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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.

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.

I am still struggling to understand how come both together are sufficient? What is common in both the answer choices that makes c a correct choice?

Here is the logic for C:

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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13 Jun 2010, 11:41

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gsaxena26 wrote:

x/y>1

But x can be both +ve and negative. But if x>y then answer choice will be true as it can only be positive when x >y. Is my understanding correct?

Your reasoning is almost correct. Statement II doesn't say x>y, it only says x/y>1... Be careful there is a difference, this is the mistake you made initially

Taking this statement alone, it means x and y are both same sign (either both + or both -) and |x|>|y|. Thus, it is insufficient and you need to combine it with Statement I.

From statement I, you know x=y+1/2, hence x>y. If you know x>y and also |x|>|y| this means both are positive.

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

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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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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.

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).
_________________

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)
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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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my doubt is regarding x/y - 1 = (x - y)/y (can we do this even if do not know the sign of y)

Please advise!

Yes. We are simply subtracting one fraction (y/y) from another (x/y). Again, we are not multiplying or dividing the inequality by anything.
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We need to determine whether x and y are both positive.

Statement One Alone:

2x – 2y = 1

Simplifying statement one we have:

2(x – y) = 1

x – y = ½

The information in statement one is not sufficient to determine whether x and y are both positive. For instance if x = 1 and y = ½, x and y are both positive; however if x = -1/2 and y = -1, x and y are not both positive. We can eliminate answer choices A and D.

Statement Two Alone:

x/y > 1

Using the information in statement two, we see that x and y can both be positive or both be negative. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using the information in statements one and two we know that x – y = ½ and that x/y > 1. Isolating x in the equation we have: x = ½ + y. We can now substitute ½ + y for x in the inequality x/y > 1 and we have:

(1/2 + y)/y > 1

(1/2)/y + y/y > 1

1/2y + 1 > 1

1/2y > 0

Thus, y must be greater than zero. Also, since x/y is greater than one, x also must be greater than zero.

Answer: C
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GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

(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.

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