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) 4:1 | B) 5:3 |

C) 16:1 | D) 1:16 |

Explanation:

Time taken by 8th tap = 2 x 2 x 2= 8 hours

Time taken by 12th tap = 2x (1/2) x (1/2) = 1/2 hour

Ratio of time taken by 8th tap and 12th tap = 8 : 1/2 =16:1

Therefore, Ratio of efficiencies of 8th tap and 12th tap =1:16

A) 3 hours | B) 4 hours |

C) 5 hours | D) None of these |

Explanation:

Rate of leakage = 8.33% per hour

Net efficiency = 50 - (16.66 + 8.33)= 25%

Time required = 100/25 = 4 hours

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

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

Explanation:

Efficiency of P= 100/20= 5% per hour

Efficiency of Q= 100/25= 4% per hour

Efficiency of R= 100/40= 2.5% per hour

Efficiency of S=100/50= 2% per hour

Cistern filled till 10 am by P, Q and R

$\left.\begin{array}{c}\mathrm{Till}10.00\mathrm{am}\mathrm{Pipe}\mathrm{P}\mathrm{filled}20\%\\ \mathrm{Till}10.00\mathrm{am}\mathrm{Pipe}\mathrm{Q}\mathrm{filled}8\%\\ \mathrm{Till}10.00\mathrm{am}\mathrm{Pipe}\mathrm{R}\mathrm{filled}2.5\%\end{array}\right\}30.5\%$

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

Rest of cistern to be filled = 100 - 30.5 = 69.5%

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

= 69.5 / (5+4+2.5+2) = 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:

Let the number of workers be x.

Now, Using work equivalence method,

X + (X-1) + (X-2)+ . . . . + 1 = X *55% of X

=> [X * (X+1)] / 2 = X * (55X/100) [because, Series is in AP. Sum of AP = {No. of terms (first term+ last term)/2} ]

Therefore, X = 10

A) 10 | B) 5 |

C) 15 | D) 4 |

Explanation:

Since 20% i.e 1/5 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 = $\frac{25x}{175x}\times 100$ % = $14\frac{2}{7}\%$