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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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30 May 2012, 10:37

harikattamudi wrote:

I'm still not clear why X and Y has to be positive when X/Y > 1. Can you please explain the way you combined taking both X and Y to be positive and also X and Y as negative. Since in either case X/Y will be > 1.

Thanks -H

Statement 1. x-y=1/2. think of 1/2 as the distance between two points x and y on a number line marked with -1,0 and 1. any two points you pick would have to be 1/2 unit apart. can u do that? of course, you can do this on both sides of 0. so turns out x and y don't have to be both positive and can both be negative. Insufficient.

Statement 2: x/y>1 Now the reason x/y>1 is not necessarily x>y is because of the freaking sign that has to complicate everything. Here is an example. we are told x divided by y yields a number greater than one, so we know the end result is positive, but is this result positive because both x and y were positive? or because both were negative and minus and minus cancel out to give a positive number. we are not sure, so you have to consider both options. x and y both >0 and x and y both <0. first case, if both greater than 0. for example, take x=3 y=2. x/y>1 means 3>2. that's correct. second case, if both less than 0, take x=-3 and y=-2. now if u were to follow the same step, your -3>-2. But is this correct? Is -3 greater than -2? No. that's not correct. the sign has to be flipped to say -3<-2. Now you can see why we can't just blindly say that x/y>1 is the same as x>y. it is if the variables are both positive but not so if both negative. For the above reason, statement 2 is not sufficient either. But if you take the two together, you can see that in both cases they can be positive. Hence answer C is sufficient.

I think someone like Bunuel and Karishma would arrive at the answer to this in their head. I have been reading their posts and trying to think and reason like them. I would suspect their line of mental reasoning would be something close to.... Are both x and y negative? both. 1. 2x-2y=1. that's x-y=1/2. if 1/2 is all u have to get, u can get it from both of them being negative. insufficient. 2. x/y>1. x and y have to be both positive or both negative. insufficient. but 1 and 2 overlap at x and y both positive which is what the question asks. c.

Bam! Next button. New question! Try harder, computer.

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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01 Jun 2012, 05:07

Bunuel wrote:

Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

As quoted - clearly 1 and 2 alone are insufficient.

Another approach for combined: 2x-2y=1 x-y=1/2

x/y>1 if both -ve, then x < y ..... x - y < 0

if both +ve, then x > y ..... x - y > 0 This holds good as statement 1 says x - y =1/2

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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30 Sep 2012, 20:06

Bunuel wrote:

Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hello Bunuel,

I am preparing for GMAT and you are a master in quantitatives,I salute you. However this problem I did not understand why it is C.I got E by plugging in numbers. Considerin Stmt (1) 2x-2y=1 Means x-y=1/2 Now i used 2 numbers x=2,y=1.5 and x= -1.5 and y= -2 So (1) is insufficient as both x and y can be both +ve or -ve. Stmt(2) x/y > 1 Means x >y If i use the above numbers x=2,y=1.5 and x= -1.5 and y= -2 then this holds true but cannot determine if x and y are both positive or negetive. Combining (1) and (2) I still cannot determine if x and y are both positive or negetive . So for me answer is E . Please let me know what am I missing here.

Thanks
_________________

arijitb1980 Its nice to day dream sometimes,if you chase it the rest of the times.

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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01 Oct 2012, 03:22

arijitb1980 wrote:

Bunuel wrote:

Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hello Bunuel,

I am preparing for GMAT and you are a master in quantitatives,I salute you. However this problem I did not understand why it is C.I got E by plugging in numbers. Considerin Stmt (1) 2x-2y=1 Means x-y=1/2 Now i used 2 numbers x=2,y=1.5 and x= -1.5 and y= -2 So (1) is insufficient as both x and y can be both +ve or -ve. Stmt(2) x/y > 1 Means x >y If i use the above numbers x=2,y=1.5 and x= -1.5 and y= -2 then this holds true but cannot determine if x and y are both positive or negetive. Combining (1) and (2) I still cannot determine if x and y are both positive or negetive . So for me answer is E . Please let me know what am I missing here.

