Question 1: \displaystyle  A can do a piece of work in \displaystyle  15 days while \displaystyle  B can do it in \displaystyle  10 days. How long will they take together to do it?

Answer:

\displaystyle \text{ A's 1 Day }   = \frac{1}{15}

\displaystyle \text{ B's 1 Day }   = \frac{1}{10}

\displaystyle \text{ A's + B's 1 Day Work }   ( \frac{1}{15} + \frac{1}{10} )= \frac{5}{30} = \frac{1}{6}

Therefore both can finish the work in 6 Days.

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Question 2: \displaystyle  A, B and \displaystyle  C can do a piece of work in \displaystyle  12 days, \displaystyle  15 days and \displaystyle  10 days respectively. In what time will they all together finish it?

Answer:

\displaystyle \text{ A's 1 Day }   = \frac{1}{12}

\displaystyle \text{ B's 1 Day }   = \frac{1}{15}

\displaystyle \text{ C's 1 Day }   \frac{1}{10}

\displaystyle \text{ A's + B's  + C's 1 Day Work }   = ( \frac{1}{12} + \frac{1}{15} + \frac{1}{10} ) = \frac{15}{60} = \frac{1}{4}

Therefore all three can finish the work in 4 Days.

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Question 3: \displaystyle  A and \displaystyle  B together can do a piece of work in \displaystyle  35 days, while \displaystyle  A alone can do it in \displaystyle  60 days. How long would \displaystyle  B alone take to do it?

Answer:

\displaystyle \text{ A's 1 Day }   = \frac{1}{60}

\displaystyle \text{ B's 1 Day }   = \frac{1}{x}

\displaystyle \text{ A's + B's 1 Day Work }   = ( \frac{1}{60} + \frac{1}{x} ) = \frac{1}{35}

Solving for \displaystyle  x = 84 Days

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Question 4: \displaystyle  A can do a piece of work in \displaystyle  20 days while \displaystyle  B can do it in \displaystyle  15 days. With the help of \displaystyle  C , they finish the work in \displaystyle  5 days. In what time would \displaystyle  C alone do it?

Answer:

\displaystyle \text{ A's 1 Day }   = \frac{1}{20}

\displaystyle \text{ B's 1 Day }   = \frac{1}{15}

\displaystyle \text{ C's 1 Day }   = \frac{1}{x}

\displaystyle \text{ A's + B's  + C's 1 Day Work }   = (\frac{1}{20} + \frac{1}{15} + \frac{1}{x} ) = \frac{1}{5}

Solving for \displaystyle  x = 12 Days

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Question 5: \displaystyle  A can do a piece of work in \displaystyle  12 days and \displaystyle  B alone can do it in \displaystyle  16 days. They worked together on it for \displaystyle  3 days and then \displaystyle  A left. How long did \displaystyle  B take to finish the remaining work?

Answer:

\displaystyle \text{ A's 1 Day }   = \frac{1}{12}

\displaystyle \text{ B's 1 Day }   = \frac{1}{16}

\displaystyle \text{ A's + B's 1 Day Work }   = ( \frac{1}{12} + \frac{1}{16} ) = \frac{7}{48}

 \displaystyle \text{The amount of work that is completed in 3 days }  = \frac{3\times7}{48} = \frac{7}{16}

\displaystyle \text{Amount of work left for B to complete }  = 1 - \frac{7}{16} = \frac{9}{16}

 \displaystyle \text{Therefore the number of days that B will take to finish the work }   = \frac {\frac{9}{16}} {\frac{1}{16} }  = 9 \text{ days }

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Question 6: \displaystyle  A can do \displaystyle  \frac{1}{4} of a work in \displaystyle  5 days, while \displaystyle  B can do \displaystyle  \frac{1}{5} of the work in \displaystyle  6 days. In how many days can both do it together?

Answer:

If A can do \displaystyle  \frac{1}{4} of a work in \displaystyle  5 days, then A can do the entire work in \displaystyle  20 days.

\displaystyle \text{Therefore A's 1 Day Work  } = \frac{1}{20}

If B can do \displaystyle  \frac{1}{5} of a work in \displaystyle  6 days, then B can do the entire work in \displaystyle  30 days.

\displaystyle \text{ Therefore  B's 1 Day }   = \frac{1}{30}

\displaystyle \text{ A's + B's 1 Day Work }   = (\frac{1}{20} + \frac{1}{30} ) = \frac{1}{12}

Therefore both can do the work in \displaystyle  12 days.

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Question 7: \displaystyle  A can dig a trench in \displaystyle  6 days while \displaystyle  B can dig it in \displaystyle  8 days. They dug the trench working together and received \displaystyle  1120 for it. Find the share of each in it.

Answer:

\displaystyle \text{ A's 1 Day }   = \frac{1}{6}

\displaystyle \text{ B's 1 Day }   = \frac{1}{8}

\displaystyle \text{Therefore the ratio of work  } = \frac {\frac{1}{6}} {\frac{1}{8} } = \frac{8}{6}

\displaystyle \text{Therefore A's share } = \frac{8}{14} \times 1120 = 640

\displaystyle  \text{Therefore B's share  }= \frac{6}{14} \times 1120 = 480

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Question 8: \displaystyle  A can mow a field in \displaystyle  9 days; \displaystyle  B can mow it in \displaystyle  12 days while \displaystyle  C can mow it in \displaystyle  8 days. They all together mowed the field and received \displaystyle  1610 for it. How will the money be shared by them?

Answer:

\displaystyle \text{ A's 1 Day }   = \frac{1}{9}

\displaystyle \text{ B's 1 Day }   = \frac{1}{12}

\displaystyle \text{ C's 1 Day }   = \frac{1}{18}

\displaystyle \text{Therefore the ratio of their one day's work  } = \frac{1}{9} \colon \frac{1}{12} \colon \frac{1}{18} = 8 \colon 6 \colon 9

\displaystyle  \text{Hence A's share  } = \frac{8}{23} \times 1120 = 560

\displaystyle  \text{Hence B's share  } = \frac{6}{23} \times 1120 = 420

\displaystyle  \text{Hence C's share  } = \frac{9}{23} \times 1120 = 630

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Question 9: \displaystyle  A and \displaystyle  B can do a piece of work in \displaystyle  30 days; \displaystyle  B and \displaystyle  C in \displaystyle  24 days; \displaystyle  C and \displaystyle  A in \displaystyle  40 days. How long will it take them to do the work together? In what time can each finish it, working alone?

Answer:

\displaystyle \text{ A's + B's 1 Day Work }   = \frac{1}{30}

\displaystyle  \text{(B's + C's) 1 Day work  } = \frac{1}{24}

\displaystyle \text{(C's + D's) 1 Day work  } = \frac{1}{40}

\displaystyle  \text{Adding the above three  } 2\times(A + B + C) \text{ day work }=   (\frac{1}{30} + \frac{1}{24} + \frac{1}{40} ) = \frac{1}{10}

Therefore if they all work together, they will take 20 days to finish the work.

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Question 10: \displaystyle  A can do a piece of work in \displaystyle  80 days. He works at it for \displaystyle  10 days and then \displaystyle  B alone finishes the remaining work in \displaystyle  42 days. In how many days could both do it?

Answer:

\displaystyle \text{ A's 1 Day }   = \frac{1}{80}

\displaystyle  \text{Work finished by A in 10 days } = \frac{1}{80} \times 10 =  \frac{1}{8}

\displaystyle \text{B finished the remainder of work  } (1 - \frac{1}{8} ) = \frac{7}{8} \text{ in } 42\text{ days }

\displaystyle \text{Therefore 1 Days work for B  } = \frac{\frac{7}{8}}{42} = \frac{1}{48}

Hence B can do the work in 48 days

\displaystyle \text{ A's + B's 1 Day Work }   = ( \frac{1}{80} + \frac{1}{48} )= \frac{1}{30}

Therefore both can do the work in 30 days.

