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

6

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

58% (02:08) correct
42% (01:16) wrong based on 605 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, 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]
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]
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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