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Andy, George and Sally are a team of consultants working on Project Alpha. They have an eight hour deadline to complete the project. The team members work at constant rates throughout the eight hour period. If the team of three has to begin work now and no one else can work on this project, will Project Alpha be completed by the deadline? (1) Sally can finish the project alone in \(4k+7\) hours, where \(k\) is a positive integer with a minimum value of 1 and a maximum value of 5. (2) Working alone, George will take \(2k+1\) hours and Andy will take \(3+2k\) hours, where \(k\) is a positive integer with a minimum value of 1 and a maximum value of 5
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Re M0207 [#permalink]
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16 Sep 2014, 00:17
Official Solution: Statement (1) by itself is not sufficient. From S1, we know Sally can finish the project in \(4k+7\) hours, but we know nothing about George's or Andy's time. Check if Sally can finish the work by herself. If \(k\) is the smallest value (1), Sally will need 11 hours to finish the project. We need to know George's and/or Andy's rate. Statement (2) by itself is sufficient. From S2, we know that George will take \(2k+1\) hours and Andy will take \(2k+3\) hours. Check whether George and Andy can finish the project by themselves. Check to see if the maximum value of \(k\) is sufficient for these two together to finish the project. If the maximum value of \(k = 5\), then George needs 11 hours and Andy needs 13 hours to finish the project alone. Working together for one hour, George and Andy will finish: \(\frac{1}{11} + \frac{1}{13} = \frac{11+13}{11*13} = \frac{24}{143} \gt \frac{1}{8}\) of the work. Thus, working together, George and Andy finish a greater portion of work per hour than the required \(\frac{1}{8}\). As a result, they will easily finish the project by themselves. It is not necessary to know Sally's time to answer the question. Alternative explanation: To complete the job on time, George and Andy have to complete \(\frac{1}{8}\) of the job each hour together or \(\frac{1}{16}\) each (50% of \(\frac{1}{8}\)). Because they complete \(\frac{1}{11}\) and \(\frac{1}{13}\) respectively, that's all we need to know. Another method is to perform a rough calculation: \(\frac{1}{11}+\frac{1}{13} \approx \frac{1}{12}+\frac{1}{12}=\frac{2}{12}=\frac{1}{6} \gt \frac{1}{8}\). Answer: B
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Re: M0207 [#permalink]
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13 Dec 2014, 00:05
Sorry, just 1 question. Lets say S2 gave a value more than 8 hours. Then it would be insufficient, but S1+S2 would be sufficient, right?



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Re: M0207 [#permalink]
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23 Mar 2015, 05:46
Deeuk, Firstly consider the below cases: 1. say, k =5 (Maximum) Then also, on further simplification, 1/11 + 1/13 > 1/8 2. say, k =1 (Minimum) Then also, on further simplification, 1/3 + 1/5 > 1/8 Both cases, its sufficient to decipher that both person sufficient to do alpha within the schedule of 8 hours. Hence B is the answer to the said question. Yes, coming to your question, yes if we could find different solution to both cases, then we would have to consider the first option, then in that case, C would have been the answer. Hope its clear.
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Re: M0207 [#permalink]
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08 Feb 2016, 20:01
A really easy way to look at this is as follows:
for section (2) put 1 into the equation ( the minimum work value) for George and andy and you get G = 3 and A = 5. This add up to 8 so together they can complete it in 8 hours which is already enough to meet the deadline. therefore it doesn't even matter what Sallys rate is. Not sure if this is the correct rational but it worked for me.



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mpcroaming wrote: A really easy way to look at this is as follows:
for section (2) put 1 into the equation ( the minimum work value) for George and andy and you get G = 3 and A = 5. This add up to 8 so together they can complete it in 8 hours which is already enough to meet the deadline. therefore it doesn't even matter what Sallys rate is. Not sure if this is the correct rational but it worked for me. hi, your method is wrong on two counts.. George will take 2k+1 hours and Andy will take 3+2k hours, 1) firstly by substituting k as 1, you are taking the minimum time they will take. so G alone can finish in 3 hrs, and he doesn't require anyone to finish before 8 hrs.... SO, you look at the max time they take to be certain that even at the lowest speed they can finish the work in <8 hrs.. 2) second fault is if G can do it in 3 hrs and A can do in 5 hrs... both combined will not do it in 3+5 hrs, but less than 3 hrs.. if some one alone was doing in 3 hrs, and he gets help, he will do it in lesser time.. Hope the concepts have become slightly clearer..
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Re: M0207 [#permalink]
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03 May 2016, 06:54
chetan2u wrote: mpcroaming wrote: A really easy way to look at this is as follows:
for section (2) put 1 into the equation ( the minimum work value) for George and andy and you get G = 3 and A = 5. This add up to 8 so together they can complete it in 8 hours which is already enough to meet the deadline. therefore it doesn't even matter what Sallys rate is. Not sure if this is the correct rational but it worked for me. hi, your method is wrong on two counts.. George will take 2k+1 hours and Andy will take 3+2k hours, 1) firstly by substituting k as 1, you are taking the minimum time they will take. so G alone can finish in 3 hrs, and he doesn't require anyone to finish before 8 hrs.... SO, you look at the max time they take to be certain that even at the lowest speed they can finish the work in <8 hrs.. 2) second fault is if G can do it in 3 hrs and A can do in 5 hrs... both combined will not do it in 3+5 hrs, but less than 3 hrs.. if some one alone was doing in 3 hrs, and he gets help, he will do it in lesser time.. Hope the concepts have become slightly clearer.. Thank you very much! Your explanation really helped me out.



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For 2), I did it slightly differently although it is basically a same solution as the explanation.
Set k = 5 (maximum)
time = 8 hrs
So the equation to verify would be
8/11 + 8/13 > 1
Which it clearly is. Therefore 2) is sufficient. I think this way is much easier to tell at first glance without doing any further calculation rather than comparing fractions.



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Re: M0207 [#permalink]
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10 Oct 2016, 05:58
hi, why can't it be done in this way George will take 2k+1 hours and Andy will take 3+2k hours taking minimum value (1) 2(1) +1 & 3+2(1)= 3+8
wprk done in one hour = 1/3 +1/5 5/15+3/15=8/15 in one hour by this we can easily state that that work can be done in less than 8 hours.



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Re: M0207 [#permalink]
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17 Aug 2017, 03:02
Very nice question. All that has been asked is can the worked be finished on time. And if even one can do it, then we won't require rest.
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