Thanks

For x = -1.5 and y = -2, the ratio x/y = 3/4 < 1, so the condition in (2) isn't fulfilled.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hello Bunuel,

I am preparing for GMAT and you are a master in quantitatives,I salute you. However this problem I did not understand why it is C.I got E by plugging in numbers. Considerin Stmt (1) 2x-2y=1 Means x-y=1/2 Now i used 2 numbers x=2,y=1.5 and x= -1.5 and y= -2 So (1) is insufficient as both x and y can be both +ve or -ve. Stmt(2) x/y > 1 Means x >y If i use the above numbers x=2,y=1.5 and x= -1.5 and y= -2 then this holds true but cannot determine if x and y are both positive or negetive. Combining (1) and (2) I still cannot determine if x and y are both positive or negetive . So for me answer is E . Please let me know what am I missing here.

Thanks

As stated above, x= -1.5 and y= -2 does not satisfy the second statement, hence you cannot use these values when plugging numbers.
_________________

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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26 Mar 2013, 13:56

1

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Here is what I did, and maybe it will help out those not using (or struggling with) the algebra method mentioned earlier in Bunuel's post. It looks long, but I tried to be complete with the thought and reasoning process

1) \(2x-2y = 1\)

First, test numbers: If \(x\) is a positive number, then \(y\) must be a positive for the math to work. I chose \(x=2\) to evaluate this, which leads to \(y = \frac{3}{2}\) \(2(2) - 2(\frac{3}{2}) = 1\) Are \(X\) and \(Y\) both positive? Yes

If x is a negitive number, then y must be a negitive for the math to work. I chose \(x=-2\) to evaluate this, which leads to \(y= \frac{-5}{2}\) \(2(-2) - 2(\frac{-5}{2}) = 1\) Are \(X\) and \(Y\) both positive? No

From these examples, we can also see that \(x\) is always \(\frac{1}{2}\) more than \(y\) -----> \(x=y+\frac{1}{2}\) We have also learned that both variables must be the same sign

1) is not sufficient

2) \(\frac{x}{y} > 1\)

First thing this tells me is that \(x\) and \(y\) are both the same sign because the result is positive It also tells me that \(|x| >|y|\), because the absolute numerator MUST be larger than the absolute denominator to be grater than 1.

So far this tells me nothing, but let's throw in some numbers to see what is happening: If \(x\) is a positive number, than \(y\) must be positive. I picked \(x = 4\), so y must be a positive number less than \(4\); I chose \(y = 2\). \(\frac{4}{2} = 2\) which is greater than \(1\) Are \(X\) and \(Y\) both positive? Yes

If \(x\) is negitive number, than \(y\) must be negitive. I picked \(x = -4\), so \(y\) must be a negitive number whose ABSOLUTE value is less than \(4\). I chose \(y = -2\) \(\frac{-4}{-2} = 2\) which is greater than \(1\) Are \(X\) and \(Y\) both positive? No

2) is not sufficient

1+2) We have three conditions established from 1) and 2): c1 - The signs of \(x\) and \(y\) must be the same c2 - The absolute value of \(x\) is larger than the absolute value of \(y\) c3 - \(x\) is always \(\frac{1}{2}\) more than \(y\) ----> \(x=y+\frac{1}{2}\)

We also have some examples used thus far when evaluating statement 1 by itself, let's see what meets the criteria once the conditions are considered simultaneously:

- Both positive; \(x = 2\) and \(y = \frac{3}{2}\) c1 - pass, signs are the same c2 - pass, \(2 > \frac{3}{2}\) c3 - pass, \(2 = \frac{3}{2} + \frac{1}{2}\) 1) \(2x-2y=1\); yes, \(2(2) - 2(\frac{3}{2})=1\) 2) \(\frac{x}{y} > 1\); yes, \(\frac{2}{(3/2)} = \frac{4}{3}\); \(\frac{4}{3}>1\)

- Both negitive; No values will fit this condition when considering both statements: This idea violates c2 (\(|x|\) is not larger than \(|y|\)); when a negitive value for \(x\) is \(\frac{1}{2}\) greater than \(y\), the absolute value of the numerator is always going to be smaller than the denominator , making \(\frac{x}{y} < 1\). Therefore, both being positive is the only option to fit both statements.

