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Intern  Joined: 28 Dec 2014
Posts: 9
GMAT 1: 710 Q47 V41 If |x/y| < 1, then which of the following must be true?  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 54% (01:59) correct 46% (01:59) wrong based on 198 sessions

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If $$|\frac{x}{y}| < 1$$, then which of the following must be true?

(I) $$\frac{|x+1|}{|y+1|}< 1$$

(II)$$\frac{|x^2+1|}{|y^2+1|} < 1$$

(III)$$\frac{|x-1|}{|y-1|} < 1$$

A. I only
B. II only
C. III only
D. I & II only
E. II & III only

Originally posted by aazt on 25 Jun 2017, 12:57.
Last edited by Bunuel on 06 Jul 2017, 09:02, edited 3 times in total.
Edited the question.
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Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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1
I presume you meant $$\frac{|x|}{|y|} < 1$$
So, either x and y could be both positive or both negative,
x is negative, y positive or x is positive and y, negative.

However it is clear that the magnitude of y will definitely be greater than x(it will be farther away from the origin)

If x=1,y=-3, the expression $$\frac{|x+1|}{|y+1|}$$ has value 1, which is not lesser than 1.
While evaluating condition II, x^2 and y^2 will both become positive and added to 1, will have a ratio similar to that of $$\frac{|x|}{|y|}$$.
If x=-1,y=3, the expression $$\frac{|x-1|}{|y-1|}$$ has value 1, which is not lesser than 1.

Hence Option B(II only) is the only choice, for which the condition holds true.
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Intern  Joined: 28 Dec 2014
Posts: 9
GMAT 1: 710 Q47 V41 Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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1
Thanks!

Sorry, the gmatclub formatting makes it look weird. The question was actually |x/y| $$< 1$$. I presume this changes things?

aazt

pushpitkc wrote:
I presume you meant $$\frac{|x|}{|y|} < 1$$
So, either x and y could be both positive or both negative,
x is negative, y positive or x is positive and y, negative.

However it is clear that the magnitude of y will definitely be greater than x(it will be farther away from the origin)

If x=1,y=-3, the expression $$\frac{|x+1|}{|y+1|}$$ has value 1, which is not lesser than 1.
While evaluating condition II, x^2 and y^2 will both become positive and added to 1, will have a ratio similar to that of $$\frac{|x|}{|y|}$$.
If x=-1,y=3, the expression $$\frac{|x-1|}{|y-1|}$$ has value 1, which is not lesser than 1.

Hence Option B(II only) is the only choice, for which the condition holds true.
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Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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If $$|\frac{x}{y}| < 1$$, then which of the following must be true?

(I) $$\frac{|x+1|}{|y+1|}< 1$$

(II)$$\frac{|x^2+1|}{|y^2+1|} < 1$$

(III)$$\frac{|x-1|}{|y-1|} < 1$$

A. I only
B. II only
C. III only
D. I & II only
E. II & III only

Let's plug in numbers and work strategically

(1/2), (-1,-2), (-1,2), (1, -2)...We need to consider all possibilities that satisfy the $$|\frac{x}{y}| < 1$$

Start with I, let's examine either (1/2) or (1, -2) (because by eyes, other pairs make fraction less than one: 0 < 1)

(1, -2): $$\frac{|1+1|}{|-2+1|}< 1$$.........$$\frac{|2|}{|-1|}< 1$$..........Eliminate A & D

Let's check III as there is possibility to eliminate E and get II as only solution.

let's examine either (-1/2) or (-1, -2) (because by eyes, other pairs make fraction less than one: 0 < 1)

(-1/2): $$\frac{|-1-1|}{|2-1|}< 1$$.........$$\frac{|-2|}{|1|}< 1$$..........Eliminate C & E

No need to check as we have only one answer left.

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Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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Why can't we take x, y = 0 ??
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Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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Prashant420 wrote:

Why can't we take x, y = 0 ??

You can take x =0 and y anything except 0 because 0/0 is undefined value.

You CAN NOt take x = any number and y =0 because Number/0 is undefined value.

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Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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_________________ Re: If |x/y| < 1, then which of the following must be true?   [#permalink] 27 Aug 2018, 06:04
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