A) 60 | B) 70 |

C) 80 | D) 90 |

Explanation:

Let A's 1 day's work=x and B's 1 day's work=y

Then x+y = and 20x+60y=1

Solving these two equations , we get : x= and y=

B's 1 day work =

Hence,B alone shall finish the whole work in 80 days

A) 8/15 | B) 7/9 |

C) 6/13 | D) 4/11 |

Explanation:

P's 1 day's work = | 1 | |

15 |

Q's 1 day's work = | 1 | |

20 |

(P + Q)'s 1 day's work = | 1 | + | 1 | = | 7 | |||

15 | 20 | 60 |

(P + Q)'s 4 day's work = | 7 | x 4 | = | 7 | |||

60 | 15 |

Therefore, Remaining work = | 1 - | 7 | = | 8 | . | ||

15 | 15 |

A) 1/3 | B) 2/3 |

C) 1/6 | D) 5/6 |

Explanation:

Given X can do in 10 days

=> 1 day work of X = 1/10

Y can do in 15 days

=> 1 day work of Y = 1/15

1day work of (X + Y) = 1/10 + 1/15 = 1/6

Given they are hired for 5 days

=> 5 days work of (X + Y) = 5 x 1/6 = 5/6

Therefore, **Unfinished work = 1 - 5/6 = 1/6 **

A) 3 days | B) 4 days |

C) 5 days | D) 6 days |

Explanation:

Given 10 men take 15 days to complete a work

=> Total mandays = 15 x 10 = 150

Let the work be 150 mandays.

=> Now 37 men can do 150 mandays in 150/37 =~ 4 days

A) 9 hrs | B) 7 hrs |

C) 13 hrs | D) 11 hrs |

Explanation:

Given,

P can fill in 12 hrs

Q can fill in 15 hrs

R can fill in 20 hrs

=> Volume of tank = LCM of 12, 15, 20 = 60 lit

=> P alone can fill the tank in 60/12 = 5 hrs

=> Q alone can fill the tank in 60/15 = 4 hrs

=> R alone can fill the tank in 60/20 = 3 hrs

Tank can be filled in the way that

(P+Q) + (P+R) + (P+Q) + (P+R) + ....

=> Tank filled in 2 hrs = (5+4) + (5+3) = 9 + 8 = 17 lit

=> In 6 hrs = 17 x 6/2 = 51 lit

=> In 7th hr = 51 + (5+4) = 51 + 9 = 60 lit

=> So, total tank will be filled in **7 hrs**.

A) 56 days | B) 54 days |

C) 60 days | D) 64 days |

Explanation:

(P+Q)'s 1 day work = 1/24

P's 1 day work = 1/32

=> Q's 1 day work = 1/24 - 1/32 = 1/96

Work done by (P+Q) in 8 days = 8/24 = 1/3

Remainining work = 1 - 1/3 = 2/3

Time taken by Q to complete the remaining work = 2/3 x 96 = 64 days.