Bill downloads the movie "Revenge of the Avengers" to his : GMAT Problem Solving (PS)
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27 Jul 2012, 17:11
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70% (03:22) correct 30% (02:23) wrong based on 195 sessions

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A. 6 hours and 40 minutes
B. 15 minutes
C. 12 minutes
D. 10 minutes
E. 3 minutes
[Reveal] Spoiler: OA
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27 Jul 2012, 20:22
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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.

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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28 Jul 2012, 00:38
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superpus07 wrote:

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.

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15 Nov 2012, 04:15
$$\frac{1}{A}+\frac{1}{C}=\frac{1}{2.5}=\frac{2}{5}$$
$$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{4}$$

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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26 Dec 2013, 05:29
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superpus07 wrote:

A. 6 hours and 40 minutes
B. 15 minutes
C. 12 minutes
D. 10 minutes
E. 3 minutes

1/A + 1/B + 1/C = 2/5
1/A + 1/C = 1/4

Then 1/B = 3/20

Now 40 times smaller then 20/3/40 = 20/120 = 1/6 hrs

10 minutes

So D is your friend today

Cheers!
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01 Feb 2014, 03:08
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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01 Feb 2014, 04:27
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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01 Feb 2014, 19:19
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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02 Feb 2014, 05:22
Yes Bunuel ,It was a felony I made not being understand the simple equation...
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08 Aug 2016, 10:24
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