Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: If x and y are both negative and xy < y^2, which of the [#permalink]
Show Tags
22 Oct 2013, 08:19
... 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.
Re: If x and y are both negative and xy < y^2, which of the [#permalink]
Show Tags
22 Oct 2013, 09:07
2
2
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 [#permalink]
Show Tags
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 [#permalink]
Show Tags
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 \(x > y\).
Status: Please do not forget to give kudos if you like my post
Joined: 19 Sep 2008
Posts: 107
Location: United States (CA)
Re: If x and y are both negative and xy < y^2, which of the [#permalink]
Show Tags
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.
_________________
Re: If x and y are both negative and xy < y^2, which of the [#permalink]
Show Tags
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 [#permalink]
Show Tags
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?
_________________
Re: If x and y are both negative and xy < y^2, which of the [#permalink]
Show Tags
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
_________________
Please contact me for super inexpensive quality private tutoring
My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876
Re: If x and y are both negative and xy < y^2, which of the [#permalink]
Show Tags
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
_________________
Re: If x and y are both negative and xy < y^2, which of the [#permalink]
Show Tags
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 [#permalink]
Show Tags
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.)
Answer: C
_________________
Jeffery Miller Head of GMAT Instruction
GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions