pirate wrote:Let the amount of work 1 man does in 1 day be U.S-H wrote:Hi Can someone help me with the following question, thank you!

5 men are hired to complete a job. If one more man is hired, the job can be completed 8 days earlier. Assuming that all the men work at the same rate, how many more men should be hired so that the job can be completed 28 days earlier?

Amount of work 5 men does in 1 day = 5U

Amount of work 6 men does in 1 day = 6U

Let the no. days required by 5 men to complete the job be D

Amount of work required to complete the job = 5U x D

5U x D = 6U x (D - 8)

5UD = 6UD - 48U

48U = UD

48 = D

No. of days required by 5 men to complete the job = 48 days

Amount of work required to complete the job = 48 x 5U = 240U

To complete the job 28 days earlier = completing the job in 48 - 28 = 20 days

240U / 20 = 12U

No. of men required to complete the job in 20 days is 12

Therefore 12 - 5 = 7 more men should be hired to complete the job 28 days earlier.

Hi pirate,

Thank you for your help!