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While working alone at their respective constant rates, [#permalink]
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09 Feb 2013, 13:37
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Question Stats:
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While working alone at their respective constant rates, server X uploads 480 files in 4 hours and server Y uploads 480 files in 8 hours. If all files uploaded by these servers are the same size, how long would it take the two servers, working at the same time and at their respective constant rates, to process a total of 480 files? (A) 2 hr (B) 2hr 20 min (C) 2hr 40 min (D) 5 hr 40 min (E) 6hr Source: Gmat Hacks 1800
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Re: While working alone at their respective constant rates, [#permalink]
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09 Feb 2013, 15:54
megafan wrote: While working alone at their respective constant rates, server X uploads 480 files in 4 hours and server Y uploads 480 files in 8 hours. If all files uploaded by these servers are the same size, how long would it take the two servers, working at the same time and at their respective constant rates, to process a total of 480 files?
(A) 2 hr (B) 2hr 20 min (C) 2hr 40 min (D) 5 hr 40 min (E) 6hr
Source: Gmat Hacks 1800 Server X processes 480/4 files per hour = 120 per hour Server Y processes 180/8 files per hour = 60 per hour Total files processed per hour when X and Y work together = 120+60 per hour = 180 files per hour 480/180 = 2 2/3 hours



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Re: While working alone at their respective constant rates, [#permalink]
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09 Feb 2013, 16:18
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hitman5532 wrote: Server X processes 480/4 files per hour = 120 per hour Server Y processes 180/8 files per hour = 60 per hour Total files processed per hour when X and Y work together = 120+60 per hour = 180 files per hour
480/180 = 2 2/3 hours
While this is good, there is a faster way to do this—If you realize that the # of files is the same. Then you just use the combined rate formula: \(\frac{(4)(8)}{4+8}\) = \(2\frac{2}{3}\)
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Re: While working alone at their respective constant rates, [#permalink]
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09 Feb 2013, 16:50
megafan wrote: hitman5532 wrote: Server X processes 480/4 files per hour = 120 per hour Server Y processes 180/8 files per hour = 60 per hour Total files processed per hour when X and Y work together = 120+60 per hour = 180 files per hour
480/180 = 2 2/3 hours
While this is good, there is a faster way to do this—If you realize that the # of files is the same. Then you just use the combined rate formula: \(\frac{(4)(8)}{4+8}\) = \(2\frac{2}{3}\) If the numbers were more odd/complex the formula always helps me, I did the other way in my head in about 8 seconds because the numbers are simple.



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Re: While working alone at their respective constant rates, [#permalink]
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10 Feb 2013, 01:10
megafan wrote: While working alone at their respective constant rates, server X uploads 480 files in 4 hours and server Y uploads 480 files in 8 hours. If all files uploaded by these servers are the same size, how long would it take the two servers, working at the same time and at their respective constant rates, to process a total of 480 files?
(A) 2 hr (B) 2hr 20 min (C) 2hr 40 min (D) 5 hr 40 min (E) 6hr
Source: Gmat Hacks 1800 Original question is from OG: Quote: Machine A produces bolts at a uniform rate of 120 every 40 seconds, and Machine B produces bolts at a uniform rate of 100 every 20 seconds. If the two machines run simultaneously, how many seconds will it take for them to produce a total of 200 bolts?
(A) 22 (B) 25 (C) 28 (D) 32 (E) 56 Discussed here: machineaproducesboltsatauniformrateof120every143745.html
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Re: While working alone at their respective constant rates, [#permalink]
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11 Feb 2013, 23:27
Server X : Work done in 1 hour is =1/4 Server Y : Work done in 1 hour is = 1/8 Work done by X and Y togather in 1 hour = 1/4+1/8= 6/16 Hence time required to complete upload= 16/6 hours = 2 hour 40 min. Hence option C



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Re: While working alone at their respective constant rates, [#permalink]
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29 Nov 2017, 06:39
megafan wrote: hitman5532 wrote: Server X processes 480/4 files per hour = 120 per hour Server Y processes 180/8 files per hour = 60 per hour Total files processed per hour when X and Y work together = 120+60 per hour = 180 files per hour
480/180 = 2 2/3 hours
While this is good, there is a faster way to do this—If you realize that the # of files is the same. Then you just use the combined rate formula: \(\frac{(4)(8)}{4+8}\) = \(2\frac{2}{3}\) megafanplease, what's the original formula?



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Re: While working alone at their respective constant rates, [#permalink]
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01 Dec 2017, 06:44
megafan wrote: While working alone at their respective constant rates, server X uploads 480 files in 4 hours and server Y uploads 480 files in 8 hours. If all files uploaded by these servers are the same size, how long would it take the two servers, working at the same time and at their respective constant rates, to process a total of 480 files?
(A) 2 hr (B) 2hr 20 min (C) 2hr 40 min (D) 5 hr 40 min (E) 6hr The rate of X is 480/4 = 120 and the rate of Y is 480/8 = 60. Since the combined rate is 180, it will take 480/180 = 48/18 = 8/3 = 2 2/3 hours = 2 hours and 40 minutes. Answer: C
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