Is x/y 0 (2) y

Question

Is  \frac{x}{y} < xy  ?

(1)  xy > 0

(2)  y < -1

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chetan2u · Answer

\frac{x}{y} < xy xy-\frac{x}{y} >0 \frac{xy^2-x}{y}>0 \frac{x(y-1)(y+1)}{y}>0 This means both the numerator x(y-1)(y+1) and the denominator y have same sign. We have to check for the following cases A) When the denominator or y<0 in \frac{x(y-1)(y+1)}{y}>0 , the numerator x(y-1)(y+1)<0, and we get two cases (i) 0>y>-1, and (y-1)(y+1)<0, Now x*(y-1)(y+1)<0, that is x*(negative)<0. This means x>0....(I) (ii) y<-1, and (y-1)(y+1)>0, Now x*(y-1)(y+1)<0, that is x*(positive)<0. This means x<0.......(II) B) When the denominator or y>0 in \frac{x(y-1)(y+1)}{y}>0 , the numerator x(y-1)(y+1)>0, and we get two cases (i) 00, that is x*(negative)>0. This means x<0....(III) (ii) y>1, and (y-1)(y+1)>0, Now x*(y-1)(y+1)<0, that is x*(positive)>0. This means x>0.......(IV) Our statements should clearly give us one of the 4 options as given above. (1) xy > 0 Both x and y have same sign. So case II and IV possible. But we cannot say whether y falls in the ranges y<-1 or y>1. Insufficient (2) y < -1 Nothing about x. Combined xy>0 and y<-1. Exactly as per Case IV above. Answer is yes. C

TarunKumar1234 · Answer

Is x/y0 Either, x>0 and (y-1/y) >0, it means, x>0 and y>1 or, x<0 and (y-1/y) <0, it means, x<0 and 00 It means, either, x>0 and y>0 or, x<0 and y<0, which doesn't have always, x/yxy. Sufficient. So, I think C.

akadiyan · Answer

Is x/y < xy ?

(1) xy>0
We know that both x and y are of same sign. Both x and y can be positive or both negative.

consider x=2, y=1, then x/y = xy
consider x=1, y=2 , then x/y< xy

we get 2 different answers, so Option 1 - Not sufficient.

(2) y<−1
We do not have any information about X, so Option 2 - Not sufficient.

lets consider 1 and 2 together

From statement 2 we know that y < -1, now since xy >0, we know x is also < 0
we know that both x and y are negative and y < -1. So x/y will always be lesser than xy.

Option 1 and 2 together - SUFFICIENT

Ans C

swatib28 · Answer

chetan2u

Can you please help explain , how did you get into the below

We have to check for the following cases
A) y<0
0>y>-1........(y-1)(y+1)<0, so x>0....(I)
y<-1........(y-1)(y+1)>0, so x<0.......(II)
B) y>0
0<y<1........(y-1)(y+1)<0, so x<0.....(III)
y>1.......(y-1)(y+1)>0, so x>0........(IV)

chetan2u · Answer

Hi

The fraction x(y-1)(y+1)/y>0.
This means both numerator and denominator have same sign. 
Four options are thereafter analysed 
I have added few details in the original post. Please go through it.

swatib28 · Answer

Thank you so much  chetan2u  . I got it now.Just wondering, how are we going to understand the 4 cases which we bifurcated to be analyzed in exam?

GMATGuruNY · Answer

Since both statements indicate that y is NONZERO, y^2>0 , implying that we can safely multiply the question stem by y^2 : \frac{x}{y}*y^2 < xy*y^2 xy < xy^3 Question stem, rephrased: Is xy < xy^3 ? Statement 1: xy > 0 Case 1: x=y=1 In this case, xy = xy^3 , so the answer to the rephrased question stem is NO. Case 2: x=1 and y=2 In this case, xy < xy^3 , so the answer to the rephrased question stem is YES. Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT. Statement 2: y < -1 No information about x. INSUFFICIENT. Statements combined: xy > 0 and y < -1 Since xy > 0, we can safely divide the rephrased question stem by xy: \frac{xy}{xy} < \frac{xy^3}{xy} 1 < y^2 The question stem becomes: Is y^2 > 1 ? Since y < -1, the answer is YES. SUFFICIENT. C

Rickooreo · Answer

gmatophobia   chetan2u 
 
Can you please share your approach. Also, (@cheatan2u) can please make me understand why and how to make the two cases as mentioned in the solution provided

gmatophobia · Answer

Here is how I would approach this question.

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Is x/y < xy ? (1) xy > 0 (2) y < -1

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