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# If |x/y| < 1, then which of the following must be true?

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

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25 Nov 2019, 02:04
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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 and II only
E. II and III only

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

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25 Nov 2019, 03:13
1
check with 2/3 , -2/3 and 2/-3
only $$\frac{|x^2+1|}{|y^2+1|} < 1$$
will be valid
IMO B

Bunuel wrote:
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 and II only
E. II and III only

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Manager
Joined: 10 Dec 2017
Posts: 178
Location: India
Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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25 Nov 2019, 23:11
1
Bunuel wrote:
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 and II only
E. II and III only

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lxl<lyl
lxl^2<lyl^2
x^2<y^2
x^2+1<y^2+1
$$lx^2+1l<ly^2+1l$$
$$[lx^2+1l/ly^2+1l ]<1$$
Since we do not know about x and y (x>y, x<y) we can't say about rest of the options.
if x=-2, y=-5
lxl<lyl
lx+1l/ly+1l=1/4<1 (True)
But x=3, y=-5
1<1(absurd)
B:)
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Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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14 Dec 2019, 04:50
1
Can ny one help me with a proper explanation.
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Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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01 Jan 2020, 00:46
1
ScottTargetTestPrep

Can you help explain this please?

My approach:

I picked x=2,-2 and y= 3, -3

When I make the calculations, I am getting I, II and III as correct answers.
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If |x/y| < 1, then which of the following must be true?  [#permalink]

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05 Jan 2020, 19:56
2
sarahfiqbal wrote:
ScottTargetTestPrep

Can you help explain this please?

My approach:

I picked x=2,-2 and y= 3, -3

When I make the calculations, I am getting I, II and III as correct answers.

The question is asking for the statements which "must" be true, meaning the statement should be true for ANY value of x and y satisfying |x/y| < 1. As far as I understood, you found values which make statements I, II and III hold, but the question is not asking for that.

For statement I, if you take x = 2 and y = -3, then |(x + 1)/(y + 1)| becomes |3/-2|, which is greater than 1. That single example is enough to show that it is not true that statement I MUST be true. There can be values of x and y where statement I is true, but even one counter example is enough to show that it is not always true.

Similarly, for statement III, if you take x = -2 and y = 3, then |(x - 1)/(y - 1)| becomes |-3/2|, which is greater than 1. For the same reasons as above, we conclude that it's not true that statement III must be true either.

While we actually know the answer at this point (there is only one answer choice which does not involve statement I or III), let's show that statement II must be true. Notice that you can't just show that |(x^2 + 1)/(y^2 + 1)| < 1 by picking values for x and y; you can find a million pairs of x and y which satisfy statement II, but it does not prove that the statement is true for all x and y with |x/y| < 1. Instead, you proceed as follows:

If x/y > 0, then |x/y| = x/y. In this case, we have 0 < x/y < 1. Since the square of a number between 0 and 1 is also between 0 and 1, we have x^2/y^2 < 1 as well. Then, x^2 < y^2. Adding 1 to each side, we get x^2 + 1 < y^2 + 1; thus (x^2 + 1)/(y^2 + 1) < 1. Since (x^2 + 1)/(y^2 + 1) is positive, |(x^2 + 1)/(y^2 + 1)| < 1.

If x/y < 0, then |x/y| = -x/y. We have 0 < -x/y < 1. As above, the square of a number between 0 and 1 is again between 0 and 1; thus 0 < x^2/y^2 < 1. We follow the same steps as above to conclude that |(x^2 + 1)/(y^2 + 1)| < 1 in this case as well.

Just as a side note, if the question was "Which of the following can be true?", then your approach and your answer would have been correct.
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Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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05 Feb 2020, 03:50
ScottTargetTestPrep wrote:
sarahfiqbal wrote:
ScottTargetTestPrep

Can you help explain this please?

My approach:

I picked x=2,-2 and y= 3, -3

When I make the calculations, I am getting I, II and III as correct answers.

The question is asking for the statements which "must" be true, meaning the statement should be true for ANY value of x and y satisfying |x/y| < 1. As far as I understood, you found values which make statements I, II and III hold, but the question is not asking for that.

