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# If x and y are non-zero numbers such that x>y, which of the following

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Manager
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If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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28 Jan 2019, 00:36
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If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

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If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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28 Jan 2019, 09:59
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

IMO D

Just test values, Given x > y
If x and y are non-zero numbers
Test values : 2> 1 or -1 > -2

A) $$\frac{1}{x} < \frac{1}{y}$$, To avoid any confusion, always do this.

$$\frac{1}{y} > \frac{1}{x}$$, this works for me

C1: 1 > 0.5, Yes
C2: -1 > -0.5 No

B) $$\frac{x}{y} > 1$$

C1: 2 > 1 Yes
C2: 0.5 > 1 No

C) $$|x| > |y|$$
This will give multiple values

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$, same here as well

$$\frac{1}{x^2y} > \frac{1}{xy^2}$$

For any set of values, Here i am always getting a No.

E) $$\frac{x}{y} < \frac{y}{x}$$[/quote], same here as well

$$\frac{y}{x} > \frac{x}{y}$$[/quote]

Here you can get a Yes and a No
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Re: If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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30 Jan 2019, 03:55
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

Can someone tell me why C is wrong
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Re: If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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30 Jan 2019, 04:04
AlN wrote:
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

Can someone tell me why C is wrong

If you take x as -2 and y as -3. We can see that x > y as -2 > -3. But |x| = 2 and |y| = 3. And |x| is not greater than |y|.

Hence C is wrong

Posted from my mobile device
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Re: If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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30 Jan 2019, 08:19
KanishkM wrote:
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

IMO D

Just test values, Given x > y
If x and y are non-zero numbers
Test values : 2> 1 or -1 > -2

A) $$\frac{1}{x} < \frac{1}{y}$$, To avoid any confusion, always do this.

$$\frac{1}{y} > \frac{1}{x}$$, this works for me

C1: 1 > 0.5, Yes
C2: -1 > -0.5 No

B) $$\frac{x}{y} > 1$$

C1: 2 > 1 Yes
C2: 0.5 > 1 No

C) $$|x| > |y|$$
This will give multiple values

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$, same here as well

$$\frac{1}{x^2y} > \frac{1}{xy^2}$$

For any set of values, Here i am always getting a No.

E) $$\frac{x}{y} < \frac{y}{x}$$
, same here as well

$$\frac{y}{x} > \frac{x}{y}$$[/quote]

Here you can get a Yes and a No[/quote]

If D is giving a NO with every set of values of x and y. How come is D the right answer?
Isnt te question stem asking for an option which is always true??
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Re: If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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30 Jan 2019, 12:24
hannan4644 wrote:
KanishkM wrote:
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

IMO D

Just test values, Given x > y
If x and y are non-zero numbers
Test values : 2> 1 or -1 > -2

A) $$\frac{1}{x} < \frac{1}{y}$$, To avoid any confusion, always do this.

$$\frac{1}{y} > \frac{1}{x}$$, this works for me

C1: 1 > 0.5, Yes
C2: -1 > -0.5 No

B) $$\frac{x}{y} > 1$$

C1: 2 > 1 Yes
C2: 0.5 > 1 No

C) $$|x| > |y|$$
This will give multiple values

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$, same here as well

$$\frac{1}{x^2y} > \frac{1}{xy^2}$$

For any set of values, Here i am always getting a No.

E) $$\frac{x}{y} < \frac{y}{x}$$
, same here as well

$$\frac{y}{x} > \frac{x}{y}$$

Here you can get a Yes and a No[/quote]

If D is giving a NO with every set of values of x and y. How come is D the right answer?
Isnt te question stem asking for an option which is always true??[/quote]

Though i am not an expert, but i believe if we look at the rest of the options, we are able to negate this option completely with the same test cases.

What one can try is to look for other options which will satisfy D, for different test cases.

But i doubt that will happen, under the stipulated time limit of trying to solve a question in 2-3 minutes.

