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(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.

Re: Are X and Y both positive? GMAT PREP CAT [#permalink]

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20 Jul 2010, 12:41

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I found this one easiest to solve by drawing a graph. Clearly 1) and 2) alone are not sufficient as discussed, so what remains to be seen is if 2) adds enough information to 1) to determine if both x and y are positive.

Drawing a quick graph of the line y=x-1/2 we find that the x-intercept of the line is (0.5,0) and the y-intercept is (0,-0.5). From this graph we can clearly see that we don't need to worry about anything in the 4th quadrant (+x/-y is not >1) or the 3rd quadrant (|x|<|y|, therefore x/y is not >1). All that is left is the 1st quadrant, in which x and y are both positive.

Sufficient.
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(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.

Statement (1): x-y = 1/2. We can have x=1,y=1/2. Can also have x=0,y=-1/2. Insufficient. Statement (2): x/y>1. We can have x=3,y=2. Can also have x=-3,y=-2. Insufficient.

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

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04 Jan 2012, 00:28

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C is the answer. Question: Is x > 0 AND y > 0?

Statement 1: 2x - 2y = 1 => 2(x - y) = 1 => x - y = 1/2 This just tells us that the difference is positive. But this can be true for cases when both x and y are positive, and when both x and y are negative. For instance, x = 1.5, y = 1 => x - y = 0.5; also, x = -1, y = -1.5 => x - y = 0.5. Thus, INSUFFICIENT.

Statement 2: x/y > 1 This just tells us that x and y have the same sign. That is, both are positive or both are negative. INSUFFICIENT.

Combining these statements, we can use the same numbers used in Statement 1 to find out that both the cases together do not work for negative numbers. For instance, x = -1, y = -1.5 => x - y = 0.5. However, x/y < 1. This violates statement 2.

Thus, the combination of the given statements tells us that x and y both have to be positive. => x > 0 AND y > 0. SUFFICIENT.
_________________

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

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14 Sep 2013, 22:39

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imhimanshu wrote:

Hello Bunuel, Request you to please provide your comments on the doubt posted here-

Usually, whenever I see combining an inequality and equation, I substitute the value of one of the variable in the inequality and then analyze the effect. So, going by that approach;

x-y=1/2 ---(1) x/y>1 --(2) Substituting the value of x in equation(2)

(y+1/2)/y>1

Lets assume that y is positive-

(y+1/2) > y

1/2>0 --This means that our assumption is true since 1/2 is greater than Zero. Hence, y > 0

Now, Lets assume that y is negative-

Now, here I'm stuck, I know that multiplying by a negative number changes the sign of the inequality. I'm sure that the sign will be changed but what would be the resulting equation. I mean, do we need to replace y with "-y" in the whole equation. Please clarify. Which of the following would be correct then

a) y+1/2 <y b) y+1/2 < -y c) -y+1/2 < -y

Please help. Thanks

Refer to the highlighted portion : Actually you don't have to take 2 cases at this point: The expression you have is : \(\frac{y+0.5}{y}>1 \to 1+\frac{0.5}{y}>1 \to \frac{1}{y}>0\)--> Hence, y>0.

As for your doubt, if y is negative, we cross-multiply it and get : \(y+0.5<y \to 0>0.5\), which is absurd.

If y is negative, then -y would be positive, and for multiplying a positive quantity, you don't need to flip signs. So , yes expression a is correct.
_________________

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

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24 Sep 2015, 16:21

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jitendra31 wrote:

What happens when y = o? X = 0.5 and y = 0, satisfies i and x/y is indeed > 0 since anything divided by 0 is infinity. But, despite this, y is not positive as it '0'. '

It may seem that 0.5/0 = infinity, but this is not the case. If we approach 0 from the positive side, then it looks like 0.5/0 is a REALLY BIG POSITIVE NUMBER 0.5/0.1 = 5 0.5/0.01 = 50 0.5/0.001 = 500 0.5/0.0001 = 5000 0.5/0.00001 = 50000 etc.

But what if we approach 0 from the NEGATIVE side: 0.5/(-0.1) = -5 0.5/(-0.01) = -50 0.5/(-0.001) = -500 0.5/(-0.0001) = -5000 0.5/(-0.00001) = -50000 Here it looks like 0.5/0 will be a REALLY BIG NEGATIVE NUMBER

I guessed and got it right with a 50/50 guess at the end.

