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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.

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

Hope it helps.

Sorry for the bump but could you elaborate on the last part where you go from x/y>1 to (x-y)/y>0 to 1/y>0 ..?

I don't quite follow this algebra

\(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>\) --> \(\frac{x-y}{y}>0\). Now, substitute \(x=y+\frac{1}{2}\) there to get \(\frac{1}{2y}>0\), which further simplifies to \(\frac{1}{y}>0\).

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

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

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

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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

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 example, if x = 1 and y = ½, x and y are both positive; however, if x = -1/2 and y = -1, x and y are both negative. 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. Since x = ½ + y, x also must be greater than zero.

Answer: C
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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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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.

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 .

Check Number Theory chapter of Math Book for more on number properties (link in my signature).
_________________

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: \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>\) --> \(\frac{x-y}{y}>0\).
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