A) 2 hours | B) 8 hours |

C) 6 hours | D) 4 hours |

Explanation:

Efficiency of Inlet pipe A = 4.16%

Efficiency of Inlet pipe B = 5.83%

Efficiency of A and B together = 100 %

Now, if the efficiency of outlet pipe be x% then in 10 hours the capacity of tank which will be filled = 10 * (10 - x)

Now, since this amount of water is being emptied by C at x% per hour, then

x = 8%

Therefore, in 10 hours 20% tank is filled only. Hence, the remaining 80% of the capacity will be filled by pipes A and B in 80/10 = 8 hours

A) 47/7 days | B) 59/6 days |

C) 48/5 days | D) 57/5 days |

Explanation:

Amount of work K can do in 1 day = 1/16

Amount of work L can do in 1 day = 1/12

Amount of work K, L and M can together do in 1 day = 1/4

Amount of work M can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 – 1/12 = 5/48

=> Hence M can do the job on 48/5 days = 9 (3/5) days

A) 24 1/2 days | B) 25 3/2 days |

C) 24 2/3 days | D) 26 2/3 days |

Explanation:

1/3 ---- 8

1 -------?

Hari can do total work in = 24 days

As satya is 60% efficient as Hari, then

Satya = 1/24 x 60/100 = 1/40

=> Satya can do total work in 40 days

1 ----- 40

2/3 ---- ? => 26 2/3 days.

A) 4 days | B) 8 days |

C) 3 days | D) 6 days |

Explanation:

4/10 + 9/x = 1

=> x = 15

Then both can do in

1/10 + 1/15 = 1/6

=> 6 days

A) 2 days | B) 2.5 days |

C) 2.25 days | D) 3 days |

Explanation:

1 man's 1 day work = 1/108

12 men's 6 day's work = 1/9 x 6 = 2/3

Remaining work = 1 - 2/3 = 1/3

16 men's 1 day work = 1/108 x 16 = 4/27

4/27 work is done by them in 1 day.

1/3 work is done by them in 27/4 x 1/3 = 9/4 days.

A) 5 (2/3) | B) 6 (3/4 ) |

C) 4 (1/2) | D) 3 |

Explanation:

Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.