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?
heed my call* :p
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.
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
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.
CAT-holics 
nice way sir
ans-24 days
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,
Adorini
Patience! Those, too, will come. I will put at least 3-4 more posts on this topic, over the next few Wednesdays.
regards
J
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
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?
heed my call* :p
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.
ok this too is wrong.
Is the answer 17 days?
That certain sounds like a much more plausible answer 🙂
regards
J
Is the Answer to last King-Temple question is 9+9+1 =19 days(45+45+10 = 100 mandays)?
Yes it is.
regards
J
approach
approach please
can you please explain this?
Manual calculation. Total man-days = 100. So we find when 1 + 1 + 3 + 3 + 5 + … reaches 100.
regards
J
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
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
Oh okay. Got your point. Thank you, J.
Sorry..it is 75/5 = 25 days