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

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

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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
[Reveal] Spoiler: OA
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]

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New post 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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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.
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Re: If x and y are both negative and xy < y^2, which of the [#permalink]

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New post 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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New post 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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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If x and y are both negative and xy < y^2, which of the [#permalink]

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New post 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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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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New post 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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Re: If x and y are both negative and xy < y^2, which of the [#permalink]

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New post 21 Nov 2015, 22:22
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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New post 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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New post 21 Dec 2015, 00:58
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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New post 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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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x and y are both negative and xy < y^2, which of the [#permalink]

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New post 27 Aug 2017, 04:13
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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Re: If x and y are both negative and xy < y^2, which of the [#permalink]

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New post 27 Aug 2017, 04:18
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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] 27 Aug 2017, 04:18
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