If x and y are both negative and xy

Question

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

Bunuel · Accepted Answer

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.

Bunuel · Answer

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.

vogelleblanc · Answer

... 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?

bulletpoint · Answer

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?

pradeepss · Answer

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.

aeglorre · Answer

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.

bytatia · Answer

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

longfellow · Answer

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"

shapla · Answer

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?

NoHalfMeasures · Answer

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

chetan2u · Answer

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

profileusername · Answer

I'm completely dumbfounded here. The approach I tried: xy0 and x>y As y<0, only the first possibility stands and the second one is ruled out. This suggests that x

chetan2u · Answer

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..

BrentGMATPrepNow · Answer

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 Answer: C RELATED VIDEO https://www.youtube.com/watch?v=Od98BOicIq4

Nived · Answer

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

JeffTargetTestPrep · Answer

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

Tanisha_shasha · Answer

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?

Bunuel · Answer

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.

Probus · Answer

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

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

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