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## Arithmetic: Time and Work Test-5

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Question 1 |

If it takes A four days to dig a certain ditch, whereas B can dig it in 8 days and A, B, C together can dig it in $ \displaystyle 2\frac{2}{7}$ days, how long C alone would take to dig it?

8 | |

4 | |

6 | |

16 |

Question 1 Explanation:

Let the ditch take 16 units to dig up.

A does 4 units per day and B does 2 units per day.

A+B+C can do $ \displaystyle \frac{16}{\frac{16}{7}}=7\,units/day$

C does 7-4-2 = 1 unit of work per day.

Therefore, C digs the ditch in 16/1 = 16 days.

A does 4 units per day and B does 2 units per day.

A+B+C can do $ \displaystyle \frac{16}{\frac{16}{7}}=7\,units/day$

C does 7-4-2 = 1 unit of work per day.

Therefore, C digs the ditch in 16/1 = 16 days.

Question 2 |

A sum of Rs.25 was paid for a work which A can do in 32 days, B in 20 days, Band C in 12 days and D in 24 days. How much did C receive if all the four work together?

Rs.15/3 | |

Rs.14/3 | |

Rs.13/3 | |

Rs.16/3 |

Question 3 |

A cistern can be filled by two pipes filling separately in 12 and 16 min respectively. Both pipes are opened together for a certain time but being clogged, only 7/8 of full quantity water flows through the former and only 5/6 through the latter pipe. The obstructions, however, being suddenly removed, the cistern is filled in 3 min from that moment. How long was it before the full flow began?

4.5 min | |

2.5 min | |

3.5 min | |

5.5 min |

Question 3 Explanation:

Let the total quantity be 96p liters.

A fills @ 8p l/min. and B fills 6p l/min.

Rate of A+B /min= 8p + 6p = 14p l/min.

Now due to obstruction,

A’s rate is 8p X 7/8 = 7p l/min. and B’s rate is 6p x 5/6 = 5p l/min.

Total rate of A+B /min = 7+6 = 12p l/min.

Let the time for which the clog is there be t mins.

For 3 mins the pipes had a combined rate of 14p l/min.

In 3 mins pipes filled up = 14p x 3= 42p l/min.

Left (96 - 42)p l/min= 54p l/min.

Time when clogged pipes supplied the water is = 54p/12p = 4.5 mins.

A fills @ 8p l/min. and B fills 6p l/min.

Rate of A+B /min= 8p + 6p = 14p l/min.

Now due to obstruction,

A’s rate is 8p X 7/8 = 7p l/min. and B’s rate is 6p x 5/6 = 5p l/min.

Total rate of A+B /min = 7+6 = 12p l/min.

Let the time for which the clog is there be t mins.

For 3 mins the pipes had a combined rate of 14p l/min.

In 3 mins pipes filled up = 14p x 3= 42p l/min.

Left (96 - 42)p l/min= 54p l/min.

Time when clogged pipes supplied the water is = 54p/12p = 4.5 mins.

Question 4 |

The work done by a woman in 8 h is equal to the work done by a man in 6 h and by a boy in 12 h. If working 6 h per day 9 men can complete a work in 6 days, then in how many days can 12 men, 12 women and 12 boys together finish the same working 8 h per day?

$ \displaystyle 2\frac{1}{2}$ | |

$ \displaystyle 1\frac{1}{2}$ | |

$ \displaystyle 3\frac{1}{2}$ | |

None of these |

Question 4 Explanation:

Let the work be of 72 units.

1 woman does 72/8=9 units of work per hour.

1 man does 72/6=12 units of work per hour.

1 child can do 72/12=6 units of work /hour.

At 6 hours per day 9 men can do 9 x 6 x 12 units of work in 1 day.

Therefore, at 6 hours per day 9 men can do 9 x 6 x 6 x 12 units of work in 6 day.

12 men +12 women +12 children can do 12 (9+12+6) = 12 x 27 units of work in 1 hour 1 day.

Total number of hours required $ \frac{9\times 6\times 6\times 12}{12\times 27}=12\,\,hours$ The group can work for 8 hours per day.

Therefore it will take 12/8 = 1 ½ days to finish the work.

1 woman does 72/8=9 units of work per hour.

1 man does 72/6=12 units of work per hour.

1 child can do 72/12=6 units of work /hour.

At 6 hours per day 9 men can do 9 x 6 x 12 units of work in 1 day.

Therefore, at 6 hours per day 9 men can do 9 x 6 x 6 x 12 units of work in 6 day.

12 men +12 women +12 children can do 12 (9+12+6) = 12 x 27 units of work in 1 hour 1 day.

Total number of hours required $ \frac{9\times 6\times 6\times 12}{12\times 27}=12\,\,hours$ The group can work for 8 hours per day.

Therefore it will take 12/8 = 1 ½ days to finish the work.

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