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Machines X and Y work at their respective constant rates. How many [#permalink]
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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 twothirds 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! Attachment:
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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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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: \(yx=?\) (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.
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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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19 Sep 2010, 17:52
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: \(yx=?\)
(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^23}{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?
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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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19 Sep 2010, 23:26
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: \(yx=?\)
(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 \(\frac{1}{y}=\frac{2x^23}{3x}\) 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.
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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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30 Nov 2010, 10:14
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.
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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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03 Nov 2013, 15:07
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: \(yx=?\)
(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 Bunuel, Can you please see why my logic is not correct? my equation was this: (X+Y)*2*(Z/3) = XZ where X is the rate of machine X, Y the rate of machine Y and Z is the time it takes for machine X to fill the order alone. In the end, I get an equation that is opposite from what you got: 2Y=X. What's wrong with that?



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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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04 Nov 2013, 01:53
ronr34 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: \(yx=?\)
(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 Bunuel, Can you please see why my logic is not correct? my equation was this: (X+Y)*2*(Z/3) = XZ where X is the rate of machine X, Y the rate of machine Y and Z is the time it takes for machine X to fill the order alone. In the end, I get an equation that is opposite from what you got: 2Y=X. What's wrong with that? Please read solutions carefully.
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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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04 Dec 2013, 12:15
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: \(yx=?\)
(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 be 1/X + 1/Y = 2/3X It gives a degree 3 equation but I am not sure where I am going wrong in logic ?



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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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04 Dec 2013, 21:27
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: \(yx=?\)
(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 be 1/X + 1/Y = 2/3X It 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/3 Rates are 1/X and 1/Y which are additive. The combined rate is 3/2X 1/X + 1/Y = 3/2X Also 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.
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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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04 Dec 2013, 23:07
VeritasPrepKarishma wrote: 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: \(yx=?\)
(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 be 1/X + 1/Y = 2/3X It 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/3 Rates are 1/X and 1/Y which are additive. The combined rate is 3/2X 1/X + 1/Y = 3/2X Also 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. Thanks Karishma .. The point you made about time really makes sense and I missed a small but very important point



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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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05 May 2014, 00:29
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: \(yx=?\)
(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.



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Re: Machines X and Y work at their respective constant rates. How many [#permalink]
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05 May 2014, 01:36
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: \(yx=?\)
(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 xy=0. Does this make sense?
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