Gmater-1 Posted September 11, 2008 Share Posted September 11, 2008 How to solve this inequality within 2 mins :hmm: Is xy 1). (X^3 * Y^5)/ (X*Y^2) 2). |X| - |y| OA will be provided later Quote Link to comment Share on other sites More sharing options...
namitb Posted September 11, 2008 Share Posted September 11, 2008 Is the answer E Quote Link to comment Share on other sites More sharing options...
al_abc_def Posted September 11, 2008 Share Posted September 11, 2008 (1) => x^2*y*3 => y (2) => x>0;y0 (not suff) together: => x>0;y Quote Link to comment Share on other sites More sharing options...
ankit0uc Posted September 11, 2008 Share Posted September 11, 2008 1 gives you (XY)^2 8 Y Implies Y not suff 2. This one is tricky, use number line concept. This inequality will never hold if both x and y are positive. This will hold if and only if. either x is + or Y is negative. Or when both x and y are negative with the condition that x> y. Implies more info needed, and we dont get this info even if we combine two conditions. Ans E Quote Link to comment Share on other sites More sharing options...
cin45220 Posted September 11, 2008 Share Posted September 11, 2008 Ans is E as per ankit's explaination (but not sure why he said Ans is B is the first line) -BusBuy Quote Link to comment Share on other sites More sharing options...
gmatissimple Posted September 11, 2008 Share Posted September 11, 2008 Is xy 1). (X^3 * Y^5)/ (X*Y^2) 2). |X| - |y| Stem1: x^2*y^3 x^2 is always positive. And hence, y^3 is -ve, which leads to the -ve value of y. So, xy Stem2: |X| - |y| =>|x-0|-|y-0| Meaning that difference between the distance between 0 and x and between 0 and y is less than the distance between x and y. It can be shown in the number line as below: -3(x)........-2..........-1(y)...........0............1(y).........2................3(x). We are not sure where is the position of x and y. NOT SUFF. Answer: A. Quote Link to comment Share on other sites More sharing options...
prep in progress Posted September 12, 2008 Share Posted September 12, 2008 1. simplifies to x^2y^3 which is not suff since x could be either +ve or -ve. 2. is a property of mod variables that |x| - |y| so it doesn't give any information, since there is no other relevant information about x and y... So my pick is E Quote Link to comment Share on other sites More sharing options...
lesfeaves Posted September 12, 2008 Share Posted September 12, 2008 A for me Quote Link to comment Share on other sites More sharing options...
lesfeaves Posted September 12, 2008 Share Posted September 12, 2008 A for me Quote Link to comment Share on other sites More sharing options...
Retake_GMAT Posted September 12, 2008 Share Posted September 12, 2008 Is xy We need to check if x and y have opposite signs? St I) (X^3 * Y^5)/ (X*Y^2) Hence x^2*y^3 St II) |X| - |y| Try same sign and opposte signs of x and y. Let x=4, y=-3 |4|-|-3| = 4 -3 =1 |4-(-3)|= 7 Let x=4, y=3 |4|-|3| = 4 -3 =1 |4-3|=1 Hence A Quote Link to comment Share on other sites More sharing options...
Gmater-1 Posted September 12, 2008 Author Share Posted September 12, 2008 A is NOT the answer guys. Gurus: Makumajon,Kroviddy,Retake pls step in. Quote Link to comment Share on other sites More sharing options...
Retake_GMAT Posted September 12, 2008 Share Posted September 12, 2008 I am surprised how A is not the Answer. Let's wait for Makumajon or Queen to comment Quote Link to comment Share on other sites More sharing options...
chandak_anand Posted September 12, 2008 Share Posted September 12, 2008 I am surprised how A is not the Answer. Let's wait for Makumajon or Queen to comment Answer cannot be A because in 1) after removing common factors (xy^2) we are left with x^2y^3 Quote Link to comment Share on other sites More sharing options...
Retake_GMAT Posted September 12, 2008 Share Posted September 12, 2008 Answer cannot be A because in 1) after removing common factors (xy^2) we are left with x^2y^3 As x is raised to an even power,whether x is an integer or fraction or +ve or -ve, it is always +ve. Comment? Quote Link to comment Share on other sites More sharing options...
