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Re: If x and y are both negative and xy < y^2, which of the
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22 Oct 2013, 09:07

3

5

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.

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

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

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?

Re: If x and y are both negative and xy < y^2, which of the
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23 Oct 2013, 00:30

1

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 \(x > y\).

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

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
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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.

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

Re: If x and y are both negative and xy < y^2, which of the
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12 Feb 2018, 17:07

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

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.)

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

Answer: C.

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
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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.

Answer: C.

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 \(x > y\).

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

Answer: C.

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 Probus
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