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Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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27 Jul 2012, 18:11
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Bill downloads the movie "Revenge of the Avengers" to his computer in 2.5 hours, using a download manager that downloads from 3 sources marked A, B and C. Each source provides download at a constant rate, but the rates of different sources are not necessarily identical. If the movie was downloaded from sources A and C alone, it would take 4 hours to complete the download. The next day, source B is available, but the other sources are inactive. How long will it take to download the trailer of the movie, a file that is 40 times smaller, from source B alone? A. 6 hours and 40 minutes B. 15 minutes C. 12 minutes D. 10 minutes E. 3 minutes
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Re: Bill downloads the movie "Revenge of the Avengers" to his co [#permalink]
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27 Jul 2012, 21:22
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Let's consider the movie's size is 100Mb
It takes 4 hours (240 min) to download the movie using only A and C. It takes 2.5 hours (150 min) to download the movie using A, B and C.
Consider the download rate to be: Rate = Data/Time And Va, Vb and Vc are the download rates. So, we have:
(Va+Vb+Vc) = 100/150 = 2/3 Mb/min
Also:
(Va+Vc) = 100/240 = 5/12 Mb/min
Consequently:
(Va+Vc) + Vb = 2/3 Vb=2/3  5/12 = 1/4 Mb/min
The question states that the trailer is 40 times smaller than the movie. So, 100Mb / 40 = 2.5 Mb
Again:
Vb = Data / Time 1/4 = 2.5 / Time Time = 2.5 / 0.25 = 10 minutes
Hope it helps!



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Re: Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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28 Jul 2012, 01:38
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superpus07 wrote: Bill downloads the movie "Revenge of the Avengers" to his computer in 2.5 hours, using a download manager that downloads from 3 sources marked A, B and C. Each source provides download at a constant rate, but the rates of different sources are not necessarily identical. If the movie was downloaded from sources A and C alone, it would take 4 hours to complete the download. The next day, source B is available, but the other sources are inactive. How long will it take to download the trailer of the movie, a file that is 40 times smaller, from source B alone?
A. 6 hours and 40 minutes B. 15 minutes C. 12 minutes D. 10 minutes E. 3 minutes Say the time needed to download the movie from source A alone be \(a\) minutes, from source B alone be \(b\) minutes, and from source C alone be \(c\) minutes. Next, since to download the movie from all three sources 2.5 hours (150 minutes) are needed then we have that: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{150}\) (we can sum the rates to get the combined rate of 3 sources). Also, since to download the movie from A and C alone 4 hours (240 minutes) are needed then we have that: \(\frac{1}{a}+\frac{1}{c}=\frac{1}{240}\). Subtract the second equation from the first one: \(\frac{1}{b}=\frac{1}{150}\frac{1}{240}\) > \(\frac{1}{b}=\frac{1}{400}\) > \(b=400\). So, to download the movie from source B alone 400 minutes are needed, therefore to download the trailer, which is 40 time smaller than the movies 400/40=10 minutes are needed. Answer: D.
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Re: Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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15 Nov 2012, 05:15
\(\frac{1}{A}+\frac{1}{C}=\frac{1}{2.5}=\frac{2}{5}\) \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{4}\) Combine the two equations to get \(\frac{1}{B}\) \(\frac{1}{B}=\frac{1}{4}\frac{2}{5}=\frac{3}{20}==>\frac{1}{B}=\frac{3}{20}\) The time it takes to normally download is \(\frac{20}{3}hrs\) Since the file is 40 times smaller, divide the time by 40. \(t=\frac{20}{(3)(40)}=\frac{1}{6}hrs=10min\)
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Re: Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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26 Dec 2013, 06:29
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superpus07 wrote: Bill downloads the movie "Revenge of the Avengers" to his computer in 2.5 hours, using a download manager that downloads from 3 sources marked A, B and C. Each source provides download at a constant rate, but the rates of different sources are not necessarily identical. If the movie was downloaded from sources A and C alone, it would take 4 hours to complete the download. The next day, source B is available, but the other sources are inactive. How long will it take to download the trailer of the movie, a file that is 40 times smaller, from source B alone?
A. 6 hours and 40 minutes B. 15 minutes C. 12 minutes D. 10 minutes E. 3 minutes Solution Here, please follow 1/A + 1/B + 1/C = 2/5 1/A + 1/C = 1/4 Then 1/B = 3/20 Time taken to download in B is 20/3. Now 40 times smaller then 20/3/40 = 20/120 = 1/6 hrs 10 minutes So D is your friend today Cheers! J



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Re: Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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01 Feb 2014, 04:08
Hi Bunuel
I came down to 400 but wont answer. After ur explanation I am bit confused y u divide with 40...????
"So, to download the movie from source B alone 400 minutes are needed, therefore to download the trailer, which is 40 time smaller than the movies 400/40=10 minutes are needed".
Rgds Prasannajeet



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Re: Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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01 Feb 2014, 05:27



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Re: Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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01 Feb 2014, 20:19
Let the movie size be 400 u. Given,A+C=4 hrs. A+C=100 u/hr And A+B+C=2.5 hrs or 400/2.5=160 u/hr B alone=160100=60 u/hr Trailer=40 times smaller or400/40=10 u B will take 10/60 hrs or 10 minutes. Ans D
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Re: Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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02 Feb 2014, 06:22
Yes Bunuel ,It was a felony I made not being understand the simple equation...



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Re: Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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15 Sep 2017, 13:52
Can someone please expalin why my reasoning is wrong?
R(a+c) = 1/4 1=1/4Rb(5/2) Rb=1/2
To calculate a file 40 times smaller:
1/40 = 1/2T
T=1/20x60 = 3 minutes



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Bill downloads the movie "Revenge of the Avengers" to his [#permalink]
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15 Sep 2017, 16:24
MvArrow wrote: Can someone please expalin why my reasoning is wrong?
R(a+c) = 1/4 1=1/4Rb(5/2) Rb=1/2
To calculate a file 40 times smaller:
1/40 = 1/2T
T=1/20x60 = 3 minutes MvArrow , 1=1/4Rb(5/2)I'm not sure how you calculated Rb? Rb should be 3/20. R(a+b+c) = 2/5 (time is 5/2, invert for rate) R(a+c) = 1/4 Rb: (2/5  1/4) = 3/20 = Rb Then \(\frac{\frac{1}{40}}{\frac{3}{20}}\) = \(\frac{1}{6}\) * 60 = 10 minutes Also, I think this part isn't quite right either: 1/40 = 1/2T Time would be 20/3: then 1/40 * 20/3 is 1/6 * 60 = 10 minutes Hope that helps. If not, would you please explain how you got Rb?
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