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Machines X and Y work at their respective constant rates

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Machines X and Y work at their respective constant rates [#permalink] New post 23 Feb 2012, 08:05
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Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?

(1) Machines X and Y, working together, fill a production order of this size in two-thirds the time that machine X, working alone, does.
(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does.
[Reveal] Spoiler: OA
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Re: DATA12 [#permalink] New post 23 Feb 2012, 08:13
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Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?

Let \(x\) and \(y\) be the times needed for machines X and Y respectively working alone to fill a production order of this size.

Question: \(y-x=?\)

(1) Machines X and Y, working together, fill a production order of this size in 2/3 the time that machine X, working alone, does --> general relationship: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{total \ time}\) --> Total time needed for machines X and Y working together is \(total \ time=\frac{xy}{x+y}\) (general formula) --> given \(\frac{xy}{x+y}=x*\frac{2}{3}\) --> \(2x=y\). Not sufficient

(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does --> \(2x=y\), the same info. Not sufficient

(1)+(2) Nothing new. Not Sufficient.

Answer: E.

Hope it helps.
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 05 Mar 2012, 05:35
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Just a question regarding this problem.
I choose E because I decided that there's no concrete value indicating the hours. The values they give from both statements indicate relative values. Is this a good approach or did I just get lucky?
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 05 Mar 2012, 07:08
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calvin1984 wrote:
Just a question regarding this problem.
I choose E because I decided that there's no concrete value indicating the hours. The values they give from both statements indicate relative values. Is this a good approach or did I just get lucky?


No that's not correct. We have two exactly the same linear equations from both statements, that's why we can not solve for x and y. But if we had two distinct linear equations then we would be able to solve.

For example if either of statement were: the time needed for Machines X working alone to fill a production order of this size plus the time needed for Machines Y working alone to fill a production order of double the size is 10 hours --> x+2y=10.

So, for (1)+(2) we would have x+2y=10 and 2x=y --> x=2 and y=4 --> y-x=2.

Generally if you have n distinct linear equations and n variables then you can solve for them. "Distinct linear equations" means that no equation can be derived with the help of others or by arithmetic operation (multiplication, addition).

For example:
\(x+y=2\) and \(3x+3y=6\) --> we do have two linear equations and two variables but we cannot solve for \(x\) or \(y\) as the second equation is just the first one multiplied by 3 (basically we have only one distinct equation);
OR
\(x+y=1\), \(y+z=2\) and \(x+2y+z=3\) --> we have 3 linear equations and 3 variables but we can not solve for \(x\), \(y\) or \(z\) as the third equation can be derived with the help of first two if we sum them (basically we have only two distinct equation).

Hope it's clear.
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 07 Aug 2012, 03:59
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Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?

Let x and y be the times needed for machines X and Y respectively working alone to fill a production order of this size.

Question: y-x=?

(1) Machines X and Y, working together, fill a production order of this size in 2/3 the time that machine X, working alone, does --> general relationship: \frac{1}{x}+\frac{1}{y}=\frac{1}{total \ time} --> Total time needed for machines X and Y working together is total \ time=\frac{xy}{x+y} (general formula) --> given \frac{xy}{x+y}=x*\frac{2}{3} --> 2x=y. Not sufficient

(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does --> 2x=y, the same info. Not sufficient

(1)+(2) Nothing new. Not Sufficient.

Answer: E.

Hope it helps.

Above is the solution posted by Bunnel- (sorry I always misspell your name)

When we solve statement 1 we really need to discard one solution x =0 (when you solve you get two solutions) Can we safely discard X =0 in rate. time questions since 0 time does not make sense; 0 time spent means machine did not work at all. Question says Machine x did work so X =0 is not a possible solution and that is why we can discard that solution keeping Y =2x
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Re: DATA12 [#permalink] New post 25 Sep 2013, 05:58
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Bunuel wrote:
Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?

Let \(x\) and \(y\) be the times needed for machines X and Y respectively working alone to fill a production order of this size.

Question: \(y-x=?\)

(1) Machines X and Y, working together, fill a production order of this size in 2/3 the time that machine X, working alone, does --> general relationship: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{total \ time}\) --> Total time needed for machines X and Y working together is \(total \ time=\frac{xy}{x+y}\) (general formula) --> given \(\frac{xy}{x+y}=x*\frac{2}{3}\) --> \(2x=y\). Not sufficient

(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does --> \(2x=y\), the same info. Not sufficient

(1)+(2) Nothing new. Not Sufficient.

