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Raghavachari, Viswanathan and Venkatraman together finished
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18 Mar 2013, 08:45
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59% (02:49) correct 41% (02:45) wrong based on 169 sessions
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Raghavachari, Viswanathan and Venkatraman together finished 4/5 th of a certain task . Had only Raghavachari and Viswanathan worked on the task, it would have taken 20 days to complete that part of the task. If Venkatraman is the fastest of the three workers, which of the following could be the total number of days taken to complete the whole task assuming all the three work together throughout? A. 16 B. 18 C. 20 D. 22 E. 24
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Re: Raghavachari, Viswanathan and Venkatraman together finished
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18 Mar 2013, 09:40
I found this exercise too complex to be a real GMAT problem. For simplicity , let's call them Rag, Vis & Ven. We know Rag and Vis working together spend 20 days for 4/5 of the job > this means 25 days for the entire job, Rag and Vis working together, at that pace. (1) And we have the formula used for work problems: the entire job time T, with the three guys working together, will be (2): \(\frac{1}{T}=\frac{1}{Rag}+\frac{1}{Vis}+\frac{1}{Ven}\) where: Rag is the time Rag needs to complete the work alone, Vis is the time Vis needs to complete the work alone, and Ven is the time Ven needs to complete the work alone. Now, because of (1), we know that (3): \(\frac{1}{25}=\frac{1}{Rag}+\frac{1}{Vis}\) Therefore, we can substitute (3) on (2), to get (4): \(\frac{1}{T}=\frac{1}{25}+\frac{1}{Ven}\) Now, it said that Ven is faster that the other guys with long names. This was the difficult part for me: if Rag and Vis can do the entire work in 25 days, reordering (3) this means that (5): \(25=\frac{Rag*Vis}{Rag+Vis}\) There are at least 5 options for this to be 25 (see attached graph): Rag=50 and Vis=50 Rag=30 and Vis=150 Rag=26 and Vis=650 Rag=150 and Vis=30 Rag=650 and Vis=26. But we have to analyze to choices they give us. We get (4) and solve for Ven depending on T (6): \(Ven=\frac{25*T}{25T}\) If we plug T=16 (option A), we get that Ven=44 days approx. If we plug T=18 (option B), we get that Ven=64 days approx. If we plug T=20 (option C), we get that Ven=100 days approx. and so on With the choices we are given, and knowing that Ven has to be the fast guy, we can only opt for the answer A, assuming that Rag and Vis take each 50 days to complete the task by their own. Answer A
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Re: Raghavachari, Viswanathan and Venkatraman together finished
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19 Mar 2013, 00:18
SravnaTestPrep wrote: Raghavachari, Viswanathan and Venkatraman together finished 4/5 th of a certain task . Had only Raghavachari and Viswanathan worked on the task, it would have taken 20 days to complete that part of the task. If Venkatraman is the fastest of the three workers, which of the following could be the total number of days taken to complete the whole task assuming all the three work together throughout?
A. 16 B. 18 C. 20 D. 22 E. 24 Solution: Let us call the three men as R, VI and VE 1) Time taken by R and VI to complete 4/5 of the task is 20 days 2) VE is the fastest of the three workers 3) From (1), Time that would be taken by R and VI to complete the whole task is 25 days 4) We may assume that R and VI take 50 days each to complete the task 5) From (2) and (4) we have VE taking less than 50 days or \(1/VE > 1/50\) 6) From (4) and (5), we have , \(1/Total time = 1/R + 1/VI + 1/VE > 1/50 + 1/50+ 1/50\) or\(1/ Total Time\) > 3/50 or Total time taken < 16.67 days. The only choice that fits the above value is choice A. Note: The number of days taken to complete the task cannot be greater than the above because VE being the fastest worker the maximum number of days that VE could possibly take is 49. Lesser values would lessen the total time taken.
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Re: Raghavachari, Viswanathan and Venkatraman together finished
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27 Sep 2013, 22:36
Not a typical GMAT question IMHO !



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Re: Raghavachari, Viswanathan and Venkatraman together finished
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27 Sep 2013, 23:07
Assume Task = Writing 50 books In 20 days, Ra and Vi finish writing 40 books Assuming they both work at equal speed (since we want to minimize Ve's rate), each of them writes at 1 book per day. If Ve worked at the same rate as Ra and Vi, the rate would be 3 books per day and the task would be completed in 50/3 = 16.6 days. Since Ve works the fastest, the work should be completed in less than 16.6 days. A is the only option available.
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Re: Raghavachari, Viswanathan and Venkatraman together finished
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21 May 2014, 08:48
Dude why complicate things and use those weird names. KISS (Keep it simple Sravna)
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Raghavachari, Viswanathan and Venkatraman together finished
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Updated on: 01 Mar 2016, 19:50
let 1/25=rate of R+Vi 1/x=rate of Ve d=number of days taken by all three to complete task d(1/25+1/x)=1 x=25d/(25d) plugging in 16 as d, x=400/9 1/x=9/400 1/25=16/400 1/25+1/x=25/400=1/16 16 days(1/16 combined daily rate)=1 completed task 16 works as possible number of days R, Vi, and V worked together to complete task
Originally posted by gracie on 24 Nov 2015, 08:04.
Last edited by gracie on 01 Mar 2016, 19:50, edited 1 time in total.



