Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

23 Feb 2012, 08:05

8

This post received KUDOS

33

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

57% (02:08) correct
43% (01:17) wrong based on 1308 sessions

HideShow timer Statistics

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

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

Show Tags

05 Mar 2012, 05:35

1

This post received KUDOS

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?

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]

Show Tags

07 Aug 2012, 03:59

1

This post received KUDOS

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

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]

Show Tags

04 Oct 2014, 09:58

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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 [#permalink]

Show Tags

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]

Show Tags

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]

Show Tags

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]

Show Tags

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]

Show Tags

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....

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...