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One computer can upload 100 megabytes worth of data in 6 [#permalink]
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20 Mar 2012, 12:17
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84% (01:26) correct 16% (03:11) wrong based on 360 sessions
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One computer can upload 100 megabytes worth of data in 6 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 42 seconds. How long would it take for the second computer, working on its own, to upload 100 megabytes of data? (A) 6 (B) 7 (C) 9 (D) 11 (E) 13 This is how I am trying to solve this question: Let say A be the computer 1 and B be the computer 2 Computer A 1 second work : 100/6 Computer A + Computer B 1 second work : 1300/42 1/A + 1/B = AB/A+B We have to find 1/B. I struggle to complete this question.
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Re: Two computers [#permalink]
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20 Mar 2012, 12:22
Quote: This is how I am trying to solve this question:
Let say A be the computer 1 and B be the computer 2
Computer A 1 second work : 100/6 Computer A + Computer B 1 second work : 1300/42
1/A + 1/B = AB/A+B
We have to find 1/B. I struggle to complete this question. You are on the right track.. So Computer B's rate = 1300/42100/6 = 600/42 So computer B can upload 600MB in 42 seconds.. divide numerator and denominator by 6 and you get 100/7 so 2nd computer takes 7 seconds to upload 100MB



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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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20 Mar 2012, 14:53
enigma123 wrote: One computer can upload 100 megabytes worth of data in 6 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 42 seconds. How long would it take for the second computer, working on its own, to upload 100 megabytes of data?
(A) 6 (B) 7 (C) 9 (D) 11 (E) 13
This is how I am trying to solve this question:
Let say A be the computer 1 and B be the computer 2
Computer A 1 second work : 100/6 Computer A + Computer B 1 second work : 1300/42
1/A + 1/B = AB/A+B
We have to find 1/B. I struggle to complete this question. Since the first computer can upload 100 megabytes worth of data in 6 seconds then in 6*7=42 seconds it can upload 7*100=700 megabytes worth of data, hence the second compute in 42 seconds uploads 1300700=600 megabytes worth of data. The second computer can upload 600/6=100 megabytes of data in 42/6=7 seconds. Answer: B.
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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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07 Apr 2012, 05:56
enigma123 wrote: This is how I am trying to solve this question:
Let say A be the computer 1 and B be the computer 2
Computer A 1 second work : 100/6 Computer A + Computer B 1 second work : 1300/42
1/A + 1/B = AB/A+B
We have to find 1/B. I struggle to complete this question.
A needs 6sec for 100 ; A+B need 42 sec for 1300 or 100*42/1300=42/13 sec for 100 so now we have rate of A for 100 is 1/6 rate of B for 100 is 1/x rate of (a+b) for 100 is 13/42 1/6+1/x=13/42 x=7 sec hope it helps
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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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13 Apr 2012, 03:52
RT=W (100/6+X)*42=1300 > 42X=600 > X=600/42 rate of the second machine 600/42*t=100 > t=7 sec. needs the second machine to upload data



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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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15 Nov 2012, 23:29
Setup the rate equations. This always help. \(ratefirst=\frac{1}{A}=\frac{100}{6sec}\) \(ratetogether=\frac{1}{A}+\frac{1}{B}=\frac{1300}{42}\) ratetogether  ratefirst = ratesecond\(\frac{1}{A}+\frac{1}{B}\frac{1}{A}=\frac{1300}{42}\frac{100}{6}=\frac{600}{42}=\frac{100}{7secs}\) Answer: B
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One computer can upload 100 megabytes worth of data in 6 [#permalink]
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04 Mar 2013, 07:10
One computer can upload 100 megabytes worth of data in 6 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 42 seconds. How long would it take for the second computer, working on its own, to upload 100 megabytes of data?(A) 6 (B) 7 (C) 9 (D) 11 (E) 13 Source: Gmat Hacks 1800
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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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04 Mar 2013, 07:25



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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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04 Mar 2013, 07:28
The given computer can upload 100 MB of data in 6 seconds => In 42 seconds, it will upload 100 MB x 7 = 700 MB of data In 42 seconds, the two computers working together upload 1300 MB of data => The other computer uploads 600 MB (1300700 MB) of data in 42 seconds => The other computer uploads 100 MB of data in 42/6 = 7 seconds Option B
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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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04 Mar 2013, 11:57



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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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27 Sep 2013, 23:41
Shouldn't be a 700 level problem. Rate (1 computer) = 100/6 Rate ( 2 computers) = 1300/42 Rate (Computer in question) = 1300/42  100/6 = 100/7 Rate * Time = Work 100/7 * Time =100 Time =7
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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]
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22 Jul 2016, 10:33
enigma123 wrote: One computer can upload 100 megabytes worth of data in 6 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 42 seconds. How long would it take for the second computer, working on its own, to upload 100 megabytes of data?
(A) 6 (B) 7 (C) 9 (D) 11 (E) 13
This is how I am trying to solve this question:
Let say A be the computer 1 and B be the computer 2
Computer A 1 second work : 100/6 Computer A + Computer B 1 second work : 1300/42
1/A + 1/B = AB/A+B
We have to find 1/B. I struggle to complete this question. Well, you have to understand the fundamental difference between time and rate. TIME is not an additive quantity but RATE is. Meaning you can add rates directly to get a valid new rate. (This never happens in time; remember those tricky average speed problems ) So here first computer's rate (R1) is 100/6 eq1 and rate of both computer (R1+R2)= 1300/42 eq2 so if you subtract eq 2 from eq 1 you will get the rate of second computer alone 1300/42100/6= 700/42 R2 is 700/42 R2*T= work R2*T=100 700/42*T=100 T=6 ANSWER IS A
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One computer can upload 100 megabytes worth of data in 6 [#permalink]
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18 Aug 2017, 17:24
enigma123 wrote: One computer can upload 100 megabytes worth of data in 6 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 42 seconds. How long would it take for the second computer, working on its own, to upload 100 megabytes of data?
(A) 6 (B) 7 (C) 9 (D) 11 (E) 13
This problem's solution is very quick if you realize that 6 seconds is a multiple of 42 seconds. Just multiply M1 rate's numerator and denominator by 7. Machine 1, M1 = \(\frac{100Mb}{6secs}\) M1 + M2 = \(\frac{1300Mb}{42secs}\) M1 = \(\frac{100Mb}{6secs}\) * \(\frac{7}{7}\)=\(\frac{700Mb}{42secs}\) So M2 can do \(\frac{(1300  700)}{42}\) = \(\frac{600Mb}{42secs}\) \(\frac{600}{42}\) = \(\frac{100Mb}{x}\) 4,200 = 600x x = 7 Answer B
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One computer can upload 100 megabytes worth of data in 6
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