Answer: E.

Hope it helps.


Is xy/x+y the total time or the combined rate?
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Re: DATA12 [#permalink] New post 25 Sep 2013, 07:51
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Skag55 wrote:
Bunuel wrote:
Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?

Let \(x\) and \(y\) be the times needed for machines X and Y respectively working alone to fill a production order of this size.

Question: \(y-x=?\)

(1) Machines X and Y, working together, fill a production order of this size in 2/3 the time that machine X, working alone, does --> general relationship: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{total \ time}\) --> Total time needed for machines X and Y working together is \(total \ time=\frac{xy}{x+y}\) (general formula) --> given \(\frac{xy}{x+y}=x*\frac{2}{3}\) --> \(2x=y\). Not sufficient

(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does --> \(2x=y\), the same info. Not sufficient

(1)+(2) Nothing new. Not Sufficient.

Answer: E.

Hope it helps.


Is xy/x+y the total time or the combined rate?


Total time.

Check here: machines-x-and-v-produced-identical-bottles-at-different-104208.html#p812628

Hope it helps.
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Re: DATA12 [#permalink] New post 26 Sep 2013, 02:40
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Bunuel wrote:
Skag55 wrote:
Bunuel wrote:
Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?

Let \(x\) and \(y\) be the times needed for machines X and Y respectively working alone to fill a production order of this size.

Question: \(y-x=?\)

(1) Machines X and Y, working together, fill a production order of this size in 2/3 the time that machine X, working alone, does --> general relationship: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{total \ time}\) --> Total time needed for machines X and Y working together is \(total \ time=\frac{xy}{x+y}\) (general formula) --> given \(\frac{xy}{x+y}=x*\frac{2}{3}\) --> \(2x=y\). Not sufficient

(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does --> \(2x=y\), the same info. Not sufficient

(1)+(2) Nothing new. Not Sufficient.

Answer: E.

Hope it helps.


Is xy/x+y the total time or the combined rate?


Total time.

Check here: machines-x-and-v-produced-identical-bottles-at-different-104208.html#p812628

Hope it helps.


I see where I got it fundamentally wrong:

I confused
1 job / x hours which gives rate = 1/x
x jobs(say printed pages) / 1 hour which gives rate = x
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 04 Oct 2014, 09:58
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 11 May 2015, 03:49
Let us suppose that machine X takes x hours, while machine Y takes y hours, to fill production order. According to question, we have to find (y-x)

So, in 1 hour, X finishes 1/x while Y finishes 1/y production order

(1) says: Machines X and Y, working together, fill a production order of this size in two-thirds the time that machine X, working alone, does.

Working together, X and Y complete (1/x+1/y) production order in 1 hour

=> Working together, X and Y complete production order in 1/(1/x+1/y) hours

But, (1) says that 1/(1/x+1/y) = (2/3) x

Solving this, we get: y = 2x

So, clearly not sufficient for us to say what is (y – x)

(2) says that Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does.
=> y = 2x

So, clearly not sufficient for us to say what is (y – x)

Combining the two statements, again, both actually say the same thing (y=2x) and so, this is not sufficient for us to say what is (y – x).

Hence, E.
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 11 May 2015, 07:20
E

let us suppose rate of machine x is Rx and rate of machine y is Ry
given to us is Rx>Ry

statement 1.

let us suppose it takes t hours to fill an order

so Rx=1/t

given Rx + Ry = 3/2t

substituing value of Rx in the above eq

Ry = 1/2t

we dont know the value of t. So insufficient

statement 2.

let us suppose here also that it takes t hours to fill and order

given Ry = 1/2t
and Rx = 1/t

we dont have the value of t so ,insufficient.


Combining both

either of the statements above imply same and value of t still remains unsaid.


SO E


hope it helps
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 12 May 2015, 02:35
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The rates of Machine X and Machine Y can be 1/A and 1/B, respectively. A and B represent the number of hours to complete the task. The question is asking for B-A.
Statement 1 tells you that (1/B) + (1/A) = (2/3)(1/A). There are still 2 unknowns, so eliminate A, D.
Statement 2 tells you that (1/B) = (1/2A) or B=2A. Still, we have 2 unknowns. Eliminate B.
No new information can be obtained by combining to the two statements. Therefore E is the answer.
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Machines X and Y work at their respective constant rates [#permalink] New post 05 Jul 2015, 20:24
If y = 2x, then I can't simply leave the answer as y - x = 2x - x = x?