What I have done here is this

1) 2x - 2y = 1 hence x - y = \frac{1}{2} {Dividing both side by 2} In sufficient

2) x > y Alone in sufficient

When (1) + (2) We can say that if X is greater than y than x-y will yield a positive result.

Please correct me if I am wrong

First of all: the question is "are x and Y both positive?" not whether "x-y will yield a positive result".

Next, 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 \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

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

See the complete solution of this problem in my previous post.

Re: Are X and Y both positive? GMAT PREP CAT [#permalink]

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30 Sep 2010, 07:15

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Bunuel wrote:

zerotoinfinite2006 wrote:

Manbehindthecurtain wrote:

Are x and Y both positive?

1) 2X-2Y = 1 2) (x/y) > 1

I guessed and got it right with a 50/50 guess at the end.

What I have done here is this

1) 2x - 2y = 1 hence x - y = \frac{1}{2} {Dividing both side by 2} In sufficient

2) x > y Alone in sufficient

When (1) + (2) We can say that if X is greater than y than x-y will yield a positive result.

Please correct me if I am wrong

First of all: the question is "are x and Y both positive?" not whether "x-y will yield a positive result".

Next, 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 \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

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

See the complete solution of this problem in my previous post.

Hope it helps.

I can clearly see how much weak I am in DS . I have no idea how to improve it. I am extremely weak in number system , including these kind of question. And day by day I am getting demoralize that I can't solve these kind of questions.

Anyways, Thanks a lot for your explanation Bunuel. You are genius as always. +1 more .
_________________

I don't want kudos.. I want to see smile on your face if I am able to help you.. which is priceless.

Just to add we can multiply y to both numerator and denominator of x/y the advantage is that the denominator becomes a square i.e in this case \(y^2\) so now we can safely cross multiply in \(xy/y^2>1\) since square of a no. is always +ve \(xy>y^2\) or \(y(x-y)>0\) This is a general method.

More usual way of doing this would be: \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>\) --> \(\frac{x-y}{y}>0\).
_________________

(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.

I was trying to solve this question by plugging in numbers. I too agree that staements A and B both alone are insufficient. SO now by taking both the statements together x>y so let us take x=3/2 and y=1 and plugging this value we can satisfy the equation 2x-2y = 1. Now let us take x=-1 and y=-3/2 and again plugging this value we can satisfy the equation 2x-2y = 1.

So the answer must be E. Please correct me where I am going wrong.

x=-1 and y=-3/2 don't satisfy the second statement: x/y=(-1)/(-3/2)=2/3<1.
_________________

Re: Are X and Y both positive? GMAT PREP CAT [#permalink]

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02 Oct 2012, 19:42

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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.

Bunuel i would like tto know how \(\frac{1}{y}>0\) have this : if I have ( y + 1 - y / 2 ) / y > 0 the result should be \(\frac{1}{2y}>0\) and not \(\frac{1}{y}> 0\)

can you please explain ??'

thanks

@edited ............I have seen the explanation in another answer by you ) Ok
_________________

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

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17 Jan 2013, 04:55

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Manbehindthecurtain wrote:

Are x and y both positive?

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

1. x-y = 1/2 This means that the distance between x and y is 1/2 unit and that x is greater than y. But x and y could be positive such as x=5 and y=4.5, OR x and y could be both negative such as x=-4 and y=-4.5

INSUFFICIENT.

2. x/y > 1 This shows that x and y must be positive meaning they are either both (+) or both (-). ex) x/y = 5/2 OR x/y = -5/-2 = 5/2 still > 1

INSUFFICIENT.

Combine. Let x = 5 and y=9/2: 5/(9/2) = 10/9 > 1 - This means when x and y are both positive it could be a solution to x/y > 1 Let x = -4 and y=-9/2: -4/(-9/2) = 8/9 < 1 - This means when x and y are negative it could not be a solution to x/y > 1

Thus, SUFFICIENT that x and y are both positive.

Answer: C
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Re: Are x and y both positive? (1) 2x-2y = 1 (2) x/y > 1
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