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Re: Raghavachari, Viswanathan and Venkatraman together finished
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27 Feb 2016, 10:45
I got confused of the wording.. does R and V need 20 days to complete 4/5 of the job or 1/5 of the job?



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Raghavachari, Viswanathan and Venkatraman together finished
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27 Feb 2016, 12:55
SravnaTestPrep wrote: Raghavachari, Viswanathan and Venkatraman together finished 4/5 th of a certain task . Had only Raghavachari and Viswanathan worked on the task, it would have taken 20 days to complete that part of the task. If Venkatraman is the fastest of the three workers, which of the following could be the total number of days taken to complete the whole task assuming all the three work together throughout?
A. 16 B. 18 C. 20 D. 22 E. 24 This one took me by surprise. Anyway, here goes: Prelims: \(R\): rate of Raghavachari \(V_i\): rate of Viswanathan \(V_e\): rate of Venkatraman Task = rate * time (days) \(T=r * t\) From the narrative: \((R+V_i)*20days=\frac{4}{5}T \implies (R+V_i)=\frac{1}{25} Task/day\) If either \(R\) or \(V_i\) does nothing, maximum \(V_e\) is \(\frac{1}{25}\) If \(R = V_i\) (i.e. half the rate of either one), minimum \(V_e\) is \(\frac{1}{50}\) In other words, \(\frac{1}{50}<V_e<\frac{1}{25}\) The time it takes to complete the task: \(t=\frac{1}{(R+V_i) + V_e}\) Considering \(V_e\) boundaries: \(\frac{1}{\frac{1}{25}+\frac{1}{25}} <t<\frac{1}{\frac{1}{25}+\frac{1}{50}}\) \(\implies 12.5days <t< 16.7days\) The only answer meeting this range is A.



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Re: Raghavachari, Viswanathan and Venkatraman together finished
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03 Mar 2016, 04:32
johnwesley wrote: \(\frac{1}{T}=\frac{1}{Rag}+\frac{1}{Vis}+\frac{1}{Ven}\)
where: Rag is the time Rag needs to complete the work alone, Vis is the time Vis needs to complete the work alone, and Ven is the time Ven needs to complete the work alone. I can't get my head around this one. As it received a Kudos from OP, it had to make sense. I would like to understand the successful solution, so I learn. How does one arrives at the reciprocal relationship above? Thanks for any help.



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Re: Raghavachari, Viswanathan and Venkatraman together finished
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03 Mar 2016, 06:16
SravnaTestPrep wrote: Raghavachari, Viswanathan and Venkatraman together finished 4/5 th of a certain task . Had only Raghavachari and Viswanathan worked on the task, it would have taken 20 days to complete that part of the task. If Venkatraman is the fastest of the three workers, which of the following could be the total number of days taken to complete the whole task assuming all the three work together throughout?
A. 16 B. 18 C. 20 D. 22 E. 24 Hi every one, there have been very long solutions for this Q.. But a very simple and totally valid solution would be
All three do 4/5 of work  this doesn't have any value apart from helping in the preceding task.. Two of them can do same amount of work in 20 days.. so 4/5th work is done in 20 days.. complete work is done by the two in 20*5/4=25 days..
Now the third is fastest, but lets take his speed the least possible.. for that the two others should be of same speed and the fastest third slightly more or nearly same.. so if two of them did the work in 25 days, each at same speed would do it in 50 days.. let the fastest do it in slightly less than 50 or say 50 itself.. so all three will do the work in 1/50 + 1/50 +1/50= 3/50 or 50/3 days = 16 2/3 days.. so the total time in no way can be more than 16 2/3.. only A is left
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Re: Raghavachari, Viswanathan and Venkatraman together finished
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21 Feb 2018, 12:47
A  raghavachari, B  viswanathan, C  venkatraman Given: A + B can finish 4/5 task in 20 days => so A + B can finish the whole task in 25 days if A work at same speed as B, then A alone can finish the task in 50 days.
worst case, if C were to work at speed of A and B, then total time taken to finish by three altogether = 50/3 = 16.67 days but given C is definitely better than A and B, then total time taken to finish must be less than 16.67, only option (A) stands.




Re: Raghavachari, Viswanathan and Venkatraman together finished &nbs
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21 Feb 2018, 12:47






