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# Machines X and Y produced identical bottles at different

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Director
Joined: 20 Aug 2007
Posts: 851
Location: Chicago
Schools: Chicago Booth 2011
Machines X and Y produced identical bottles at different [#permalink]

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28 Jan 2008, 13:38
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Machines X and Y produced identical bottles at different constant rates. Machine X filled part of a lot working alone for 4 hours. Then Machine Y worked alone for 3 hours and filled the rest of the lot. How long would it have taken Machine X to do the entire job alone?

1) Machine X produced 30 bottles per minute
2) Machine X producted twice as many bottles in 4 hours as Machine Y did in 3 hours
Director
Joined: 20 Aug 2007
Posts: 851
Location: Chicago
Schools: Chicago Booth 2011
Re: DS - Machine X and Y [#permalink]

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28 Jan 2008, 14:16
walker wrote:
B

t=4h+4/2h=6h

Director
Joined: 14 Oct 2007
Posts: 753
Location: Oxford
Schools: Oxford'10
Re: DS - Machine X and Y [#permalink]

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28 Jan 2008, 14:29
$$r_x = \frac{T-y}{4}$$ where y = number of bottles that y processed in the job (T)
$$r_y = \frac{y}{3}$$

1) $$r_x = 60$$
doesn't tell us anything about y's rate or the total size of the job. so insuff

2)tell us that T-y = 2y
3y = Total

therefore $$T-y = \frac{2}{3}$$

solve for $$r_x = \frac{1}{6}$$

so it would take x 6 hours for the whole job.
CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: DS - Machine X and Y [#permalink]

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28 Jan 2008, 14:49
Sorry guys, I just relax a little bit.
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Director
Joined: 14 Oct 2007
Posts: 753
Location: Oxford
Schools: Oxford'10
Re: DS - Machine X and Y [#permalink]

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28 Jan 2008, 14:55

can't wait for them after GMAT is over...
Manager
Joined: 10 Dec 2007
Posts: 54
Re: DS - Machine X and Y [#permalink]

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28 Jan 2008, 15:27
sorry, could you please elaborate further on how you started with that equation?
Director
Joined: 14 Oct 2007
Posts: 753
Location: Oxford
Schools: Oxford'10
Re: DS - Machine X and Y [#permalink]

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28 Jan 2008, 15:34
AVakil wrote:
sorry, could you please elaborate further on how you started with that equation?

me? sure..

$$rate = \frac{distance}{Time}$$ (in this case the distance = producing a whole heap of Stellas)

so $$r_x = \frac{Total - y}{4hours}$$
CEO
Joined: 29 Mar 2007
Posts: 2560
Re: DS - Machine X and Y [#permalink]

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29 Jan 2008, 12:37
sonibubu wrote:
Machines X and Y produced identical bottles at different constant rates. Machine X filled part of a lot working alone for 4 hours. Then Machine Y worked alone for 3 hours and filled the rest of the lot. How long would it have taken Machine X to do the entire job alone?

1) Machine X produced 30 bottles per minute
2) Machine X producted twice as many bottles in 4 hours as Machine Y did in 3 hours

We have x/4 and y/3

x+y=z x=z-y

so we have (z-y/4) and y/3.

S1: z-y/4=1800bottles/hr, but doesnt really help us b/c we don't know what fraction of the whole thing this is. insuff.

S2; z-y=2y thus z=3y

X's rate is thus 3y-y/4 --> y/2 So: t=3y/y/2 ---> t=6hrs. B
Senior Manager
Joined: 26 Jan 2008
Posts: 263
Re: DS - Machine X and Y [#permalink]

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29 Jan 2008, 13:19
sonibubu wrote:
Machines X and Y produced identical bottles at different constant rates. Machine X filled part of a lot working alone for 4 hours. Then Machine Y worked alone for 3 hours and filled the rest of the lot. How long would it have taken Machine X to do the entire job alone?

1) Machine X produced 30 bottles per minute
2) Machine X producted twice as many bottles in 4 hours as Machine Y did in 3 hours

The answer is (B). I think others have provided clear and elaborate explanations. Here's another take :-

(1) This doesn't provide any information about Machine Y. Nothing in the question statement has information on Machine Y, other than the fact that it takes 3 hours to fill its portion of the lot. Due to a lack of comparison between the rates or times of machine X and machine Y, this is insufficient.

(2) We now have a comparison point. From this, we can deduce that the lot was full in 4 hours and Machine X filled twice as many bottles as Machine Y. Which means that Machine X filled 2/3rd of the lot in 4 hours. This is sufficient to calculate the rest.

In the actual test, there is no need to calculate the total, but here's how I would do it:
time taken by Machine X to fill 2/3rd of the lot = 4 hours
time taken by Machine X to fill the entire lot = [4 * (3 / 2)] hours = 6 hours
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Re: DS - Machine X and Y   [#permalink] 29 Jan 2008, 13:19
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