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Is x/y > 3? (1) 3x > 9y (2) y > 3y

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Bunuel
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Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   27 Oct 2016, 05:04
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Is x/y > 3?

(1) 3x > 9y

(2) y > 3y
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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   27 Oct 2016, 08:09
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Bunuel wrote:
Is x/y > 3?

(1) 3x > 9y

(2) y > 3y


With these types of problems, one must always consider the sign of the variables before you multiply them. If y>0, then the question is asking if \(x>3y\). If y<0, then the question is asking if \(x<3y\).

(1) 3x > 9y. Divide by 3 since it's always positive. \(x>3y\). We don't know if y is positive or negative. Insufficient.

(2) y > 3y. This tells us that y<0. We still can't say anything about the relationship between x and y. Obviously insufficient.

Taking them both together, we know that y<0 and thus that the question is asking if \(x<3y\). Statement 1 says that \(x>3y\). Thus \(x/y <3\). Sufficient.

Answer: C
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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   03 Nov 2016, 03:08
OE is E
1) x>3y
8>-6 (3*-2) or 8>6 (3*2). Not sufficient

2) y>3y
y is negative. But not sufficient

1)+2) - x=8, 8>-ve y
x=-2 > -ve y (say 3y=-6).

Nice question. Thanks.
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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   04 Nov 2016, 14:03
speedilly wrote:
Bunuel wrote:
Is x/y > 3?

(1) 3x > 9y

(2) y > 3y


With these types of problems, one must always consider the sign of the variables before you multiply them. If y>0, then the question is asking if \(x>3y\). If y<0, then the question is asking if \(x<3y\).

(1) 3x > 9y. Divide by 3 since it's always positive. \(x>3y\). We don't know if y is positive or negative. Insufficient.

(2) y > 3y. This tells us that y<0. We still can't say anything about the relationship between x and y. Obviously insufficient.

Taking them both together, we know that y<0 and thus that the question is asking if \(x<3y\). Statement 1 says that \(x>3y\). Thus \(x/y <3\). Sufficient.

Answer: C


I agree with @speedilly. How can the answer be E?? Can someone give algebraic explanation??


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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   05 Nov 2016, 01:54
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acegmat123 wrote:
speedilly wrote:
Bunuel wrote:
Is x/y > 3?

(1) 3x > 9y

(2) y > 3y


With these types of problems, one must always consider the sign of the variables before you multiply them. If y>0, then the question is asking if \(x>3y\). If y<0, then the question is asking if \(x<3y\).

(1) 3x > 9y. Divide by 3 since it's always positive. \(x>3y\). We don't know if y is positive or negative. Insufficient.

(2) y > 3y. This tells us that y<0. We still can't say anything about the relationship between x and y. Obviously insufficient.

Taking them both together, we know that y<0 and thus that the question is asking if \(x<3y\). Statement 1 says that \(x>3y\). Thus \(x/y <3\). Sufficient.

Answer: C


I agree with speedilly. How can the answer be E?? Can someone give algebraic explanation??


Sent from my iPhone using GMAT Club Forum mobile app


The correct answer is C.

Is x/y > 3?

(1) 3x > 9y --> if y is positive, then x/y > 3 (answer YES) but if y is negative, we should flip the sign when dividing by negative value and we'll get x/y < 3 (answer NO). Not sufficient.

(2) y > 3y --> this implies that y < 0. Not sufficient.

(1)+(2) When combining we have that y < 0 and thus the second case from (1) x/y < 3. Sufficient.

Answer: C.

Hope it's clear.
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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   08 Nov 2016, 04:03
Hi Bunuel , why are we not considering the sign of x? What AM I missing?
Thanks

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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   08 Nov 2016, 06:30
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rakaisraka wrote:
Hi Bunuel , why are we not considering the sign of x? What AM I missing?
Thanks

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We are not multiplying/dividing by x anywhere in the solution, so it's irrelevant.
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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   25 Dec 2016, 06:10
Bunuel, dont both x and y have to be of same sign for x/y>3? Hence shouldnt we be also bothered about x's sign?

Thank you
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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   03 Sep 2020, 07:37
Bunuel wrote:
acegmat123 wrote:
speedilly wrote:
Bunuel wrote:
Is x/y > 3?

(1) 3x > 9y

(2) y > 3y


With these types of problems, one must always consider the sign of the variables before you multiply them. If y>0, then the question is asking if \(x>3y\). If y<0, then the question is asking if \(x<3y\).

(1) 3x > 9y. Divide by 3 since it's always positive. \(x>3y\). We don't know if y is positive or negative. Insufficient.

(2) y > 3y. This tells us that y<0. We still can't say anything about the relationship between x and y. Obviously insufficient.

Taking them both together, we know that y<0 and thus that the question is asking if \(x<3y\). Statement 1 says that \(x>3y\). Thus \(x/y <3\). Sufficient.

Answer: C


I agree with speedilly. How can the answer be E?? Can someone give algebraic explanation??


Sent from my iPhone using GMAT Club Forum mobile app


The correct answer is C.

Is x/y > 3?

(1) 3x > 9y --> if y is positive, then x/y > 3 (answer YES) but if y is negative, we should flip the sign when dividing by negative value and we'll get x/y < 3 (answer NO). Not sufficient.

(2) y > 3y --> this implies that y < 0. Not sufficient.

(1)+(2) When combining we have that y < 0 and thus the second case from (1) x/y < 3. Sufficient.

Answer: C.

Hope it's clear.


But by combining both, y is negative then in x>3y, x could be either positive or positive. If x is positive, then x/y will be negative, but x if is negative then x/y will be positive giving us two different answers. So could answer be (C)? Where am I missing?
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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink] New post   04 May 2023, 04:53
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Re: Is x/y > 3? (1) 3x > 9y (2) y > 3y [#permalink]
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