Hussain15 wrote:

It takes 6 technicians a total of 10 hours to build and program a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11:00 AM, and one technician per hour is added beginning at 5:00 PM, at what time will the server be complete?

A. 7:30 PM

B. 7:45 PM

C. 8:00 PM

D. 9:00 AM

E. 10:00 PM

We are given that 6 technicians can complete a job in 10 hours. Since rate = work/time, the rate of the 6 technicians is 1/10. Thus, the rate of 1 technician is (1/10)/6 = 1/60.

We see that 6 technicians work from 11:00 AM to 5:00 PM, or 6 hours. Since rate x time = work, those technicians complete 1/10 x 6 = 6/10 = 3/5 of the job.

If another technician is added to the job at 5:00 PM, the new rate for the technicians is (1/60) x 7 = 7/60, and thus all 7 technicians complete 7/60 of the job in the hour from 5:00 PM to 6:00 PM.

Thus, 3/5 + 7/60 = 36/60 + 7/60 = 43/60 of the job will be completed.

If another technician is added to the job at 6:00 PM, the new rate for the technicians is (1/60) x 8 = 8/60, and thus all 8 technicians complete 8/60 of the job in the hour from 6:00 PM to 7:00 PM.

Thus, 43/60 + 8/60 = 51/60 of the job will be completed.

If another technician is added to the job at 7:00 PM, the new rate for the technicians is (1/60) x 9 = 9/60, and thus all 9 technicians complete 9/60 of the job in the hour from 7:00 PM to 8:00 PM.

Thus, 51/60 + 9/60 = 60/60 = 1 whole job will be completed at 8:00 PM.

Answer: C

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