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It takes 6 technicians a total of 10 hours to build and

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

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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

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Originally posted by Hussain15 on 12 Apr 2010, 05:41.
Last edited by Bunuel on 09 Feb 2012, 12:38, edited 1 time in total.
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Re: Work Rate Problem  [#permalink]

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New post 12 Apr 2010, 07:22
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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


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

5:00 PM - 6:00 PM 7 man/hour;
6:00 PM - 7:00 PM 8 man/hour;
7:00 PM - 8:00 PM 9 man/hour.

7+8+9=24.

Answer: C.
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Re: Work Rate Problem  [#permalink]

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New post 12 Apr 2010, 07:20
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Ans: C

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

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New post 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

(1/10)t+(1/60)[t+(t+1)+...+(t+n)]=(4/10)
(1/10)t+(1/60)[Summation formula: ((tX(t+1))/2)]=4/10

Because t cannot be negative; therefore t=3
3+5=8

Could someone please verify that this method will work for any rate question presented. Thanks.
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Re: It takes 6 technicians a total of 10 hours to build and  [#permalink]

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New post 28 Jun 2013, 12:58
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6 person can complete the job in 10 hours.
Rate of work per person per hour = x
W=R*T
1=6x*10
x=1/60

Work done by 6 persons for 6 hours at rate of (1/60)=
6*6*(1/60) = 6/10

Remaining Work = 1-6/10 = 4/10

4/10 = Y(1/60)

Y=24; at 5 pm no of persons were 7, at 6pm: 8, at 7pm :9persons, 7+8+9=24

Answer C

it took 3 hours after 5pm, to complete the task.
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Re: It takes 6 technicians a total of 10 hours to build and  [#permalink]

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New post 28 Jun 2013, 20:12
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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



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....

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

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New post Updated on: 17 Aug 2013, 04:23
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


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

The answer is choice C.
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Originally posted by SravnaTestPrep on 29 Jun 2013, 04:50.
Last edited by SravnaTestPrep on 17 Aug 2013, 04:23, edited 2 times in total.
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Re: Work Rate Problem  [#permalink]

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New post 17 Aug 2013, 03:52
Bunuel wrote:
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


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

5:00 PM - 6:00 PM 7 man/hour;
6:00 PM - 7:00 PM 8 man/hour;
7:00 PM - 8:00 PM 9 man/hour.

7+8+9=24.

Answer: C.


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

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New post 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.

so 5+3 = 8pm (Answer)
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Re: It takes 6 technicians a total of 10 hours to build and  [#permalink]

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New post 25 Sep 2013, 22:03
2
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
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Re: It takes 6 technicians a total of 10 hours to build and  [#permalink]

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

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New post 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

Hence answer is (C)

Hope it helps
Cheers!
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Re: It takes 6 technicians a total of 10 hours to build and  [#permalink]

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New post 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

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Chanakya

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

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New post Updated on: 08 Jul 2016, 14:01
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
ratio of technician hours (t) needed to actual hours (h) needed→
t/h=(h+13)/2
24/h=(h+13)/2
h=3
8:00pm

Originally posted by gracie on 21 Nov 2015, 15:57.
Last edited by gracie on 08 Jul 2016, 14:01, edited 1 time in total.
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It takes 6 technicians a total of 10 hours to build and  [#permalink]

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

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New post 19 Apr 2016, 00:19
total time needed o finish= 6 peeps * 10 hours = 60 hours

6 workers working together ---> starting from 11am to 5pm i.e., 6 hours = 6*6 = 36 hours work done

remaining = 24 hours work
7 hours work in 1st hour
8 in 2nd hr
9 in 3rd hr
-------------
24 in 3hrs --> 5pm + 3hrs --> 8pm
-------------
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Re: It takes 6 technicians a total of 10 hours to build and  [#permalink]

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New post 20 Apr 2016, 00:26
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on work/rate problems: work-word-problems-made-easy-87357.html
All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66


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

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New post 08 Jul 2016, 09:35
So I generally solve a Man-work question using the following method however I having an issue following that approach. Can somebody egmat help?

11 am to 5 pm is 6 hours

Man--Time--Work

6--10h--w
1--10h--w/6
1--1h--w/60
1--6h--(w/60)*6= w/10
6--6h--(w/10)*6= (3w/5)

So now 2/5th of work is remaining. As per the approach explained by the egmat course when the total work is finite ( as in this case) we multiply and divide the work( by number of men and time) as required by the question and then finally equate that to total work "W". I am unable to apply that approach here as with each hour the number of men is increasing.

After a lot of struggle I have finally been able to find a method that works for me which is why I want to follow this one and change my approach again.

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Re: It takes 6 technicians a total of 10 hours to build and &nbs [#permalink] 08 Jul 2016, 09:35

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