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 Author: ravitejapandiri [ 18 Sep 2010, 20:13 ] Post subject: Machines X and Y work at their respective constant rates. How many 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, doesCan you explain this one Bunuel plz?At the end,we are having a definite quantity "X"..Right?So I still feel the answer is D.Because there is no other value/variable affecting the outcome except for the "X".Please clarify if I am going badly wrong somewhere!Attachment: DS-3 (1).jpg [ 143.42 KiB | Viewed 43091 times ]

 Author: Bunuel [ 23 Feb 2012, 08:13 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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.

 Author: Bunuel [ 18 Sep 2010, 20:31 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many 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.P.S. Please post one question per topic. Also you can attach image files directly so no need to attach zip files.

 Author: vigneshpandi [ 19 Sep 2010, 16:52 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many 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.P.S. Please post one question per topic. Also you can attach image files directly so no need to attach zip files.Pardon me if this is a stupid approach. but still I want to get it clarified.From 1, as you said, \frac{1}{x}+\frac{1}{y}=\frac{2x}{3}threfore \frac{1}{y}=\frac{2x}{3}-\frac{1}{x}this gives \frac{1}{y}=\frac{2x^2-3}{3x}Now using the value y=2x, substitute in the above equation and u get 2x^2=\frac{9}{4}hence x=\frac{3}{2}with this we can even find y.Hence answer is C.What is wrong in this approach?

 Author: Bunuel [ 19 Sep 2010, 22:26 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many vigneshpandi 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.P.S. Please post one question per topic. Also you can attach image files directly so no need to attach zip files.Pardon me if this is a stupid approach. but still I want to get it clarified.From 1, as you said, ...$$\frac{1}{x}+\frac{1}{y}=\frac{2x}{3}$$threfore $$\frac{1}{y}=\frac{2x}{3}-\frac{1}{x}$$this gives $$\frac{1}{y}=\frac{2x^2-3}{3x}$$Now using the value y=2x, substitute in the above equation and u get 2x^2=\frac{9}{4}hence x=\frac{3}{2}with this we can even find y.Hence answer is C.What is wrong in this approach?What you are basically saying is that you can solve 1 equation $$\frac{xy}{x+y}=x*\frac{2}{3}$$ with 2 unknowns $$x$$ and $$y$$. Though it's not generally impossible (for example: 2x+y=y+4) this is not the case here. Next, we have $$total \ time=\frac{xy}{x+y}=x*\frac{2}{3}$$ and not $$\frac{1}{x}+\frac{1}{y}=\frac{2x}{3}$$ as you wrote (the calculation in red is not correct, it should be: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{total \ time}=\frac{1}{x*\frac{2}{3}}=\frac{3}{2x}$$). So if you substitute $$y$$ by $$2x$$ in $$total \ time=\frac{xy}{x+y}=x*\frac{2}{3}$$ (which by the way gives this relationship) you don't get the $$x=\frac{3}{2}$$, you'll get $$x*\frac{2}{3}=x*\frac{2}{3}$$ --> $$\frac{2}{3}=\frac{2}{3}$$ as $$x$$ will cancel out.Hope it's clear.

 Author: KarishmaB [ 30 Nov 2010, 09:14 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many In this question, I would like to discuss the use of logic.Ques: How many more hours does it take machine Y than it does machine X.So I am looking for a number like 2 hrs or something.Neither of the statements gives me a number of hours for anything. Only relative time taken. So we can straight away say the answer is (E).Also, how to deal with a statement like without getting into equations and variables: Machines X and Y, working together, fill an order in 2/3 the time that machine X, working alone, does.Together, they take 2/3 the time taken by machine X. i.e. if machine X took 6 hrs, together they took 4 hrs. The 2 hrs were saved because machine Y was also working for those 4 hrs. In 4 hrs machine Y did what machine X would have done in 2 hrs. So time taken by machine Y alone will be twice the time taken by machine X alone.

 Author: calvin1984 [ 05 Mar 2012, 05:35 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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?

 Author: Bunuel [ 05 Mar 2012, 07:08 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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 cannot 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 cannot 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.

