A) 10 min. 20 sec. | B) 11 min. 45 sec. |

C) 12 min. 30 sec. | D) 14 min. 40 sec. |

Explanation:

Part filled in 4 minutes =

Remaining part =

Part filled by B in 1 minute =

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec

A) 85 min | B) 95 min |

C) 105 min | D) 100 min |

Explanation:

Work done by the third pipe in 1 min = 1/50 - (1/60 + 1/75) = - 1/100.

[-ve sign means emptying]

The third pipe alone can empty the cistern in 100 min.

A) 360 lit | B) 480 lit |

C) 320 lit | D) 420 lit |

Explanation:

1/k - 1/20 = -1/24

k = 120

120 x 4 = 480

Therefore, the capacity of the cistern is 480 liters.

A) 85 min | B) 92 min |

C) 187 min | D) 144 min |

Explanation:

Let the slower pipe alone fill the tank in x min.

A) 24.315 minutes | B) 26.166 minutes |

C) 22.154 minutes | D) 24 minutes |

Explanation:

Upto first 5 minutes I, J and K will fill => 5[(1/20)+(1/30)+(1/40)] = 65/120

For next 6 minutes, J and K will fill => 6[(1/30)+(1/40)] = 42/120

So tank filled upto first 11 minutes = (65/120) + (42/120) = 107/120

So remaining tank = 13/120

Now at the moment filling with C and leakage @ 1/60 per minute= (1/40) - (1/70) = 3/280

So time taken to fill remaining 13/120 tank =(13/120) /(3/280) = 91/6 minutes

Hence total time taken to completely fill the tank = 5 + 6 + 91/6 = 26.16 minutes.

A) 3 minutes | B) 2 minutes |

C) 4 minutes | D) 5 minutes |

Explanation:

Let the total capacity of tank be 90 liters.

Capacity of tank filled in 1 minute by K = 3 liters.

Capacity of tank filled in 1 minute by L = 15 liters.

Therefore, capacity of the tank filled by both K and L in 1 minute = 18 liters.

Hence, time taken by both the pipes to overflow the tank = 90/18 = 5 minutes.