empty it in 6 min. The first two pipes are kept open for 5 min in the beginning and then the third pipe is also opened. Time taken to empty the cistern is
A). 38 min |
B). 22 min |
C). 42 min |
D). 45 min |
Correct Answer: 45 min
Let the number of minutes taken to empty the cistern be x min.
According to the question,
\( \Large \frac{x}{6}-\frac{x+5}{12}-\frac{x+5}{12}=0 \)
\( \Large \frac{x}{6}-\frac{x}{12}-\frac{5}{12}-\frac{x}{15}-\frac{5}{15}=0 \)
\( \Large \frac{x}{6}-\frac{x}{12}-\frac{x}{15}=\frac{5}{12}+\frac{5}{15} \)
\( \Large \frac{10x-5x-4x}{60} = \frac{25+20}{60} \)
\( \Large \frac{x}{60} = \frac{45}{60} = x = 45 min. \) |

15). A cistern has three pipes A, B and C. Pipes A and B can fill it in 3 and 4 h, respectively, while pipe C can empty the completely filled cistern in 1 h. If the pipes are opened in order at 3:00 pm, 4:00 pm and 5:00 pm, respectively, at what time will the cistern be empty?
A). 6:15 pm |
B). 7:12 pm |
C). 8:12 pm |
D). 8:35 pm |
Correct Answer: 7:12 pm
Let the cistern gets emptied in m 3:00 pm. Work done by A in m h, by (m-1)hand byC in(m-2) h=0 =9 2 + 2.1 _ (m z) = o 3 4 => 4m+3(m-1)-12(m-2)=O -~ 5m = 21
m=4h12min Required time = 7 ; 12 pm. |

16). Three pipes A, B and C can fill a tank in 30 min, 20 min and 10 min, respectively. When the tank is empty, all the three pipes are opened. If A, B and C discharge chemical solutions P, Q and R respectively, then the part of solution R in the liquid in the tank after 3 min is
A). \( \Large \frac{8}{11} \) |
B). \( \Large \frac{5}{11} \) |
C). \( \Large \frac{6}{11} \) |
D). \( \Large \frac{7}{11} \) |
Correct Answer: \( \Large \frac{6}{11} \)
Total quantity of solutions P, Q and R
from A, B and C respectively, after 3 min
= \( \Large \frac{3}{30}+\frac{3}{20}+\frac{3}{10}=3 \left(\frac{2+3+6}{60}\right) \)
= \( \Large \frac{3 \times 11}{60} = \frac{11}{20} \)
Quantity of solution R in liquid in 3 min = \( \Large \frac{3}{10} \)
Therefore, Part of solution R
= \( \Large \frac{10}{\frac{11}{20}} = \frac{3 \times 20}{10 \times 11} = \frac{6}{11} \) |

17). A tap can fill a tank in 30 minutes and another tap can fill the same tank in 60 minutes. lf both the taps are opened simultaneously when the tank is half empty, in how many minutes will the tank be full?
A). 10 minutes |
B). 18 minutes |
C). 20 minutes |
D). 15 minutes |
Correct Answer: 10 minutes
Tap A can fill the tank in 30 minutes. Therefore, it fills \( \Large \frac{1}{30} \)th of the tank every minute.
Tap B can fill the tank in 60 minutes. Therefore, it fills \( \Large \frac{1}{60} \)th of the tank every minute.
Together, the two taps will fill \( \Large \frac{1}{30}+\frac{1}{60}=\frac{2+1}{60}=\frac{3}{60}=\frac{1}{20} \)th of the tank every minute.
Therefore, when both the taps are opened simultaneously, they will fill the tank in 20 minutes. As the tank is already half full, they need to fill only half the tank.
Therefore, the tank will overflow 10 minutes after both the taps are opened. |

18). Pipe A can fill a tank in 40 minutes. Pipe B can fill the same tank in 20 minutes. A crack at the bottom of the tank can empty the tank in 120 minutes. If both the pipes are opened simultaneously when the tank is empty and water gets drained through the crack at the bottom, when will the tank be full?
A). 18 minutes |
B). 15 minutes |
C). 12 minutes |
D). 21 minutes |
Correct Answer: 15 minutes
Pipe A fills \( \Large \frac{1}{40} \)th of the tank every minute. Pipe B fills \( \Large \frac{1}{20} \)th every minute. The crack at the bottom drains \( \Large \frac{1}{120} \)th of the tank every minute.
Therefore, when both the pipes are opened and the crack is left unchecked,
\( \Large \frac{1}{40}+\frac{1}{20}=\frac{1}{120} \)th of the tank gets filled.
That is, \( \Large \frac{3+6-1}{120}=\frac{8}{120}=\frac{1}{15} \)th of the tank gets filled every minute.
Hence, the tank will be full in 15 minutes. |

19). A pipe can fill \( \Large \frac{2}{3} \)rds of a tank in 16 minutes. How long will it take to fill 3/4ths of the tank?
A). 12 minutes |
B). 15 minutes |
C). 18 minutes |
D). 24 minutes |
Correct Answer: 18 minutes
lf the pipe can fill \( \Large \frac{2}{3} \)rds of the in 16 minutes, it will take \( \Large \frac{1}{\frac{2}{3}} \times 16=\frac{3}{2} \times 16 = 24 minutes \) to fill the entire tank.
If it takes 24 minutes to fill the tank, it will take \( \Large \frac{3}{4} \times 24=18 \) minutes to fill 3/4ths of the tank. |

20). Pipe A can fill a tank at the rate of 35 gallons in 10 minutes. Pipe A will fill a tank that is \( \Large \frac{x}{3} \)rds empty in 12 minutes. What is the capacity of the tank?
A). 45 gallons |
B). 52 gallons |
C). 60 gallons |
D). 63 gallons |
Correct Answer: 63 gallons
Pipe A fills at the rate of 35 gallons per 10 minutes = 3.5 gallons per minute.
Pipe A is kept open for 12 minutes. So, the water discharged in as min. = \( \Large 12 \times 3.5 = 42 gallons \)
When pipe A was opened the tank was 2/3rd empty.
Therefore, 42 gallons = 2/3rd the total capacity of the tank.
Hence, the capacity of the tank = \( \Large \frac{3}{2} \times 42 = 63 gallons \) |