 Author: A4G [ 04 Dec 2013, 11:15 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many 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.P.S. Please post one question per topic. Also you can attach image files directly so no need to attach zip files.Hi,I am not understanding statement 1.If x & Y are the rates respectively , then 1/X and 1/Y are the time taken to complete the task.Shouldnt the equation be1/X + 1/Y = 2/3XIt gives a degree 3 equation but I am not sure where I am going wrong in logic ?

 Author: KarishmaB [ 04 Dec 2013, 20:27 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many A4G 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.P.S. Please post one question per topic. Also you can attach image files directly so no need to attach zip files.Hi,I am not understanding statement 1.If x & Y are the rates respectively , then 1/X and 1/Y are the time taken to complete the task.Shouldnt the equation be1/X + 1/Y = 2/3XIt gives a degree 3 equation but I am not sure where I am going wrong in logic ?As Bunuel noted above, X is the time taken by machine X and Y is the time taken by machine Y. Combined time taken is 2X/3Rates are 1/X and 1/Y which are additive. The combined rate is 3/2X1/X + 1/Y = 3/2XAlso note that you are trying to add individual time taken in your equation. But times are not additive, only rates are additive. e.g. if you take 2 hrs to complete a work and I take 3 hrs, together will we take 5 hrs? I hope you understand that we will take less than 2 hrs for sure because you alone can complete it in 2 hrs. So times are NOT additive.

 Author: himanshujovi [ 04 May 2014, 23:29 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many 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.P.S. Please post one question per topic. Also you can attach image files directly so no need to attach zip files.Bunuel my thought process was that - No where in the question is the absolute time mentioned for any any machine so none of the 2 points are sufficent . hence E.

 Author: Bunuel [ 05 May 2014, 00:36 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many himanshujovi 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.P.S. Please post one question per topic. Also you can attach image files directly so no need to attach zip files.Bunuel my thought process was that - No where in the question is the absolute time mentioned for any any machine so none of the 2 points are sufficent . hence E.That's not entirely correct. For example, if the first statement were: machines X and Y, working together, fill a production order of this size in 1/2 the time that machine X, working alone, does, then this would be sufficient. Because in this case we would have x=y, and hence x-y=0.Does this make sense?

 Author: sukanyar [ 11 May 2015, 03:49 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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) hoursBut, (1) says that 1/(1/x+1/y) = (2/3) xSolving this, we get: y = 2xSo, 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 = 2xSo, 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.

 Author: OptimusPrepJanielle [ 12 May 2015, 02:35 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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.

 Author: iPen [ 05 Jul 2015, 20:24 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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?

 Author: Bunuel [ 05 Jul 2015, 23:47 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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?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.

 Author: iPen [ 08 Jul 2015, 15:00 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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 = 2xTwo variables - we don't know the values of either x or y, so insufficient.(2) $$\frac{1}{y} = \frac{1}{2x}$$; y = 2xAgain, 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.

 Author: ZaydenBond [ 04 Feb 2016, 14:27 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor 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....

 Author: Bunuel [ 05 Feb 2016, 00:30 ] Post subject: Re: Machines X and Y work at their respective constant rates. How many mor ZaydenBond 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.$$\frac{xy}{x+y}=x*\frac{2}{3}$$ --> $$2x=y$$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....$$\frac{xy}{x+y}=\frac{2x}{3}$$;Cross-multiply: $$3xy=2x(x+y)$$;$$3xy = 2x^2 + 2xy$$;$$xy = 2x^2$$;Reduce by x: $$y=2x$$.Hope it's clear.

Author:  GMATinsight [ 18 Jul 2020, 20:31 ]
Post subject:  Re: Machines X and Y work at their respective constant rates. How many mor

ravitejapandiri 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?

(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

Can you explain this one Bunuel plz?

At the end,we are having a definite quantity "X"..Right?So I still feel the answer is D.

Because there is no other value/variable affecting the outcome except for the "X".Please clarify if I am going badly wrong somewhere!

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The attachment DS-3 (1).jpg is no longer available