Let me know if this helps you in anyway

Posted from my mobile device
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Quote which i can relate to.
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If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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03 Feb 2019, 07:15
hannan4644 wrote:
KanishkM wrote:
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

IMO D

Just test values, Given x > y
If x and y are non-zero numbers
Test values : 2> 1 or -1 > -2

A) $$\frac{1}{x} < \frac{1}{y}$$, To avoid any confusion, always do this.

$$\frac{1}{y} > \frac{1}{x}$$, this works for me

C1: 1 > 0.5, Yes
C2: -1 > -0.5 No

B) $$\frac{x}{y} > 1$$

C1: 2 > 1 Yes
C2: 0.5 > 1 No

C) $$|x| > |y|$$
This will give multiple values

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$, same here as well

$$\frac{1}{x^2y} > \frac{1}{xy^2}$$

For any set of values, Here I am always getting a No.

E) $$\frac{x}{y} < \frac{y}{x}$$
, same here as well

$$\frac{y}{x} > \frac{x}{y}$$

Take X=100
Y=1

how can D be the Answer?
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Re: If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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03 Feb 2019, 08:56
KanishkM wrote:
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

IMO D

Just test values, Given x > y
If x and y are non-zero numbers
Test values : 2> 1 or -1 > -2

A) $$\frac{1}{x} < \frac{1}{y}$$, To avoid any confusion, always do this.

$$\frac{1}{y} > \frac{1}{x}$$, this works for me

C1: 1 > 0.5, Yes
C2: -1 > -0.5 No

B) $$\frac{x}{y} > 1$$

C1: 2 > 1 Yes
C2: 0.5 > 1 No

C) $$|x| > |y|$$
This will give multiple values

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$, same here as well

$$\frac{1}{x^2y} > \frac{1}{xy^2}$$

For any set of values, Here i am always getting a No.

E) $$\frac{x}{y} < \frac{y}{x}$$
, same here as well

$$\frac{y}{x} > \frac{x}{y}$$[/quote]

Here you can get a Yes and a No[/quote]

In A you haven't done it correct. It should be -0.5>-1 after reciprocal which is a yes.

Also in D, it can be proven wrong if we take 5 & -5
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Re: If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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03 Feb 2019, 08:59
None of the options is correct. All give Yes or No as answers.
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Re: If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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03 Feb 2019, 09:01
1
Afc0892 wrote:
AlN wrote:
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

Can someone tell me why C is wrong

If you take x as -2 and y as -3. We can see that x > y as -2 > -3. But |x| = 2 and |y| = 3. And |x| is not greater than |y|.

Hence C is wrong

Posted from my mobile device

Thank you Got it!
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If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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18 Aug 2019, 23:37
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

Given: x and y are non-zero numbers such that x > y

Asked: Which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$
If xy>0 ; $$\frac{1}{x} < \frac{1}{y}$$
If xy<0 ; $$\frac{1}{x} > \frac{1}{y}$$
3>2 => 1/3<1/2 YES
-2>-3 => -1/2<-1/3 YES
3>-2 => 1/3<-1/2 NO
NOT NECESSARILY TRUE

B) $$\frac{x}{y} > 1$$
If y>0 x/y >1 ; 3>2 => 3/2 > 1 YES
If y<0 x/y <1 ; 3>-2 => -3/2>1 NO
NOT NECESSARILY TRUE

C) $$|x| > |y|$$
3>2 => |3|>|2| YES
-2>-3 => |-2| > |-3| => 2>3 NO
NOT NECESSARILY TRUE

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$
$$\frac{1}{xy^2} < \frac{1}{x^2y}$$
Multiplying both sides by$$x^2y^2$$
x>y TRUE
MUST BE TRUE

E) $$\frac{x}{y} < \frac{y}{x}$$
3>2 => 3/2 < 2/3 NO
NOT NECESSARILY TRUE

IMO D
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Re: If x and y are non-zero numbers such that x>y, which of the following  [#permalink]

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18 Aug 2019, 23:58
4d wrote:
If x and y are non-zero numbers such that x > y, which of the following is always true?

A) $$\frac{1}{x} < \frac{1}{y}$$

B) $$\frac{x}{y} > 1$$

C) $$|x| > |y|$$

D) $$\frac{1}{xy^2} < \frac{1}{x^2y}$$

E) $$\frac{x}{y} < \frac{y}{x}$$

This is a flawed question. None of the options is always true.

D cannot be the answer because it's never true if x > y. Check x = 2 and y = 1.

TOPIC IS LOCKED.
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Re: If x and y are non-zero numbers such that x>y, which of the following   [#permalink] 18 Aug 2019, 23:58
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