Shooter Posted September 12, 2008 Share Posted September 12, 2008 Retake I guess we are not bothered abt the sign of x^2 ! we need the sign of x. Knowing that x^2 is positive you cant judge whether x is +ve or -ve. EX: we already know that y is -ve and x^2 is +ve Lets take two cases: y is -ve ; x is -ve ( satisfies x^2 is +ve) xy>0 (II) y is -ve ; x is +ve ( satisfies x^2 is +ve) xy Quote Link to comment Share on other sites More sharing options...
Retake_GMAT Posted September 12, 2008 Share Posted September 12, 2008 Retake I guess we are not bothered abt the sign of x^2 ! we need the sign of x. Knowing that x^2 is positive you cant judge whether x is +ve or -ve. EX: we already know that y is -ve and x^2 is +ve Lets take two cases: y is -ve ; x is -ve ( satisfies x^2 is +ve) xy>0 (II) y is -ve ; x is +ve ( satisfies x^2 is +ve) xy Thanks shooter Got it!!. Quote Link to comment Share on other sites More sharing options...
finsisher Posted September 12, 2008 Share Posted September 12, 2008 STmt-1, y is -ve stmt-2 , nothing extra, cant say wheather XY> or Quote Link to comment Share on other sites More sharing options...
Gmater-1 Posted September 12, 2008 Author Share Posted September 12, 2008 Still no solid conclusion ..:hmm: Quant guru's not pouring in ?? :grad::grad:800Bob, Kroviddy, Makumajon, Queen09 ??:rolleyes: Quote Link to comment Share on other sites More sharing options...
Queen09 Posted September 12, 2008 Share Posted September 12, 2008 IMO E Stmt1: (X^3 * Y^5)/ (X*Y^2) => (x,y) can be in III or IV quadrant. so xy can be +ve or -ve. INSUFF Stmt 2: |X| - |y| (x,y) can be in any four quad. so xy can be +ve or -ve. INSUFF combining, can be in III or IV quadrant. so xy can be +ve or -ve. Hence E Quote Link to comment Share on other sites More sharing options...
namitb Posted September 12, 2008 Share Posted September 12, 2008 Hi , Queen09 can you please explain the 2nd statement , x, y can be in all the 4 quadrants , I tried some values and it seems x, y can be in all quadrants but the first quadrant. can you please explain .. Quote Link to comment Share on other sites More sharing options...
ankit0uc Posted September 12, 2008 Share Posted September 12, 2008 Those who are suggesting A. Just wondering how do you know that x is + from condition1 Quote Link to comment Share on other sites More sharing options...
Queen09 Posted September 12, 2008 Share Posted September 12, 2008 Hi , Queen09 can you please explain the 2nd statement , x, y can be in all the 4 quadrants , I tried some values and it seems x, y can be in all quadrants but the first quadrant. can you please explain .. Take (-5, -6) or (5 , -6) (5, 6) or (-5, 6) Quote Link to comment Share on other sites More sharing options...
chandak_anand Posted September 12, 2008 Share Posted September 12, 2008 As x is raised to an even power,whether x is an integer or fraction or +ve or -ve, it is always +ve. Comment? My job done by "shooter". Guess u know why B is in-correct Quote Link to comment Share on other sites More sharing options...
bose Posted September 12, 2008 Share Posted September 12, 2008 from 1 we know x -ve or +ve, and y-ve insuff from 2 we see x>0, y> 0 or x>0 y insuff 1+2 we only need to see if from 2 we can show that x>0 always since y but (-3,-5) also satisfies 2 so ans E Quote Link to comment Share on other sites More sharing options...
kool_sunny Posted September 12, 2008 Share Posted September 12, 2008 Hi Queen, Can you please explain in more detail abt how you analyzed Quadrants ??:hmm: IMO E Stmt1: (X^3 * Y^5)/ (X*Y^2) => (x,y) can be in III or IV quadrant. so xy can be +ve or -ve. INSUFF Stmt 2: |X| - |y| (x,y) can be in any four quad. so xy can be +ve or -ve. INSUFF Quote Link to comment Share on other sites More sharing options...
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