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If x and y are both negative and xy < y^2, which of the [#permalink]
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22 Oct 2013, 06:44
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If x and y are both negative and xy < y^2, which of the following must be true? a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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22 Oct 2013, 07:19
... could please someone explain this? I chose A. Because xy < y^2 and both are negative, I thought x < y. So I crossed off answers c), d), and e). And because x < y , x^2 < y^2, so I crossed off answer b) and picked answer a), which seems to be wrong. Where am I wrong?



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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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22 Oct 2013, 08:07
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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22 Oct 2013, 19:57
Bunuel wrote: bulletpoint wrote: If x and y are both negative and xy < y^2, which of the following must be true?
a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x Since \(y\) is negative then after reducing \(xy < y^2\) by \(y\) we get: \(x > y\) which is the same as \(y<x\). Both x^2 and y^2 will be greater than either x or y. Only C fits. Answer: C. is this because x*y< y^2, and we are dividing by y, and that is why we flip the equality sign after? for me, I got the wrong answer because I solved the following way: Since x and y are both negative xy is positive. Thus we divide a positive y over from xy < y^2 to become x< y^2/y, which becomes x<y. Please help me understand why this is not the case?



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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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22 Oct 2013, 23:30
bulletpoint wrote: Bunuel wrote: bulletpoint wrote: If x and y are both negative and xy < y^2, which of the following must be true?
a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x Since \(y\) is negative then after reducing \(xy < y^2\) by \(y\) we get: \(x > y\) which is the same as \(y<x\). Both x^2 and y^2 will be greater than either x or y. Only C fits. Answer: C. is this because x*y< y^2, and we are dividing by y, and that is why we flip the equality sign after? for me, I got the wrong answer because I solved the following way: Since x and y are both negative xy is positive. Thus we divide a positive y over from xy < y^2 to become x< y^2/y, which becomes x<y. Please help me understand why this is not the case? No. We know that \(y\) is negative. Divide \(xy < y^2\) by \(y\) and flip the sign because y is negative to get \(x > y\). Hope it's clear.
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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08 Nov 2013, 09:35
bulletpoint wrote: Bunuel wrote: bulletpoint wrote: If x and y are both negative and xy < y^2, which of the following must be true?
a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x Since \(y\) is negative then after reducing \(xy < y^2\) by \(y\) we get: \(x > y\) which is the same as \(y<x\). Both x^2 and y^2 will be greater than either x or y. Only C fits. Answer: C. is this because x*y< y^2, and we are dividing by y, and that is why we flip the equality sign after? for me, I got the wrong answer because I solved the following way: Since x and y are both negative xy is positive. Thus we divide a positive y over from xy < y^2 to become x< y^2/y, which becomes x<y. Please help me understand why this is not the case? This is how i understand this (btw i did the same mistake) but later realized after plugging in some number (when in doubt plug it) Say X = 1 and Y = 2, XY => 2 < 4 Check. Now divide denominators with 2. LHS => 1 < 2 but this is not true so flip the sign. 1 > 2 Check. Now the equation becomes X > Y. Ans is C.
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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16 Dec 2013, 14:25
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bulletpoint wrote: If x and y are both negative and xy < y^2, which of the following must be true?
a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x I used the exact same thought process as you but after seeing Bunuel's solution I think I understand where my/our intuition is lacking: When it says xy < y^2 you have to remember that both terms on either side of the inequality are POSITIVE. Meaning, change their sign as if they were negative (at least this helps me because I am used to changing the sign of the inequality when we have negatives). So: (value of x, which is negatie)*(value of y, which is negative) = positive, and the same goes for y^2. So simply change their sign so it says (x)*(y) < (y)*(y) > after you've moved the y from left to right you get x < y > x > y And from there, you already know that x and y are smaller numbers than their respective squares (since both are negative), thus only (C) works.