For statement I, if you take x = 2 and y = -3, then |(x + 1)/(y + 1)| becomes |3/-2|, which is greater than 1. That single example is enough to show that it is not true that statement I MUST be true. There can be values of x and y where statement I is true, but even one counter example is enough to show that it is not always true.

Similarly, for statement III, if you take x = -2 and y = 3, then |(x - 1)/(y - 1)| becomes |-3/2|, which is greater than 1. For the same reasons as above, we conclude that it's not true that statement III must be true either.

While we actually know the answer at this point (there is only one answer choice which does not involve statement I or III), let's show that statement II must be true. Notice that you can't just show that |(x^2 + 1)/(y^2 + 1)| < 1 by picking values for x and y; you can find a million pairs of x and y which satisfy statement II, but it does not prove that the statement is true for all x and y with |x/y| < 1. Instead, you proceed as follows:

If x/y > 0, then |x/y| = x/y. In this case, we have 0 < x/y < 1. Since the square of a number between 0 and 1 is also between 0 and 1, we have x^2/y^2 < 1 as well. Then, x^2 < y^2. Adding 1 to each side, we get x^2 + 1 < y^2 + 1; thus (x^2 + 1)/(y^2 + 1) < 1. Since (x^2 + 1)/(y^2 + 1) is positive, |(x^2 + 1)/(y^2 + 1)| < 1.

If x/y < 0, then |x/y| = -x/y. We have 0 < -x/y < 1. As above, the square of a number between 0 and 1 is again between 0 and 1; thus 0 < x^2/y^2 < 1. We follow the same steps as above to conclude that |(x^2 + 1)/(y^2 + 1)| < 1 in this case as well.

Just as a side note, if the question was "Which of the following can be true?", then your approach and your answer would have been correct.

Hi Scott

I just had concern regarding the highlighted part above where denominator from LHS can't go to RHS unless it is equal correct?

And only '+' & '-' signs can be shuffled from LHS to RHS or vice versa.
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Joined: 10 Dec 2019
Posts: 38
Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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05 Feb 2020, 03:54
Bunuel wrote:
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 and II only
E. II and III only

Hi Bunuel

thank you.
Are You Up For the Challenge: 700 Level Questions
Math Expert
Joined: 02 Sep 2009
Posts: 61338
Re: If |x/y| < 1, then which of the following must be true?  [#permalink]

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05 Feb 2020, 04:09
1
1
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 and II only
E. II and III only

Re-write $$|\frac{x}{y}| < 1$$ as $$\frac{|x|}{|y|} < 1$$
Multiply by |y|: $$|x| < |y|$$ (we can safely do that because |y| > 0).

$$|x| < |y|$$ means that y is further from 0 than x. We can have four cases:
----y----x----0--------------
----y---------0----x---------
---------x----0---------y----
--------------0----x----y----

I. $$\frac{|x+1|}{|y+1|}< 1$$ --> $$|x+1|<|y+1|$$. Not necessarily true, for example consider x = 1 and y = -2.

II. $$\frac{|x^2+1|}{|y^2+1|} < 1$$ --> $$|x^2+1|<|y^2+1|$$. Since both x^2 + 1 and y^2 + 1 are positive, then we have $$x^2+1<y^2+1$$ --> $$x^2<y^2$$. Taking the square root give $$|x| < |y|$$. TRUE

III. $$\frac{|x-1|}{|y-1|} < 1$$ --> $$|x-1|<|y-1|$$. Not necessarily true, for example consider x = -1 and y = 2.

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

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06 Feb 2020, 11:04
1
Anurag06 wrote:

Hi Scott

I just had concern regarding the highlighted part above where denominator from LHS can't go to RHS unless it is equal correct?

And only '+' & '-' signs can be shuffled from LHS to RHS or vice versa.

You are allowed to multiply each side of an inequality with some non-zero constant, as long as you remember to change the direction of the inequality if the constant is negative.

In the expression x^2/y^2 < 1, we know y^2 is positive; thus we can multiply each side of that inequality and obtain x^2 < y^2. If the inequality was x/y < 1 and if we had no idea whether y is positive or negative, then we would not be able to multiply each side by y and conclude that x < y.
_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
181 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: If |x/y| < 1, then which of the following must be true?   [#permalink] 06 Feb 2020, 11:04
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