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

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chetan2u

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16 Sep 2018, 21:58

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Probus

Bunuel

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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

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

You are correct till \(xy-y^2<0.....y(x-y)<0\)..
But thereafter you have gone wrong..
You are further tight that y<0..
But should be the other term x-y..
A) if x-y is negative as you have taken .. then both y and x-y are negative and -*->0
So case B
B) x-y>0 then only -*+=-

So x-y>0...x>y

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23 Jul 2020, 10:25

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but option c is not valid for value of y and x between -1 and 0 , experts please help

ayushkumar22941

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23 Jul 2020, 10:31

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Bunuel

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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

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.

what about value of x and y in between -1 and zero , that does not satisfy option c

Bunuel

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23 Jul 2020, 10:39

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ayushkumar22941

Bunuel

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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

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.

what about value of x and y in between -1 and zero , that does not satisfy option c

Please give an example of x and y, where x and y are both negative, xy < y^2 and y < x < x^2 < y^2 is not true.

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26 Nov 2020, 23:03

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given both x & y are -ve and \(xy < y^2\)
test with ( x,y) as ( -1,-3 ) & ( -1/2 ,-3/2)
plugin the values of (x,y)
we see \(y < x < x^2 < y^2\) holds valid only
option C

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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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11 Jan 2023, 16:02

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Given:
X and Y are negative integers, and \(xy<y^2.\)

We can reduce the inequality by dividing by y. Note that because \(y<0\), we must flip the sign. Therefore, we have \(y<x\). The only option that matches is C.

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27 Jun 2023, 11:18

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Given: x and y both -ve and xy < y\(^2\).

Asked: Which option Must Be True?

Solution:

Now we know, xy < y\(^2\), we can write this as y(y-x)>0

Now we know this can be true when either both "y" and "y-x" are of the same sign. Here know that both x and y are -ve.

So, as per our analysis, y<0 and y-x<0 which can be written as x>y.

Hence, we know that x is greater than y, so we assume simple values for solving. Let x = -1 and y = -2.

Only Option (C) will always be true as -> -2 < -1 < 1 < 4 is always true.

\(Answer:\) C.

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01 Oct 2024, 07:45

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x and y both are negative.

What do we need to keep in mind then?
-> Multiplying/Dividing by x or y will change the direction of the sign.

Now,

\(xy < y^2\)
\(x > y\) (Dividing both sides by y and changing the sig, coz y is negative)

We've come down to Opt C,D,E.

Now,

\(x^2 and y^2\) both will be definitely bigger than x and y....only option C agrees with us.

Hence, C.

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29 Nov 2024, 20:40

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Hello,

Initially, I did the same too.

If y is -ve then x-y => x-(-y) is +ve implies x>y
If y is +ve then X has to be <y

And since X and y are negative and we know x>y implies x^2 <y^2 ; Eg. -1>-2 => Their squares = 1<4

Thats why C

vogelleblanc

... 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?

ashKing12

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28 Jun 2025, 08:50

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\(\\
xy<y^2\\
\\
\)
\(y(x-y)<0\)

as \(y=-ve\) so for inquality to hold \( x-y>0\) thus \( x>y\)

eliminate option A and B

now as \(y = -ve\) so \(y^2= +ve\) thus \( y<y^2 \)
eliminate D and E

answer: option C

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