# Time and Work Questions

A) 12 days | B) 18 days |

C) 24 days | D) 30 days |

Explanation:

A + B = C + D

| | | |

Ratio of efficiency 10x + 5x 9x + 6x

|________| |_________|

15x 15x

Therefore , ratio of efficiency of A:C =10:9

Therefore, ratio of days taken by A:C = 9:10

Therefore, number of days taken by A = 18 days

A) 1 day | B) 2 days |

C) 5 days | D) None of these |

Explanation:

Total work = 100+50 = 150man-days

In 8 days 100 man-days work has been completed. Now on 9th and 10th day there will be 25 workers. So in 2 days they wll complete additional 50 man- days work. Thus the work requires 2 more days.

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) 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.

As the pipes are operating alternatively, thus their 2 minutes job is =

In the next 2 minutes the pipes can fill another part of cistern.

In 4 minutes the two pipes which are operating alternatively will fill

Remaining part =

Pipe A can fill of the cistern in 1 minute

Pipe A can fill of the cistern in = min

Total time taken to fill the Cistern

minutes