Example: \(x = -2\), so \(y = \frac{-5}{2}\) as per what we discovered in statement one, \(x = y+\frac{1}{2}\) When applying these values to statement 2, we can see that \(\frac{x}{y} = \frac{-2}{(-5/2)} = .8< 1\)

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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23 Aug 2013, 22:33

very nicely explained Bunuel Sir.. Appreciate

Bunuel wrote:

harikattamudi wrote:

I'm still not clear why X and Y has to be positive when X/Y > 1. Can you please explain the way you combined taking both X and Y to be positive and also X and Y as negative. Since in either case X/Y will be > 1.

Thanks -H

From (2) \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative).

When we consider two statement together:

From (1): \(2x-2y=1\) --> \(x=y+\frac{1}{2}\)

From (2): \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>0\) --> \(\frac{x-y}{y}>0\) --> substitute \(x\) from (1) --> \(\frac{y+\frac{1}{2}-y}{y}>0\)--> \(\frac{1}{2y}>0\) (we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) --> \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive.

OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too.

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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18 Nov 2013, 05:39

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I don't think we need any calculations here at all.

A) 2X-2Y=1 --> X-Y=1/2 this only means X has to be ahead of (or to the right of) Y on the number line by 1/2. Both X could be +ve or -ve or X +ve and Y -ve. Not sufficient

B) X/Y >1 means two things 1. Both have to have the same sign and 2. Abs(X)>Abs(Y). Not sufficient alone

A+B => Only on the right side of zero on the number line, both X could be ahead of Y and its absolute value greater than Y. Hence C
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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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30 May 2014, 23:41

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Are x and y both positive?

1) 2x-2y =1 2) x/y>1

Soln.:

Statement 1: 2x-2y=1 So, 2(x-y)=1 So, x – y = 1/2 So, x = y + ½ But y may still be negative. So insufficient.

Statement 2: x/y>1

(NOTE that we cannot change this to x>y, because if y is negative then inequality sign will change and we don’t know at this stage if y is +ve or –ve.)

1 will be positive only if x and y are both positive, OR if x and y are both negative. So statement 2 too is insufficient by itself.

Statements 1 & 2 together:

Whenever there is an equation and an inequality for 2 variables, try to substitute value of variable derived from the equation into the inequality.

Since x/y>1, Substituting value of x from Statement 1, (Y + ½)/y > 1 So, (2y+1)/2y > 1 So, 1 + 1/2y > 1 So, 1/2y > 0 So, (½) x (1/y) > 0 If y was negative, the above value would be less than 0 not greater than 0. Also if y was zero, the equation value would still not be greater than zero. Also, x/y>1. So, y cannot be zero. Therefore, y is positive. Since x = y + ½ Therefore, x is also positive. SUFFICIENT.

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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30 Aug 2014, 06:55

Hi guys,

After having this problem wrong (E), I though about it for a moment and wanted to validate my thinking.

(1) states that x-y=1/2. By itself it doesn't help us understand whether x and y are both positive or negative. Both are possible. (2) x/y>1. Again, both can be possible.

(1+2) From x/y>1 we know that X>Y. Using this in (1) you understand that X and Y have to be positive, because in the scenario that we considered in (1) only if Y would have been bigger than X could it compensate for the a negative X with the double negation of the negative sign. Example: X=-4, Y=-6 => (-4)-(-4,5)=0,5=1/2. Now that we know that this can't be the case, this automatically means that X and Y are both positive.
_________________

Thank you very much for reading this post till the end! Kudos?

After having this problem wrong (E), I though about it for a moment and wanted to validate my thinking.

(1) states that x-y=1/2. By itself it doesn't help us understand whether x and y are both positive or negative. Both are possible. (2) x/y>1. Again, both can be possible.