That is, machine y takes x hours longer than x? It doesn't solve for the value of x, but isn't that technically an answer?

Or, does the GMAT require a value for x, since we're still left to wonder how many hours x is... e.g. x could be 1 hour or 10 hours, or so forth? So, even if you get an answer but there's still at least one variable in the answer, the data is missing and thus insufficient?
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 05 Jul 2015, 23:47
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iPen wrote:
If y = 2x, then I can't simply leave the answer as y - x = 2x - x = x?

That is, machine y takes x hours longer than x? It doesn't solve for the value of x, but isn't that technically an answer?

Or, does the GMAT require a value for x, since we're still left to wonder how many hours x is... e.g. x could be 1 hour or 10 hours, or so forth? So, even if you get an answer but there's still at least one variable in the answer, the data is missing and thus insufficient?


Official Guide:

In data sufficiency problems that ask for the value of a quantity, the data given in the statements are sufficient only when it is possible to determine exactly one numerical value for the quantity.
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Re: Machines X and Y work at their respective constant rates [#permalink] New post 06 Jul 2015, 11:06
Bunuel wrote:
iPen wrote:
If y = 2x, then I can't simply leave the answer as y - x = 2x - x = x?

That is, machine y takes x hours longer than x? It doesn't solve for the value of x, but isn't that technically an answer?

Or, does the GMAT require a value for x, since we're still left to wonder how many hours x is... e.g. x could be 1 hour or 10 hours, or so forth? So, even if you get an answer but there's still at least one variable in the answer, the data is missing and thus insufficient?


Official Guide:

In data sufficiency problems that ask for the value of a quantity, the data given in the statements are sufficient only when it is possible to determine exactly one numerical value for the quantity.


Thanks that clears it up. Hence, the name "data sufficiency"!
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Machines X and Y work at their respective constant rates [#permalink] New post 08 Jul 2015, 11:26
Let x finish x% of the total work/min => 100% of the work will be done in 100/x mins
Let y finish y% of the total work/min => 100% of the work will be done in 100/y mins

1) 100/x+y = 2/3 * 100/x => x = 2y
2) 100/y = 2 * 100/x => x = 2y
1+2) No new info => Ans = (E)

For the sake of clarity, does it matter that both statements give me x = 2y as opposed to the y = 2x that everyone else has been getting? I think not, but would appreciate an input nonetheless - have my retake in less than 2 weeks now!

Thanks
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Machines X and Y work at their respective constant rates [#permalink] New post 08 Jul 2015, 15:00
I did it a little differently.


\(\frac{1}{x} + \frac{1}{y} = \frac{1}{h}\)

\(\frac{y}{xy} + \frac{x}{xy} = \frac{1}{h} = \frac{y + x}{xy}\)

(1) \(\frac{1}{2/3*x} = \frac{3}{2x} = \frac{y + x}{xy}\)

\(\frac{3}{2} = \frac{y + x}{y}\); 3y = 2(y + x); 3y = 2y + 2x; y = 2x

Two variables - we don't know the values of either x or y, so insufficient.

(2) \(\frac{1}{y} = \frac{1}{2x}\); y = 2x

Again, two variables remain, so it's insufficient.

(1) + (2): Two different equations, same result of y = 2x

\(\frac{3}{2} = \frac{y + x}{y}\)
\(\frac{3}{2} = 2x + \frac{x}{2x}\);\(\frac{3}{2} = 2x + \frac{1}{}2\); \(2x = 1\); \(x =\frac{1}{2}\)
\(y = 2(\frac{1}{2}) = 1\)

Together, they finish a given job in 1/3 hours.
Machine x does it in 1/2 hours.
Machine y does it in 1 hour.
y - x = 1/2 hours.

But, plugging the same y = 2x only gives us a relative difference. And, the three results above would need to be multiplied by a constant, because the equation holds true for any positive value of x (e.g. If x is 1, then y is 2, together it's 2/3, and y-x = 1). Thus, insufficient. Answer is E.
Machines X and Y work at their respective constant rates   [#permalink] 08 Jul 2015, 15:00
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