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Machines X and Y work at their respective constant rates [#permalink]
23 Feb 2012, 08:05

5

This post received KUDOS

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

56% (02:09) correct
44% (01:14) wrong based on 511 sessions

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]
05 Mar 2012, 07:08

3

This post received KUDOS

Expert's post

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).

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

Re: Machines X and Y work at their respective constant rates [#permalink]
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]
04 Oct 2014, 09:58

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