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One computer can upload 100 megabytes worth of data in 6 [#permalink]

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20 Mar 2012, 11:17

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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.

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 1300-700=600 megabytes worth of data. The second computer can upload 600/6=100 megabytes of data in 42/6=7 seconds.

One computer can upload 100 megabytes worth of data in 6 [#permalink]

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04 Mar 2013, 06:10

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?

Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]

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04 Mar 2013, 06:28

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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 (1300-700 MB) of data in 42 seconds => The other computer uploads 100 MB of data in 42/6 = 7 seconds

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

Merging similar topics. Please refer to the solutions above.
_________________

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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15 Oct 2014, 06:07

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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]

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23 Jun 2016, 03:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]

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22 Jul 2016, 09:33

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 1300/42-100/6= 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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Re: One computer can upload 100 megabytes worth of data in 6
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22 Jul 2016, 09:33

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