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# If x and y are both negative and xy < y^2, which of the

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Joined: 02 Jul 2013
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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, 07:44
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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, 09:07
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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

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.

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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, 08:19
1
... 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, 20:57
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.

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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23 Oct 2013, 00:30
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.

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, 10:35
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.

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, 15:25
1
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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22 Jan 2014, 00:40
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)

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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, 23:22
Hi,

How to get x>y? I am getting y>x.

y^2>xy
y^2-xy>0
y(y-x)>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, 10:52
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, 01:58
y^2>xy
y^2-xy>0
y(y-x)>0
Since we are given that y<0, we know that y-x<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, 02:06
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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27 Aug 2017, 05:13
I'm completely dumbfounded here.

The approach I tried:

xy<y^2
xy - y^2<0
y(x-y)<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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Posts: 7756
Re: If x and y are both negative and xy < y^2, which of the  [#permalink]

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27 Aug 2017, 05:18
1
TheMastermind wrote:
I'm completely dumbfounded here.

The approach I tried:

xy<y^2
xy - y^2<0
y(x-y)<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(x-y)<0 means ONLY one of x or x-y is NEGATIVE and other POSITIVE. Because +*-will be -or<0
So if y<0, x-y>0 or x>y..
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Re: If x and y are both negative and xy < y^2, which of the  [#permalink]

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06 Feb 2018, 15:39
Top Contributor
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

We have xy < y²
If we divide both sides by y we must REVERSE the inequality sign (because we are dividing by a NEGATIVE value).
So, we get: x > y
This allows us to ELIMINATE answer choices A and B (since they suggest that x < y)

From here, let's PLUG IN values for x and y such that they are both negative AND x > y
Let's try x = -1 and y = -2
This means x² = 1 and y² = 4
When we arrange the four values in ascending order we get: y < x < x² < y²

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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 Feb 2018, 23:32
Given that xy < y^2

Divide both sides by y^2

xy/y^2 < y^2/y^2

x/y < 1

It is given that x and y are both negative

So, x/y < 1 is possible only when y is more negative than x (for example, x = -1, y = -2)

So, then we know that y < x < 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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12 Feb 2018, 17:07
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

We see that in all the answer choices we have to compare the following four quantities: x, y, x^2 and y^2. We are given that x and y are both negative and xy < y^2.

We see that both xy and y^2 are positive since the product of two negative quantities is positive and the square of a nonzero quantity is always positive. However, we divide both sides of the inequality by y (a negative quantity), we have:

x > y

Since x^2 and y^2 are both positive, we see that y is the smallest of the four quantities. The only answer choice that has y as the smallest quantity is choice C. Thus, it’s the correct answer.

(Note: We don’t have to analyze, in this case, which is the larger quantity between x^2 and y^2 since y has to be the smallest quantity. However, if we have to, it is always true that if y < x < 0, then y^2 > x^2 > 0. For example, -3 < -2, but (-3)^2 > (-2)^2 since 9 > 4.)

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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 Sep 2018, 10:38
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.

Hello I had learnt on Gmat club itself that one shouldn't reduce a quadratic inequality, because then we end up assuming the variables to be non- zero. Is the same rule applicable for linear inequalities too?
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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 Sep 2018, 21:33
Tanisha_shasha 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.

Hello I had learnt on Gmat club itself that one shouldn't reduce a quadratic inequality, because then we end up assuming the variables to be non- zero. Is the same rule applicable for linear inequalities too?

MULTIPLYING/DIVIDING AN INEQUALITY BY A NUMBER

1. Whenever you multiply or divide an inequality by a positive number, you must keep the inequality sign.
2. Whenever you multiply or divide an inequality by a negative number, you must flip the inequality sign.
3. Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know the sign of it or are not certain that variable (or the expression with a variable) doesn't equal to zero.

Here 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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16 Sep 2018, 21:47
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.

Hi Bunuel & chetan2u,
I need your help on this .
Most of my questions i have used this property that we can move the term from one side of inequality to to other without changing the sign of the inequality.
here is what i mean
say if x<y
then we can say x-y<0
However if i use the same poperty here i don't get the result where am i going wrong?
we are given that x<0, y<0
also xy<$$y^2$$
so we can write xy- $$y^2$$<0
y(x-y)<0
so we have y<0 and or x-y<0
or we can say y<0 and or x<y
or we can say
x<y<0
then as per this we get option B .
I understand the solution given by Bunuel, but please help me correct my error in my understanding of how i solved.

Thanks
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Re: If x and y are both negative and xy < y^2, which of the   [#permalink] 16 Sep 2018, 21:47

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