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While working alone at their respective constant rates,

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While working alone at their respective constant rates,  [#permalink]

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New post 09 Feb 2013, 14:37
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Question Stats:

79% (01:44) correct 21% (02:00) wrong based on 177 sessions

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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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New post 09 Feb 2013, 16:54
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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New post 09 Feb 2013, 17:18
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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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New post 09 Feb 2013, 17:50
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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New post 10 Feb 2013, 02:10
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: machine-a-produces-bolts-at-a-uniform-rate-of-120-every-143745.html
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Re: While working alone at their respective constant rates,  [#permalink]

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New post 12 Feb 2013, 00:27
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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New post 29 Nov 2017, 07:39
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:
\(\frac{(4)(8)}{4+8}\) = \(2\frac{2}{3}\)


megafan

please, what's the original formula?
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Re: While working alone at their respective constant rates,  [#permalink]

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New post 01 Dec 2017, 07:44
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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Re: While working alone at their respective constant rates,  [#permalink]

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New post 08 Aug 2019, 02:07
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

2x = y

3x are working, so 480:3=160min = 2h 40min
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Re: While working alone at their respective constant rates,   [#permalink] 08 Aug 2019, 02:07
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