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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: Machines X and Y work at their respective constant rates [#permalink] New post 10 Sep 2013, 05:22
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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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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
Re: DATA12   [#permalink] 26 Sep 2013, 02:40
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