maihuna wrote:A contractor undertook to do a certain piece of work in 9 days. He employed certain number of men , but 6 of them being absent from the very first day, the rest could finish the work in 15 days. The no of men originally employed were:
3
4
5
6
9
A contractor employed 30 men to do a piece of work in 38 days, after 25 days he employed 5 men more and the work was finished one day earlier. how many days he would have been behind if he had not employed additional men:
1
5/4
3/2
7/4
2
For both questions, we need to employ the notion of "man days." For example, if it takes 7 men working 9 days each to complete a job, then the job takes 63 "man days" (7x9) to complete. So, we can also conclude that if there were only 3 men working on the job, then the same job would take 21 days (63/3) to complete.
Question 1:
Let x be the number of men originally hired.
Since it is not explicitly stated in the question, we must assume that these x men would have completed the job in 9 days (given the ambiguity, I'm pretty sure this is not an official question
. So, the job will take 9x man days to complete.
If 6 men do not come to work, then there are x-6 men and it takes them 15 days to complete the same job. In other words, the number of man days required is 15(x-6)
We can create the equation 9x = 15(x-6)
When we solve for x, we get x=15.
So, there must have been 15 men originally. (The answer is not provided in the choices)
Question 2:
Determine the total number of man days required to complete the job.
30 men working for 25 days = 750 man days (30x25)
35 men working for 12 days = 420 man days (35x12)
Total number of man days to complete the job = 1170
If the contractor had not hired the additional men it would have taken 1170/30 days for the thirty men to complete the job.
1170/30 = 39, so it would have taken one extra day to complete the job. The answer is A
