# Time and Work – 3

## 25 thoughts on “Time and Work – 3”

1. RAJNISH KUMAR |

nice way sir

2. siv santosh |

ans-24 days

3. Visakha Mundra |

Is the answer for the mason question 10?

Sir,
Can you post some questions related to work done on alternative days and also wages related concepts?
Regards,

• Patience! Those, too, will come. I will put at least 3-4 more posts on this topic, over the next few Wednesdays.

regards
J

5. Sliver Bane |

Is the answer to the last question 10 days. We will use the formula for sum of all odd numbers which is n^2.

• Nope, it is not 2 people joining every day but 1 person joining every second day!

regards
J

• Sliver Bane |

AHHHH! I just re-read the question! So upon realizing my mistake I now present thee with as 19 days my answer, oh humble J, head my call and tell me if me errors have brought me wisdom?

• Sliver Bane |

heed my call* :p

6. Sarika |

The sum of odd numbers is n^2 which will not give the result, We take into account that masons show up after every alternate day : so manhours look like 1+2+3+3+4+4+5 for 7 days making up 28 manhours. The rest of 120 manhours(5mason * 20days) = 120-28 = 92 are completed in 92/5 = 18.4 days.Total days required = 46.4.

7. Sarika |

ok this too is wrong.

8. Devansh Dubey |

9. That certain sounds like a much more plausible answer ðŸ™‚

regards
J

10. Marcus Aurelius |

Is the Answer to last King-Temple question is 9+9+1 =19 days(45+45+10 = 100 mandays)?

11. arvind |

12. tarun |

• Manual calculation. Total man-days = 100. So we find when 1 + 1 + 3 + 3 + 5 + … reaches 100.

regards
J

• Indhu |

When the given data is considered, we have 100 units of work and each mason can do 1 unit of work a day(considering equal strength for each mason), so the flow goes like 1+1+2+2+3+3 +.. 9+9+10 = 100 units and in 21 days.
Please suggest if this approach is wrong.

• Approach is fine, but answer is 19 not 21 right?

regards
J

13. Aish |

Hi !
I solved it considering the availability of only 5 masons to do the work. Accordingly, I got (1+1+2+2+3+3+4+4+5) = 25 for the initial 9 days. And then, all 5 workers work to complete the remaining 75 man days work in 75/15 = 15 days, totaling 24 days.

• However, the question never says that only 5 masons are available. It just said 5 masons would take 20 days.

regards
J

• Aish |

Oh okay. Got your point. Thank you, J.

14. Aish |

Sorry..it is 75/5 = 25 days