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# If 0 > x > y, then which of the following must be true?

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Math Expert
Joined: 02 Sep 2009
Posts: 58381
If 0 > x > y, then which of the following must be true?  [#permalink]

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23 Jan 2019, 02:47
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Difficulty:

55% (hard)

Question Stats:

57% (01:43) correct 43% (01:40) wrong based on 69 sessions

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If $$0 > x > y$$, then which of the following must be true?

I. $$x^2 – y^2 < 0$$

II. $$\frac{1}{x^2} < 1$$

III. $$\frac{x + y}{x} > 0$$

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

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Joined: 09 Mar 2018
Posts: 996
Location: India
Re: If 0 > x > y, then which of the following must be true?  [#permalink]

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23 Jan 2019, 03:07
Bunuel wrote:
If $$0 > x > y$$, then which of the following must be true?

I. $$x^2 – y^2 < 0$$

II. $$\frac{1}{x^2} < 1$$

III. $$\frac{x + y}{x} > 0$$

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

IMO D

If $$0 > x > y$$, this will mean x and y are -ive numbers, integers or fractions for that matter

I. $$x^2 – y^2 < 0$$

x = - 0.1 and y =- 0.2, 0.01 - 0.04 < 0, Yes

x = -1 and y = -2, 1 - 4 < 0, Yes
x will always be greater than y, in which y will have a greater value when squared

II. $$\frac{1}{x^2} < 1$$
this wont be true all the time,

x = - 0.1, 1/0.01 = 100 < 1, No

x = -2, 1/4 = 0.25 < 1, Yes

III. [m]\frac{x + y}{x} > 0

x = -1, y= -2, -1-2/-1 > 0, 3 > 0, Yes

x = -1/4, y =-1/2, Yes
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Re: If 0 > x > y, then which of the following must be true?  [#permalink]

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23 Jan 2019, 06:44
from given expression it can be quoted that both x & y are +ve

so solve for expression using values of x & y as
x= 1, 1/2 and y = 2, 3/3

we find that options 1 & 3 stand true

so
IMO D

Bunuel wrote:
If $$0 > x > y$$, then which of the following must be true?

I. $$x^2 – y^2 < 0$$

II. $$\frac{1}{x^2} < 1$$

III. $$\frac{x + y}{x} > 0$$

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
Intern
Joined: 23 Jan 2019
Posts: 1
Re: If 0 > x > y, then which of the following must be true?  [#permalink]

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23 Jan 2019, 07:13
x= -1
y= -2

Option1::
(-1)^2-(-2)^2= 1-4= -3<0

Option 2::
1/(-1)^2= 1
Is not true

Option 3::

-1-2/-1= 3>0

Posted from my mobile device
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Joined: 03 Jun 2019
Posts: 1691
Location: India
Re: If 0 > x > y, then which of the following must be true?  [#permalink]

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08 Oct 2019, 02:46
Bunuel wrote:
If $$0 > x > y$$, then which of the following must be true?

I. $$x^2 – y^2 < 0$$

II. $$\frac{1}{x^2} < 1$$

III. $$\frac{x + y}{x} > 0$$

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

I. $$x^2 – y^2 < 0$$
Since 0>x > y ; |x| < |y|; x^2 - y^2 < 0
MUST BE TRUE

II. $$\frac{1}{x^2} < 1$$
If 0>x>-1; 1/x^2 > 1
If x<-1; 1/x^2 < 1
COULD BE TRUE

III. $$\frac{x + y}{x} > 0$$
1 + y/x > 0 ; Since y/x > 1
MUST BE TRUE

IMO D
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Re: If 0 > x > y, then which of the following must be true?   [#permalink] 08 Oct 2019, 02:46
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