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Work - Rate Problem - Computer Server

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Work - Rate Problem - Computer Server

by sal_xcool » Fri Mar 04, 2011 7:59 pm
It takes 6 technicians a total of 10 hours to build and program a new server from Direct Computer, with each
technician working at the same rate. If 6 technicians start to build the server at 11:00 a.m., and one technician
per hour is added beginning at 5:00 p.m., at what time will the server be complete?


A. 6:40 p.m.
B. 7:00 p.m.
C. 7:20 p.m.
D. 7:40 p.m.
E. 8:00 p.m.

OA E

I am completely lost with this question.
Please provide some thorough answer with detailed explanation. Thanks alot in advance.
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by Anurag@Gurome » Fri Mar 04, 2011 8:58 pm
Solution:
It takes 6 technicians a total of 10 hours to build a server.
So, in 1 hour, 6 technicians do (1/10)th part of the work.
Also, it takes 1 technician a total of 60 hours to build the server.
Or, in 1 hour, 1 technician does 1/60th part of the work
So from, 11 a.m to 5 p.m which is 6 hours, 6 technicians do (1/10)*6 = (3/5)th part of the work.
Or (2/5)th part of the work is left.
Now after 5 pm, there are 7 technicians who work for 1 hour till 6 pm.
So, they complete (7/60)th part of the work.
Or 2/5 - 7/60 = (17/60)th part of the work is left.
Now 8 technicians work from 6 pm to 7 pm for 1hour.
So, they do (8/60) th part of the work.
Remaining work to be done is (17/60 - 8/60) = 9/60.
Now 9 technicians work from 7 pm to 8 pm for 1 hour.
Now 9 technicians do (9/60)th part of the work upto 8 pm.
Since (9/60) th of the work was left to be done, these 9 technicians complete the work by 8 pm.

The correct answer is (E).
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by Rich@VeritasPrep » Fri Mar 04, 2011 9:02 pm
The 6 technicians would take 10 hours to complete the job. They work from 11 to 5, which is a 6 hour period. That means that if the 6 technicians continued with the rest of the job, it would take 10-6 = 4 hours.

Since 6 technicians would complete the remaining work in 4 hours, that means 1 technician would complete the work in 24 hours. (You can see this from the work rate formula W = r*t = (6*tech rate)*4 = 24 * tech rate)

This also means 1 technician would complete 1/24 of the remaining work in one hour.

We add a seventh technician at 5pm, and the seven work for an hour. That means they get 7/24 of the remaining work done.

We add an eighth technician at 6pm, and the eight work for an hour. That means they get another 8/24 done (of the work left at 5pm).

We add a ninth technician at 7pm, and the nine work for an hour. That means they get another 9/24 (of the work left at 5pm) done at 8pm.

7/24+8/24+9/24 = 24/24 = 1 = complete job.

Answer: E
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by manpsingh87 » Fri Mar 04, 2011 10:30 pm
sal_xcool wrote:It takes 6 technicians a total of 10 hours to build and program a new server from Direct Computer, with each
technician working at the same rate. If 6 technicians start to build the server at 11:00 a.m., and one technician
per hour is added beginning at 5:00 p.m., at what time will the server be complete?


A. 6:40 p.m.
B. 7:00 p.m.
C. 7:20 p.m.
D. 7:40 p.m.
E. 8:00 p.m.

OA E

I am completely lost with this question.
Please provide some thorough answer with detailed explanation. Thanks alot in advance.
let work to be 60 units,
so from 11 to 5pm 6 technicians worked for 6 hours thus they have done 6*6 = 36 units of work. Therefore, remaining left over work is 24 units.
now at 6 pm work would be 7 as new person has joined in, similarly at 7 pm work would be 8, and at 8 work would be 9

adding these work unit we get 7+8+9 = 24.

hence the work will be completed by 8 pm.!!!
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by sanju09 » Sat Mar 05, 2011 1:10 am
sal_xcool wrote:It takes 6 technicians a total of 10 hours to build and program a new server from Direct Computer, with each
technician working at the same rate. If 6 technicians start to build the server at 11:00 a.m., and one technician
per hour is added beginning at 5:00 p.m., at what time will the server be complete?


A. 6:40 p.m.
B. 7:00 p.m.
C. 7:20 p.m.
D. 7:40 p.m.
E. 8:00 p.m.

OA E

I am completely lost with this question.
Please provide some thorough answer with detailed explanation. Thanks alot in advance.

If it takes 6 technicians a total of 10 hours to complete the said task with each technician working at the same rate, then the work rate per technician is 1/60 per hour.

From 11:00 a.m. to 5:00 p.m. the same day, there is 6 hours time, during which the unit of 6 technicians would have fixed 6/10 or 3/5 or 36/60 of the new server from Direct Computer, with only 24/60 of the work remaining. We'll have to be watchful as at what hour the sum of all numerators in the following story becomes or exceeds 24 to make certain that the task is now over.

Now, 7 technicians in another hour (i.e. by 6:00 p.m.) would fix 7/60 of the whole task, and then 8 technicians in another hour (i.e. by 7:00 p.m.) would fix 8/60 of the whole task, and then 9 technicians in another hour (i.e. by 8:00 p.m.) would fix 9/60 of the whole task...STOP, since 7 + 8 + 9 = 24, and that happened at [spoiler]8:00 pm


E[/spoiler]
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by GMATGuruNY » Sat Mar 05, 2011 6:59 am
sal_xcool wrote:It takes 6 technicians a total of 10 hours to build and program a new server from Direct Computer, with each
technician working at the same rate. If 6 technicians start to build the server at 11:00 a.m., and one technician
per hour is added beginning at 5:00 p.m., at what time will the server be complete?


A. 6:40 p.m.
B. 7:00 p.m.
C. 7:20 p.m.
D. 7:40 p.m.
E. 8:00 p.m.

OA E

I am completely lost with this question.
Please provide some thorough answer with detailed explanation. Thanks alot in advance.
When the job is undefined, plug in. Plugging in will achieve the same result as other approaches and save us from having to deal with messy fractions.

Let rate per worker = 1 unit/hour.
Every hour 6 workers produce 6*1 = 6 units.
Over 10 hours, 10*6 = 60 units are produced. This is the value of the job.
From 11am-5pm, 6 workers will produce r*t = 6*6 = 36 units.
Work remaining = 60-36 = 24 units.
Each hour thereafter another worker is added. During the next hour 7 units will be produced, then 8, then 9.
7+8+9 = 24 units.
In 3 hours, the job will have been completed.
5pm + 3 hours = 8pm.

The correct answer is E.
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by Scott@TargetTestPrep » Tue Dec 12, 2017 6:51 am
sal_xcool wrote:It takes 6 technicians a total of 10 hours to build and program a new server from Direct Computer, with each technician working at the same rate. If 6 technicians start to build the server at 11:00 a.m., and one technician per hour is added beginning at 5:00 p.m., at what time will the server be complete?


A. 6:40 p.m.
B. 7:00 p.m.
C. 7:20 p.m.
D. 7:40 p.m.
E. 8:00 p.m.

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: E

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