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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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21 Jan 2014, 23:40
bulletpoint wrote: If x and y are both negative and xy < y^2, which of the following must be true?
a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x I'm not sure if this is a correct method to use but here is how I solved it. I just replaced x and y with numbers. a) 2<1<2<1 ( no ) b) 2<1<1<2 ( 3 < 1 not sufficient) c) 2<1<1<4 ( 3<4 sufficient) The answer is C



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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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13 Sep 2015, 23:53
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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21 Nov 2015, 22:22
Hi,
How to get x>y? I am getting y>x.
y^2>xy y^2xy>0 y(yx)>0 So, y>0 or Y>x
So as per this, y>x! Can't figure out where is it going wrong!
"If x and y are both negative and xy < y^2, which of the following must be true?
a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x"



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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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19 Dec 2015, 09:52
aeglorre wrote: bulletpoint wrote: If x and y are both negative and xy < y^2, which of the following must be true?
a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x I used the exact same thought process as you but after seeing Bunuel's solution I think I understand where my/our intuition is lacking: When it says xy < y^2 you have to remember that both terms on either side of the inequality are POSITIVE. Meaning, change their sign as if they were negative (at least this helps me because I am used to changing the sign of the inequality when we have negatives). So: (value of x, which is negatie)*(value of y, which is negative) = positive, and the same goes for y^2. So simply change their sign so it says (x)*(y) < (y)*(y) > after you've moved the y from left to right you get x < y > x > y And from there, you already know that x and y are smaller numbers than their respective squares (since both are negative), thus only (C) works. Hi aeglorre, I am clear and convinced with y<x. can you please tell me about how x < x^2 < y^2 or x^2 < y^2?
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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21 Dec 2015, 00:58
y^2>xy y^2xy>0 y(yx)>0 Since we are given that y<0, we know that yx<0 > y<x. C is the only option that shows y<x hence its the answer
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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21 Dec 2015, 01:06
shapla wrote: aeglorre wrote: bulletpoint wrote: If x and y are both negative and xy < y^2, which of the following must be true?
a) x < y < x^2 < y^2 b) x < y < y^2 < x^2 c) y < x < x^2 < y^2 d) x^2 < y^2 < y < x e) y^2 < x^2 < y < x I used the exact same thought process as you but after seeing Bunuel's solution I think I understand where my/our intuition is lacking: When it says xy < y^2 you have to remember that both terms on either side of the inequality are POSITIVE. Meaning, change their sign as if they were negative (at least this helps me because I am used to changing the sign of the inequality when we have negatives). So: (value of x, which is negatie)*(value of y, which is negative) = positive, and the same goes for y^2. So simply change their sign so it says (x)*(y) < (y)*(y) > after you've moved the y from left to right you get x < y > x > y And from there, you already know that x and y are smaller numbers than their respective squares (since both are negative), thus only (C) works. Hi aeglorre, I am clear and convinced with y<x. can you please tell me about how x < x^2 < y^2 or x^2 < y^2? Hi, we know x>y and it is given that both are negative.. this will mean the numeric value(or distance from 0 on a number line ) of y>x... for example if x=3, y>3 or 4,5,.. etc since the numeric value of y>x , the square of y will be > that of x, as all integers turn positive on squaring.. hope it helped
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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27 Aug 2017, 04:13
I'm completely dumbfounded here.
The approach I tried:
xy<y^2 xy  y^2<0 y(xy)<0
Either y<0 and x<y Or y>0 and x>y
As y<0, only the first possibility stands and the second one is ruled out. This suggests that x<y !!
Where am I making the mistake?



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Re: If x and y are both negative and xy < y^2, which of the [#permalink]
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27 Aug 2017, 04:18
TheMastermind wrote: I'm completely dumbfounded here.
The approach I tried:
xy<y^2 xy  y^2<0 y(xy)<0
Either y<0 and x<y Or y>0 and x>y
As y<0, only the first possibility stands and the second one is ruled out. This suggests that x<y !!
Where am I making the mistake? Hi.. y(xy)<0 means ONLY one of x or xy is NEGATIVE and other POSITIVE. Because +*will be or<0 So if y<0, xy>0 or x>y..
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Re: If x and y are both negative and xy < y^2, which of the
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