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Question 11: \displaystyle  A and \displaystyle  B can together finish a work in \displaystyle  30 days. They worked at it for \displaystyle  20 days and then \displaystyle  B left. The remaining work was done by \displaystyle  A alone in \displaystyle  20 more days. In how many days can \displaystyle  A alone do it?

Answer:

\displaystyle \text{ A's + B's 1 Day Work }   = \frac{1}{30}

\displaystyle  \text{Amount of work finished by both in 20 days  } = 20 \times \frac{1}{30} = \frac{2}{3}

\displaystyle  \text{Work left to be finished  } = 1 - \frac{2}{3} = \frac{1}{3}

\displaystyle \text{Work done by A in 1 Day  } = \frac{\frac{1}{3}}{20} = \frac{1}{60}

Therefore A can do the work alone in 60 days.

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Question 12: \displaystyle  A can do a certain job in \displaystyle  25 days which \displaystyle  B alone can do in \displaystyle  20 days. \displaystyle  A started the work and was joined by \displaystyle  B after \displaystyle  10 days. In how many days was the whole work completed?

Answer:

\displaystyle \text{A's 1 Day work  } = \frac{1}{25}

\displaystyle \text{ B's 1 Day }   = \frac{1}{20}

\displaystyle \text{ A's + B's 1 Day Work }   = ( \frac{1}{25} + \frac{1}{20} ) = \frac{9}{100}

\displaystyle  \text{ Amount of work finished by A in 10 days  } = 10 \times \frac{1}{25}

\displaystyle  \text{Work left to be finished  } = (1- \frac{2}{5} ) = \frac{3}{5}

\displaystyle \text{Days taken by both A and B working together  } = \frac{\frac{3}{5}}{\frac{9}{100}} = 6 \frac{2}{3}

\displaystyle \text{The work got completed in  } 10 + 6 \frac{2}{3} = 16 \frac{2}{3}

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Question 13: \displaystyle  A can do a piece of work in \displaystyle  14 days, while \displaystyle  B can do in \displaystyle  21 days. They begin together. But, \displaystyle  3 days before the completion of the work, \displaystyle  A leaves off. Find the total number of days taken to complete the work.

Answer:

\displaystyle  \text{ A's 1 Day work } = \frac{1}{14}

\displaystyle \text{ B's 1 Day }   = \frac{1}{21}

\displaystyle \text{ A's + B's 1 Day Work }   = ( \frac{1}{14} + \frac{1}{21} ) = \frac{5}{42}

\displaystyle \text{ Amount of work finished by B in 3 days } = (3 \times \frac{1}{21} ) = \frac{1}{7}

\displaystyle \text{ Work left to be finished by A and B together } = (1- \frac{1}{7} ) = \frac{6}{7}

\displaystyle \text{ Days taken by A + B working together } = \frac{\frac{6}{7}}{\frac{5}{42}} = 7 \frac{1}{5}

\displaystyle \text{ The work got completed in } 3+7 \frac{1}{5} = 10 \frac{1}{5}

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Question 14: \displaystyle  A is thrice as good a workman as \displaystyle  B and \displaystyle  B is twice as good a workman as \displaystyle  C . All the three took up a job and received \displaystyle  1800 as remuneration. Find the share of each.

Answer:

\displaystyle \text{If C takes    }  x  days to complete the job

\displaystyle \text{ Then B will complete the job in  }  \frac{x}{2} 

\displaystyle \text{ And A will complete the job in  }  \frac{x}{3} \text{ days } 

\displaystyle  \text{ Therefore the ratio of 1 day's work of  }  A \colon B \colon C =  \frac{x}{3} \colon \frac{x}{2} \colon \frac{x}{1} = 3 \colon 2 \colon 1

\displaystyle \text{  Therefore the share of A  }  = 1800 \times  \frac{3}{6} = 900 \text{  Rs. }

\displaystyle  \text{  Therefore the share of B  }   = 1800 \times  \frac{2}{6} = 600 \text{  Rs. }

\displaystyle  \text{ Therefore the share of C  }   = 1800 \times  \frac{1}{6} = 300 \text{  Rs. }

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Question 15: \displaystyle  A can do a certain job in \displaystyle  12 days. \displaystyle  B is \displaystyle  60\% more efficient than \displaystyle  A . Find the number of days taken by \displaystyle  B to finish the job.

