One approach is to figure out how much work ONE man does in ONE day (the unit rate of work per person).  Once you know that, the rest falls into place easily.



6 men do the entire job in 8 days



That means 6 men do 1/8 of a job in one day



And one man does  (1/6)*(1/8) = 1/48 job in one day





How much work is done in the first three days?

3days * 6men * (1/48 job/day/man) = 18/48 job 



This leaves (48/48) - (18/48) = 30/48 of the job to do by the remaining 4 men:



Say it takes  N more days...

4 * (1/48) * N =  30/48



N = (30/48) * (48/4) = 7.5 days





Therefore, the entire job takes  3 days + 7.5 days =  days.



Check:  3men * 6days * (1/48 job/day/man) + 7.5days * 4men * (1/48 job/day/man) = 48/48 = 1 job (ok)



 

Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!
 


Here are two other ways -- among many different possible ways -- of solving the problem.


Look at different ways and find what "works" best for you....


(1) After the first 3 days, there is enough work for the 6 men for another 5 days.


When 2 men are taken off the job, there are 4 men left; the number of workers gets multiplied by a factor of 4/6=2/3.


That means the amount of time remaining to finish the job gets multiplied by a factor of 3/2.  So the number of days needed to finish the job is 5(3/2)=7.5


And so the job from start to finish takes 3+7.5 = 10.5 days.


(2) The entire job takes 6*8 = 48 man-days to complete.


The 6 men working for the first 3 days work a total of 6*3=18 man-days; there are 48-18=30 man-days of work remaining.


The number of days it takes 4 men to do the remaining 30 man-days of work is 30/4=7.5.


The total time for the job from start to finish is 3+7.5 = 10.5 days.


 

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