A) 8 hrs 45 min | B) 8 hrs 42 min |

C) 8 hrs | D) 8 hrs 34 min |

Explanation:

Number of pages typed by Adam in 1 hour = = 6

Number of pages typed by Smith in 1 hour = = 8

Number of pages typed by both in 1 hour = (6 + 8) = 14

Time taken by both to type 110 pages = (120 * 1/14) = 8 = 8 hrs 34 min.

A) 4:18 pm | B) 3:09 pm |

C) 12:15 pm | D) 11:09 am |

Explanation:

Efficiency of P= 5%

Efficiency of Q= 4%

Efficiency of R= 2.5%

Efficiency of S= 2%

%

Thus, at 10 am pipe P,Q and R filled 30.5% of the cistern.

Now, the time taken by P,Q,R and S together to fill the remaining capacity of the cistern

= = =5 Hours and 9 minutes(approx)

Therefore, total time =4 hrs + 5hrs 9 mins

= 9 hrs and 9 mins

It means cistern will be filled up at 3:09 pm

A) 50 | B) 40 |

C) 45 | D) 10 |

Explanation:

It can be solved easily through option.

55 = 55 Hence correct

**Alternatively:**

** **

** **=> n= 10

In Both cases total work is 55man-days.

A) 10 | B) 5 |

C) 15 | D) 4 |

Explanation:

Since 20% typists left the job. So, there can be any value which is multiple of 5 i.e, whose 20% is always an integer. Hence, 5 is the least possible value.

A) 10 % | B) 14 2/7 % |

C) 20 % | D) Can't be determined |

Explanation:

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.

D * x +(100- D) * 2x= 175x

=> D= 25 days

Now , the work done in 25 days = 25x

Total work = 175x

therefore, workdone before increasing the no of workers = %

A) 64000 liters | B) 56000 liters |

C) 78000 liters | D) 60000 liters |

Explanation:

Let x liter be the per day filling and v litr be the capacity of the reservoir, then

90x + v = 40000 * 90 -----(1)

60x + v= 32000 * 60 ------(2)

solving eq.(1) and (2) , we get

x = 56000

Hence , 56000 liters per day can be used without the failure of supply.