(1+2) From x/y>1 we know that X>Y. Using this in (1) you understand that X and Y have to be positive, because in the scenario that we considered in (1) only if Y would have been bigger than X could it compensate for the a negative X with the double negation of the negative sign. Example: X=-4, Y=-6 => (-4)-(-4,5)=0,5=1/2. Now that we know that this can't be the case, this automatically means that X and Y are both positive.

The red part is not correct.

\(\frac{x}{y}>1\) does not mean that \(x>y\). If both x and y are positive, then \(x>y\), BUT if both are negative, then \(x<y\).

From (2) \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative).
_________________

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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02 Oct 2014, 10:06

This is how i solved it:

Are x and y both positive? statement 1: x-y=1/2 --> x=y + (1/2). we can deduce from this expression that x is on the right of y on the number line. But x and y here can be (-,-), (+,+), or (+,-). We only know that for (x,y), (-,+) isn't possible. However, this statement is insufficient.

statement 2: x/y>1. This tells us that x>y if both positive OR x<y is both negative. Insufficient.

combined: from both, we know that (x,y) can be either (+,+) or (-,-). For (+,+), x>y, true considering statement 1 too. For (-,-), x<y, not possible considering statement 1. Hence, both x and y are positives. Option C is correct.

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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05 Aug 2015, 06:37

Bunuel wrote:

arijitb1980 wrote:

Bunuel wrote:

Are x and y both positive? (1) 2x-2y=1 (2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.

Answer: C.

Hello Bunuel,

I am preparing for GMAT and you are a master in quantitatives,I salute you. However this problem I did not understand why it is C.I got E by plugging in numbers. Considerin Stmt (1) 2x-2y=1 Means x-y=1/2 Now i used 2 numbers x=2,y=1.5 and x= -1.5 and y= -2 So (1) is insufficient as both x and y can be both +ve or -ve. Stmt(2) x/y > 1 Means x >y If i use the above numbers x=2,y=1.5 and x= -1.5 and y= -2 then this holds true but cannot determine if x and y are both positive or negetive. Combining (1) and (2) I still cannot determine if x and y are both positive or negetive . So for me answer is E . Please let me know what am I missing here.

Thanks

As stated above, x= -1.5 and y= -2 does not satisfy the second statement, hence you cannot use these values when plugging numbers.

Hi,

While I see that translating (x/y)>1 into x>y must be wrong, as clearly it lets me use a different range of numbers (and is the reason I chose E instead of C), I don't understand what is wrong with my algebra.

Why can't I multiply (x/y) * y > 1 * y and get x>y?

Remember, in inequalities, until you know the sign of the variables (y in this case) you MUST not divide or multiply by a variable.

In the given question, x/y >1 ---> you can only multiply by y if you know that y >0.

Case 1, y > 0 ---> x/y > 1 ---> x> y (correct)

Case 2, y <0 ---> x/y > 1 ---> you can not say x>y as lets say y = -2 and x = -3 then x/y =3/2 > 1 but in this case y > x . This incogruity occurs as for negative numbers , you need to switch the inequality sign when you multiply by a negative number.

This is the reason why you can not multiply or divide by a variable until you are 100% sure of the sign of the inequality (> or <0)
_________________

While I see that translating (x/y)>1 into x>y must be wrong, as clearly it lets me use a different range of numbers (and is the reason I chose E instead of C), I don't understand what is wrong with my algebra.

Why can't I multiply (x/y) * y > 1 * y and get x>y?

Please note: Multiplying a Negative Number on both sides require the Inequation sign to flip [as mentioned in Case-2]

That's where you are wrong.

I hope it helps!
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Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1 [#permalink]

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27 Nov 2015, 13:08

Can someone confirm if my thinking is correct for statement 2, as it doesn't seem to have been used in any answer so far:

x/y > 1 to me says either

Case 1: x and y < 0, and |x| > |y|

Case 2: x and y > 0, and |x| > |y|

These two conditions can then be used in conjunction with statement 1 to show x and y > 0, as if x and y < 0, then |x| must be less than |y| to satisfy statement 1, which is not possible under the two cases above.

Thanks for your help

gmatclubot

Re: Are x and y both positive? (1) 2x - 2y = 1 (2) x/y > 1
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27 Nov 2015, 13:08

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