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Victor's job requires him to complete a series of identical [#permalink]
 11 Sep 2011, 14:17
Question Stats:
48% (04:48) correct
51% (03:07) wrong based on 5 sessions
Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision? A. 34 B. 52 C. 60 D. 70 E. 92
Last edited by Bunuel on 24 Mar 2013, 02:55, edited 2 times in total.
Renamed the topic and edited the question.
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supervised not supervised
time per job t-3 t
In 144 days when victor was supervised for half of the time , he finished 36 jobs.
=> During 72 days of supervision he finished 36 jobs.
=> 36(t-3) = 72
=> t = 5
=> victor took 5 days to finish a each unsupervised job.
time taken by victor to finish 10 jobs with out supervision = 10t = 50
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Senior Manager
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Spidy001 wrote: supervised not supervised
time per job t-3 t
In 144 days when victor was supervised for half of the time , he finished 36 jobs.
=> During 72 days of supervision he finished 36 jobs.
=> 36(t-3) = 72
=> t = 5
=> victor took 5 days to finish a each unsupervised job.
time taken by victor to finish 10 jobs with out supervision = 10t = 50 OA is 60 but.... there is some mistake with your solution
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my bad. ignore the previous post. i thought the question said 36 jobs in 72 days of supervision.
Where as Questions says it took 36 jobs in 144 days out of which he worked halftime supervised.
Supervised not supervised
time per job t-3 days t days
=> jobs in 72 days of supervision + jobs in 72 days when not supervised = 36
=> 72/(t-3) + 72/t = 36
solving we get t = 1 or 6
=> t=6 ( t = 1 is not possible as t-3 would be negative)
=> time taken by victor each unsupervised job = 6 days.
=> time taken by victor to finish 10 unsupervised jobs = 10t = 60 days.
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60 days it is 72/ (x-3) + 72/x = 36 => x^2 -7x+6 =0 => x= 1 or 6 X = 1 is omitted coz x-3 = -2 if x=1 so x=6 and time taken for 10 jobs under no supervision = 10x6 = 60
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Ans 60. Found in the above method.
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Guys, I've made th same error as Spidy001. This advanced RTW problem is otherwise not that hard, but I wonder what strategies you apply in order to fully grasp the right meaning of the problem. At first, I also thought the 5 days per job scenario was right. Based on this problem, do you have any tips on that? Thanks in advance!
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bgpower wrote: Guys,
I've made th same error as Spidy001. This advanced RTW problem is otherwise not that hard, but I wonder what strategies you apply in order to fully grasp the right meaning of the problem. At first, I also thought the 5 days per job scenario was right. Based on this problem, do you have any tips on that?
Thanks in advance! bgpower, I can imagine why Spidy001 made that mistake. The question says that he was supervised for half that time, but it doesn't say that Victor did not work during the time he was unsupervised. So, what the question intended to say was, Victor worked under supervision for 72 days and without supervision for the remaining 72. What Spidy001 has done is, he hasn't taken into account the 72 days that Victor worked without supervision. Secondly, as data given to us was in the form of (days/job) and not (jobs/day), you should divide 72 by (t-3) and (t) rather than multiplying it. This sort of an error can be avoided by simply taking a step back to understand what the question says and writing the facts down, rather than trying to hold them in your head.
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Re: Victor's job requires him to complete a series of identical [#permalink]
23 Mar 2013, 23:00
siddhans wrote: Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision? I have a query R T DS (1/t) 72 (72/t)U (1/(t+3) 72 (72/(t+3))Total jobs = 36 (72/t) + [(72/(t+3)] = 36 => 2/t + 2/(t+3) = 1 => 2t + 6 + 2t = t^2 + 3t => t^2 - t - 6 = 0 t = 3 or t =-2 (can't be negative) Hence , 3 So answer is 30 Plz advice where I am wrong as I took the other way around by taking (t+3) Rgds, TGC
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Re: Victor's job requires him to complete a series of identical [#permalink]
24 Mar 2013, 03:07
targetgmatchotu wrote: siddhans wrote: Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision? I have a query R T DS (1/t) 72 (72/t)U (1/(t+3) 72 (72/(t+3))Total jobs = 36 (72/t) + [(72/(t+3)] = 36 => 2/t + 2/(t+3) = 1 => 2t + 6 + 2t = t^2 + 3t => t^2 - t - 6 = 0 t = 3 or t =-2 (can't be negative) Hence , 3 So answer is 30 Plz advice where I am wrong as I took the other way around by taking (t+3) Rgds, TGC We are told that if Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. Supervised = t days; Unsupervised = (t + 3) days. Victor to complete 10 jobs without any supervision will need 10(t + 3) = 60. Hope it helps.
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Re: Victor's job requires him to complete a series of identical [#permalink]
24 Mar 2013, 03:25
Bunuel wrote: targetgmatchotu wrote: siddhans wrote: Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision? I have a query R T DS (1/t) 72 (72/t)U (1/(t+3) 72 (72/(t+3))Total jobs = 36 (72/t) + [(72/(t+3)] = 36 => 2/t + 2/(t+3) = 1 => 2t + 6 + 2t = t^2 + 3t => t^2 - t - 6 = 0 t = 3 or t =-2 (can't be negative) Hence , 3 So answer is 30 Plz advice where I am wrong as I took the other way around by taking (t+3) Rgds, TGC We are told that if Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. Supervised = t days; Unsupervised = (t + 3) days. Victor to complete 10 jobs without any supervision will need 10(t + 3) = 60. Hope it helps. Hi Bunuel As I would follow the same method of RTW chart for any problem in TNW, it would be really helpful if you can QUOTE the error in my method . Rgds, TGC
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Re: Victor's job requires him to complete a series of identical [#permalink]
24 Mar 2013, 03:30
targetgmatchotu wrote: Bunuel wrote: targetgmatchotu wrote: I have a query
R T D
S (1/t) 72 (72/t)
U (1/(t+3) 72 (72/(t+3))
Total jobs = 36
(72/t) + [(72/(t+3)] = 36
=> 2/t + 2/(t+3) = 1
=> 2t + 6 + 2t = t^2 + 3t
=> t^2 - t - 6 = 0
t = 3 or t =-2 (can't be negative)
Hence , 3
So answer is 30
Plz advice where I am wrong as I took the other way around by taking (t+3)
Rgds,
TGC We are told that if Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. Supervised = t days; Unsupervised = (t + 3) days. Victor to complete 10 jobs without any supervision will need 10(t + 3) = 60. Hope it helps. Hi Bunuel As I would follow the same method of RTW chart for any problem in TNW, it would be really helpful if you can QUOTE the error in my method . Rgds, TGC Actually I already did. You calculated the time needed to complete 10 jobs WITH supervision (10t, less time) but we are asked to find the time needed to complete 10 jobs WITHOUT any supervision (10(t+3), more time). Hope it's clear.
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
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Re: Victor's job requires him to complete a series of identical
[#permalink]
24 Mar 2013, 03:30
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