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

Question

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 M37-48

Bunuel · Accepted Answer

Official Solution: 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 We are given that 0 > x > y , so both x and y are negative and y is further from 0 than x . Check the options: I. x^2 – y^2 < 0 : • x^2 < y^2 • |x| < |y| • The above must be true because we know that y is further from 0 than x . Or, since both x and y are negative, then |x| < |y| means -x < -y , which gives x > y and we know from the stem that this is true. II. \frac{1}{x^2} < 1 • The above is not necessarily true. For example, consider x=-\frac{1}{2} III. \frac{x + y}{x} > 0 • \frac{x}{x}+\frac{y}{x} > 0 • \frac{y}{x} > -1 . Since both x and y are negative, then \frac{y}{x} > 0 >-1 , so this option must be true. Answer: D

KanishkM · Answer

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. \frac{x + y}{x} > 0 x = -1, y= -2, -1-2/-1 > 0, 3 > 0, Yes x = -1/4, y =-1/2, Yes

Archit3110 · Answer

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

Mutaher889 · Answer

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

Kinshook · Answer

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

Paras96 · Answer

Given, 0>x>y

Let us go to the options one by one,

I x^2 - y^2 < 0
Since x and y are both negative and x>y,
x^2<y^2
=> x^2 - y^2 < 0

Hence I is must be Correct

II 1/x^2 < 1

Since 0>x
Let us first take x = -0.5, then x^2=0.25 and 1/x^2 = 4 which is not less than 1

Hence II is not necessary correct based on the case taken above

III (x+y)/x > 0

Take (x+y)/x
= (x/x) + (y/x)
= 1 + y/x

Since x and y are negative, therefore xy>0 and similarly x/y or y/x is greater than 0

therefore 1 + y/x i.e. (x+y)/x > 0

Hence III is must be Correct

hence D

youcouldsailaway · Answer

I did:

x^2  -  y^2  < 0
 x^2  <  y^2 
Taking sq. roots, 
x < y
Which is not correct as per the question stem.

So, I chose option C as it seemed to me st. 1 was incorrect.

Can someone explain where I went wrong?

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

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