One computer can upload 100 megabytes worth of data in 6

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

20 Mar 2012, 12:17

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Question Stats:

89% (02:24) correct

10% (03:19) wrong

based on 10 sessions

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.

[Reveal] Spoiler: OA

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Re: Two computers [#permalink]

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 = 1300/42-100/6 = 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]

20 Mar 2012, 14:53

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enigma123 wrote:

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.

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

07 Apr 2012, 05:56

enigma123 wrote:

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]

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]

15 Nov 2012, 23:29

Setup the rate equations. This always help.

rate-first=\frac{1}{A}=\frac{100}{6sec}
rate-together=\frac{1}{A}+\frac{1}{B}=\frac{1300}{42}

rate-together - rate-first = rate-second

\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]

04 Mar 2013, 07:10

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

04 Mar 2013, 07:25

http://www.youtube.com/watch?v=bf9ROHLJNSA
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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink]

04 Mar 2013, 07:28

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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

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

04 Mar 2013, 11:57

megafan wrote:

Merging similar topics. Please refer to the solutions above.
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Re: One computer can upload 100 megabytes worth of data in 6 [#permalink] 04 Mar 2013, 11:57

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One computer can upload 100 megabytes worth of data in 6

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One computer can upload 100 megabytes worth of data in 6

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