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While working alone at their respective constant rates, [#permalink]
09 Feb 2013, 13:37

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This post received KUDOS

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Difficulty:

25% (medium)

Question Stats:

83% (02:28) correct
17% (01:32) wrong based on 71 sessions

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

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

Re: While working alone at their respective constant rates, [#permalink]
09 Feb 2013, 16:18

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: \(\frac{(4)(8)}{4+8}\) = \(2\frac{2}{3}\) _________________

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

Re: While working alone at their respective constant rates, [#permalink]
10 Feb 2013, 01:10

Expert's post

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?

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

gmatclubot

Re: While working alone at their respective constant rates,
[#permalink]
11 Feb 2013, 23:27

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