Answer:

\displaystyle \text{Time taken to finish the job  }  = 12 \text{   }  days

\displaystyle  \text{ A's 1 Day's work  } = \frac{1}{12}

B is 60% more efficient

\displaystyle  \text{ B's 1 Day's work  } = 1.6 \times  \frac{1}{12}

\displaystyle \text{Therefore the number of days B will take to finish the job    }  = \frac{1}{\frac{1.6}{12}} = 7  \frac{1}{2} days

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Question 16: \displaystyle  A is twice as good a workman as \displaystyle  B and together they finish a piece of work in \displaystyle  14 days. In how many days can \displaystyle  A alone do it?

Answer:

\displaystyle \text{Let B take }  x  \text{days to finish the work.   }  

\displaystyle \text{B's 1 day's work    }  = \frac{1}{x}

\displaystyle  \text{ The A will take  } \frac{x}{2} \text{ days to finish the work.  }  

\displaystyle \text{A's 1 day's work    }  = \frac{2}{x}

\displaystyle \text{ (A+B) one day's work  }  = ( \frac{1}{x} + \frac{2}{x} ) = \frac{1}{14}

\displaystyle \text{Solving for    }   x=42

Therefore A will take 21 days to finish the job.

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Question 17: Two pipes \displaystyle  A and \displaystyle  B can separately fill a tank in \displaystyle  36 minutes and \displaystyle  45 minutes respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

Answer:

\displaystyle \text{A's 1 minute fill rate    }  = \frac{1}{36}

\displaystyle  \text{B's 1 minute fill rate    } = \frac{1}{45}

\displaystyle  \text{A's and B's fill rate together    } = ( \frac{1}{36} + \frac{1}{45} ) = \frac{1}{20}

Therefore if A and B are opened simultaneously, the tank will take 20 minutes to fill up.

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Question 18: One tap can fill a cistern in \displaystyle  3 hours and the waste pipe can empty the full tank in \displaystyle  5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?

Answer:

\displaystyle \text{Tap's 1 minute fill rate    }  = \frac{1}{3}

\displaystyle \text{ Waste Pipe's 1 minute empty rate  }  = \frac{1}{5}

\displaystyle \text{Therefore the net fill rate    }  = ( \frac{1}{3} - \frac{1}{5} ) = \frac{2}{15}

\displaystyle \text{Therefore if both the tap and the waste pipe are opened simultaneously then it will take    }  7\frac{1}{2}  hours to fill up.

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Question 19: Two pipes \displaystyle  A and \displaystyle  B can separately fill a cistern in \displaystyle  20 minutes and \displaystyle  30 minutes respectively , while a third pipe \displaystyle  C can empty the full cistern in \displaystyle  15 minutes. If all the pipes are opened together, in what time the empty cistern is filled?

Answer:

\displaystyle  \text{A's 1 minute fill rate    } = \frac{1}{20}

\displaystyle \text{ B's 1 minute fill rate  }  = \frac{1}{30}

\displaystyle \text{C's 1 minute empty rate    }  = \frac{1}{15}

\displaystyle \text{ Therefore the net fill rate  }  = ( \frac{1}{20} + \frac{1}{30} - \frac{1}{15} ) = \frac{1}{60}

Therefore if all the tap are opened simultaneously then it will take 60 minutes or one hour to fill up.

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Question 20: A pipe can fill a tank in \displaystyle  16 hours. Due to a leak in the bottom, it is filled in \displaystyle  24 hours. If the tank is full, how much time will the leak take to empty it?

Answer:

\displaystyle \text{ Pipe's fill rate  }  = \frac{1}{16}

\displaystyle \text{Let the leak is at a rate of    }  = \frac{1}{x}

\displaystyle \text{Therefore the net fill rate    }  = ( \frac{1}{16} - \frac{1}{x} ) = \frac{1}{24}

\displaystyle  \text{Solving for    }   x = 48  hours.