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Bill downloads the movie "Revenge of the Avengers" to his

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Bill downloads the movie "Revenge of the Avengers" to his  [#permalink]

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New post 27 Jul 2012, 18:11
15
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A
B
C
D
E

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  55% (hard)

Question Stats:

66% (02:45) correct 34% (03:01) wrong based on 242 sessions

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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  [#permalink]

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New post 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 co  [#permalink]

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New post 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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New post 15 Nov 2012, 05:15
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\(\frac{1}{A}+\frac{1}{C}=\frac{1}{2.5}=\frac{2}{5}\)
\(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{4}\)

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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New post 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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New post 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
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Re: Bill downloads the movie "Revenge of the Avengers" to his  [#permalink]

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New post 01 Feb 2014, 05:27
prasannajeet wrote:
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


It should be very easy:

To download the movie you need 400 minutes. The trailer is 40 times smaller than the movies. Hoe many minutes do you need to download the trailer?
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Re: Bill downloads the movie "Revenge of the Avengers" to his  [#permalink]

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New post 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=160-100=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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New post 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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New post 15 Sep 2017, 13:52
Can someone please expalin why my reasoning is wrong?

R(a+c) = 1/4
1=1/4-Rb(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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New post 15 Sep 2017, 16:24
MvArrow wrote:
Can someone please expalin why my reasoning is wrong?

R(a+c) = 1/4
1=1/4-Rb(5/2)
Rb=1/2

To calculate a file 40 times smaller:

1/40 = 1/2T

T=1/20x60 = 3 minutes

MvArrow ,

1=1/4-Rb(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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Re: Bill downloads the movie "Revenge of the Avengers" to his  [#permalink]

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