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Machines X and Y work at their respective constant rates [#permalink]

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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.

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

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).

Re: Machines X and Y work at their respective constant rates [#permalink]

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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?

Re: Machines X and Y work at their respective constant rates [#permalink]

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07 Aug 2012, 03:59

1

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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

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

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

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.
_________________

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

Re: Machines X and Y work at their respective constant rates [#permalink]

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04 Oct 2014, 09:58

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Machines X and Y work at their respective constant rates [#permalink]

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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?

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.
_________________

Re: Machines X and Y work at their respective constant rates [#permalink]

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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"!

Machines X and Y work at their respective constant rates [#permalink]

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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!

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]

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29 Sep 2015, 19:00

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.

Hey Bunuel, I did it by setting \(x\) = speed of Machine X and \(y\) = speed of Machine Y, which eventually led me to the equation \(x\) = 2\(y\)

However, I am trying to reconcile the logic of your working because I am still quite weak in this topic. It may seem like a stupid question but

1. Does this working mean --> Speed of X + Speed of Y = Total Speed?

2. Would you recommend that we adopt this general formula to all work/rate problems?

3. Is there any potential trap in my method?

Unrelated: 4. As you can see I have just joined, I find that when I do a search on the forum for any type of question, it always shows me questions from as long ago as 2009. Is there any way I can get access to the most recent questions? Do you think there is any difference in doing more recent questions? The way I see it is that more recent questions = more accurate to the current standard i'm up against if I were to take the GMAT during this period.

Machines X and Y work at their respective constant rates [#permalink]

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04 Feb 2016, 14:27

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.

\(\frac{xy}{x+y}=x*\frac{2}{3}\) --> \(2x=y\)

Could you go through the bolded part and explain how you derived 2x=y? I am quite confused how you got rid of the x....

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.

\(\frac{xy}{x+y}=x*\frac{2}{3}\) --> \(2x=y\)

Could you go through the bolded part and explain how you derived 2x=y? I am quite confused how you got rid of the x....

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