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It takes 6 technicians a total of 10 hours to build and [#permalink]

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12 Apr 2010, 05:41

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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

6Techs can complete the job in 10 hours. 6 Techs can do 1/10 job in 1 hour 1 Tech can do 1/(6*10) = 1/60 job in 1 hour.

Now 6 techs worked from 11.00Am tp 5.00PM means they worked for 6 hours. 6Techs worked for 6 hours = 6/10 job completed. From 5.00 PM to 6.00 Pm a tech is added, so 7 Techs worked for 1 hour = 7/60 job Total job completed at 6.00PM is = 6/10+7/60 = 43/60 job done.

From 6.00 PM to 7.00 Pm a tech is added, so 8 Techs worked for 1 hour = 8/60 job Total job completed at 7.00PM is = 43/60 + 8/60 = 51/60 job done.

From 7.00 PM to 8.00 Pm a tech is added, so 9 Techs worked for 1 hour = 9/60 job Total job completed at 7.00PM is = 51/60 + 9/60 = 60/60 job done. Job is COMPLETE.

Job is complete at 8.00PM. The total number of hours is 9 hours. _________________

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

To complete the job 6*10=60 man/hours are needed. By 5:00 PM 6*6=36 man/hour done. 24 is left.

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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13 Jun 2013, 19:45

This is how I approach it: Because 6 men started at 11am and the addition of another man occurs after 5. Therefore, the amount of work done by 5pm will be = (1/10)X(17-11)= 6/10 The remaining work will thus be = 4/10

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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28 Jun 2013, 20:12

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 should never, everrr make questions too complicated to infer and solve... 6 tech total of 10 hours... let me assume total of 60 units of work... 1 tech in 1 hr does 1 unit of work.....

11:00 am to 5:pm - 6 hours so, total work done by 6 tech in 6 hrs will be 36 unit. ( 6*6) now a new person gets added every hour after 5:00PM

11:00-5:00 - 36 unit 5:00-6:00 - 7 unit ( 6 tech were doing 6 unit in 1 hr, now since a new person has been added, the work increases to 7 unit) 6:00-7:00 - 8 unit ( another guy added) 7:00-8:00- 9 unit ( another guys added making total 9 person in the work field)

36+7+8+9 = 60.

so by 8:00PM 60 units of work was completed by the group....

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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29 Jun 2013, 04:50

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

1. Originally 6 people work for 60 man hours 2. Additional people work for 6 man hours 3. So to complete the same task since the same man hours is taken , the original 6 people work for 6 man hours less. i.e., they work for 60-6=54 man hours i.e., they work for 9 hours each 4 . In other words if they start the task at 11 A.M , they complete the task at 8:00 PM

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

To complete the job 6*10=60 man/hours are needed. By 5:00 PM 6*6=36 man/hour done. 24 is left.

Makes me giggle how Bunuel always finds a simple yet understood solution! Though first I struggle to understand the problem and the previous posters' solutions! Thanks Bunuel!

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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17 Aug 2013, 04:38

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

6man, 10 hours = 1 full work 6 hours = 3/5 work done. due = 1- 3/5 = 2/5

6 men 10 hours so 7 men 60/7 hours to complete a job

so, 7 men 1 hour = 7/60 8 men 1 hour = 8/60 9 men 1 hour = 9/60 .......................................... + 3 hours = 24/60 = 2/5 parts of total works.

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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25 Sep 2013, 22:03

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Man * Days * hours = Work done

6 people * 10 hours = 60 units

6 people * (11AM to 5 PM) 6 hours = 36

Work left 60-36 = 24

5-6 : 7 people = 7 units 6-7 : 8 people = 8 units 7-8: 9 people = 9 units Total = 24 units

hence 8PM _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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20 Nov 2013, 10:23

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

Can someone explain how this fits into the standard w=r*t formula? I feel like that learning that formula was useless, because every time I try and apply it to one of these problems, it's wrong.

"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?"

I tried this: If the rate of a technician is \(\frac{1}{T}\), and you have 6 technicians working from 9-5, then their hourly rate should be 6/T, so using w=r*t \(w=8*\frac{6}{t}\), so W=\(\frac{48}{T}\). But then what do I do after that?

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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24 Dec 2013, 13:14

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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

So we need 60 hours men to finish the work 6 technitians will work for 6 hours hence they will do 36 hours. We now have 24 hours men left to complete At 6pm - 7 hours men = 24 - 7 = 17 At 7pm - 8 hours men= 17-8 = 9 At 8pm- 9 hours men = Job Done

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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18 Jan 2015, 00:56

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Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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14 Sep 2015, 07:56

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

My take is Option C. I followed a simple approach:

6 technicians will take total of 10 hours => Total manhours required = 60 man hours 1 Technician added per hour after 5:00 PM..Hmm okay..

Total manhours completed till 5 PM = 6*^ = 36 manhours (11 am - 5 pm => 6 hoours)

Manhours from 5 PM - 6 PM = (6 + 1) man * 1 hour = 7 manhours Manhours from 6 PM - 7 PM = (7 + 1) man * 1 hour = 8 manhours Manhours from 7 PM - 8 PM = (8 + 1) man * 1 hour = 9 manhours

Total manhours till 8 PM = 36 + 7 + 8 + 9 manhours = 60 manhours. Hence option C

Re: It takes 6 technicians a total of 10 hours to build and [#permalink]

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21 Nov 2015, 15:57

it takes (6)(10)=60 technician hours to complete job (6)(6)=36 technician hours completed from 11:00am to 5:00pm 60-36=24 technician hours remaining at 5:00pm t=time required ratio of cumulative technician hours to time=(t+13)/2 24/t=(t+13)/2 t=3 hours 8:00pm

It takes 6 technicians a total of 10 hours to build and [#permalink]

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04 Mar 2016, 19:06

my approach is the one that bunnuel gave for such types of questions.. total parts to build - 60 total parts 6 build in 1 hour = 6 parts.. so for 6 hours, they build 36 parts at 6 hour, 1 technician is added. in hour 7, 7 parts were built. so 36+7 = 43 in hour 8, another technician is added, so built 8 parts = 43+8=51. in hour 9, another technician is added, so built 9 parts = 51+9=60 - total number of parts.

so it takes 9 hours co complete if they start at 11AM, then they will finish at 20:00 or 8PM.

Hi, another method, though not easy and as simple as man-hours is--

in 6 hours, \(\frac{6}{10}\)work is completed... left is \(\frac{4}{10}\).. 1 hour work of 6 is \(\frac{1}{10}\)... and 1 hr work of 1 is\(\frac{1}{60}\).. so from 5 pm onwards an AP can be formed.. \(\frac{1}{10} + \frac{1}{60} + \frac{1}{10}+\frac{2}{60} +\)... let the hour required be x.. then\(\frac{1}{10}+\frac{1}{60} +\frac{1}{10} +\frac{2}{60}+.... \frac{1}{10} + \frac{x}{60}= 1-\frac{6}{10}\) \(\frac{x}{10} + \frac{1}{60} ( 1+2+..+x) = \frac{4}{10}\).. \(\frac{x}{10} + \frac{1}{60}*\frac{x(x+1)}{2}=\frac{4}{10}\).. \(12x + x^2 + x = 48\).. \(x^2 + 13x -48 =0\).. \(x^2 + 16x - 3x -48 =0\).. \((x+16)(x-3) =0\).. so x=3 .. time = \(5+3=8pm\) _________________

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