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Machines X and Y produced identical bottles at different

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Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Machines X and Y produced identical bottles at different [#permalink]

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23 Oct 2006, 01:21
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Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone fore 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot if machine X produced twice as many bottles in 4 hours as machine Y produced in 3 hours?

... I can't seem to set-up the equation for this.

... Also as a side question, is it worth it to buy access to the \$75 GMATclub challenges? Does it come with explanations?
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Senior Manager
Joined: 19 Jul 2006
Posts: 358

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23 Oct 2006, 05:38
say Machine X has speed of x unit/ hr
say Machine Y has speed of y unit/hr

X produce in 4 hrs = 4x unit
Y produce in 3 hrs = 3y unit

its given ,
X produce in 4 hrs = 2 (Y produce in 3 hrs)

4x = 2 * 3y
or 2x = 3 y ------(1)

Machine capacity = 4x + 3y = 4x + 2x = 6x ( using 1)

So machine X will take 6 hrs for entire production alone
Director
Joined: 06 Feb 2006
Posts: 897

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23 Oct 2006, 05:40
Lets see if that makes sense....

4x=d
3x=T-d

d- work done by X
T- Total work

From the question we know:

d=2(T-d)
d=2T-2d
d+2d=2T
3d=2T
T=(3/2)d

3x=(3/2*d)-d
3x=1/2d.... which means that Y did half of the job in 3 hours.....

And X did half job in 4 hours.... So working alone Machine X would do the job in 8 hours....

Director
Joined: 06 Feb 2006
Posts: 897

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23 Oct 2006, 12:13
AK wrote:
say Machine X has speed of x unit/ hr
say Machine Y has speed of y unit/hr

X produce in 4 hrs = 4x unit
Y produce in 3 hrs = 3y unit

its given ,
X produce in 4 hrs = 2 (Y produce in 3 hrs)

4x = 2 * 3y
or 2x = 3 y ------(1)

Machine capacity = 4x + 3y = 4x + 2x = 6x ( using 1)

So machine X will take 6 hrs for entire production alone

Yeah... the use of 3x=T-d in my calculation is wrong, because you cannot assume the same work rates for both of the machines....

Manager
Joined: 07 Jun 2006
Posts: 109

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24 Oct 2006, 06:48
Code:
Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone fore 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot if machine X produced twice as many bottles in 4 hours as machine Y produced in 3 hours?

I think this can be solved without setting up the equation.

We know this
machine X produced twice as many bottles in 4 hours as machine Y produced in 3 hours..............(1)

We also know that
machine X filled the production lot for 4 hours.
AND machine Y filled the production lot for 3 hours. .......................(2)

So from (1) and (2),
Machine X filled 2/3 of the production lot (IN its 4 hours)

To fill the whole lot.......it takes 4*3/2 = 6 hours
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK

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24 Oct 2006, 08:03
Girikorat... Don't understand how you got 2/3 from (1) and (2). And why you multiplied it by 4?

I think yours is a much simpler and quicker way of solving so trying to understand and pick-up on your logic method. Hope you can explain. thanks
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Senior Manager
Joined: 01 Oct 2006
Posts: 495

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24 Oct 2006, 12:42
Let T be the total number of bottles required to fill the parking lot.
In 4 hours let X prodcuce a bottles
then in 3 hours y produce T-X bottles.

As per the given condition

x=2(T-x) thus 3x=2t..t=3x/2

For x

4 hrs..... x bottles
Y........... 3x/2

thus Y =4*(3x)/(2x)=6hrs

Yogesh
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Joined: 15 Jul 2004
Posts: 1445
Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)

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25 Oct 2006, 02:29
Lets say mc Y produced N bottles in 3 hours.

Then according to que - mc X produced 2N bottles in 4 hours.

Also given in the que is total production lot comprises 4 hrs worth of mc X work and 3 hours worth of Mc Y work.

So total lot = N + 2N = 3N.

Mc X produces 2N in 4 hours. It'll produce another N in 2 hours working at constant rate.

So total time taken by X to fill the lot is 4 + 2 = 6 hours.
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK

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26 Oct 2006, 05:33
dwivedys, you brought enlightenment again!!!

i like how you think the problems thru... thanks man
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26 Oct 2006, 05:33
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