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Is |x y| < |x| + |y|?

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Is |x y| < |x| + |y|? [#permalink] New post 08 Aug 2012, 21:08
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Is |x – y| < |x| + |y|?

(1) y < x
(2) xy < 0

Bunuel

I searched of this question on gmatclub culd not find a post please explain your method
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Re: Is |x – y| < |x| + |y|? [#permalink] New post 10 Jun 2013, 11:14
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WholeLottaLove wrote:
I understand how to reduce on both sides, I'm just wondering why one has absolute value signs (after simplifying) and why one doesn't.

Thanks!

Bunuel wrote:
WholeLottaLove wrote:
Hi.

I am wondering how you got: x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0?

I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with -2xy while on the right side you are left with 2|xy|

Thanks!




x^2-2xy+y^2<x^2+2|xy|+y^2 --> cancel x^2+y^2 in both sides: -2xy<2|xy| --> reduce by 2: -xy<|xy| --> |xy|+xy>0.

Hope it's clear.


(|x-y|)^2=(x-y)^2=x^2-2xy+y^2.
(|x| + |y|)^2=(|x|)^2+2|x|*|y|+(|y|)^2=x^2+2|xy|+y^2.
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Re: Is |x – y| < |x| + |y|? (1) y < x (2) xy < 0 [#permalink] New post 08 Aug 2012, 22:21
1. y<x implies 2 cases:
(a) y is +ve, x is +ve.
eg., y=2, x=3
|x-y| = 1
|x| = 3
|y| = 2
|x| + |y| = 5,
Hence |x-y| < |x| + |y|

(b) y is -ve, x is +ve.
eg., y=-1, x = 2
|x-y|=3
|x|+|y|=3
Here, |x-y| !< |x|+|y|.

From (a) and (b), A and D are eliminated

2. xy<0 <=> either x or y is -ve.
Hence |x-y| is always = |x| + |y|
So, here we have a definite answer to the question and hence B is the best choice.
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Re: Is |x – y| < |x| + |y|? [#permalink] New post 08 Aug 2012, 23:57
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Is |x – y| < |x| + |y|?

Since both sides of the inequality are non-negative, then we can safely square and we'll get: is x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0? Consider two cases:

If xy\leq{0}, then |xy|+xy=-xy+xy=0. So, when xy\leq{0}, then |xy|+xy>0 does NOT hold true.
If xy>0, then |xy|+xy=xy+xy=2xy>0 (because xy>0). So, when xy>{0}, then |xy|+xy>0 holds true.

As we can see the question basically asks whether xy>0.

(1) y < x. Not sufficient to say whether xy>0.

(2) xy < 0. Directly gives a NO answer to the question.

Answer: B.

Hope it's clear.
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Re: Is |x – y| < |x| + |y|? [#permalink] New post 09 Aug 2012, 03:55
venmic wrote:
Is |x – y| < |x| + |y|?

(1) y < x

(2) xy < 0

Bunuel

I searched of this question on gmatclub culd not find a post please explain your method


Useful to remember the so-called "Triangle Inequality " for the absolute value:

For any real numbers x,y \,|x+y|\leq|x|+|y|.
Equality holds if and only if xy\geq{0}, which means, at least one of the two numbers is 0 or they both have the same sign (both positive or both negative).
Or, strict inequality holds if and only if xy<0.

In our case, we can see that always |x-y|=|x+(-y)|\leq|x|+|y|, because |y|=|-y|.
The strict inequality will hold if and only if x(-y)<0 or xy>0.

(1) Either x or y can be zero, and in addition, we don't know anything about their signs.
Not sufficient.

(2) Obviously sufficient, because we can state with certainty that the given inequality does not hold.

Answer B
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Re: Is |x – y| < |x| + |y|? [#permalink] New post 09 Aug 2012, 04:00
Expert's post
EvaJager wrote:
venmic wrote:
Is |x – y| < |x| + |y|?

(1) y < x

(2) xy < 0

Bunuel

I searched of this question on gmatclub culd not find a post please explain your method


Useful to remember the so-called "Triangle Inequality " for the absolute value:

For any real numbers x,y \,|x+y|\leq|x|+|y|.
Equality holds if and only if xy\geq{0}, which means, at least one of the two numbers is 0 or they both have the same sign (both positive or both negative).
Or, strict inequality holds if and only if xy<0.

In our case, we can see that always |x-y|=|x+(-y)|\leq|x|+|y|, because |y|=|-y|.
The strict inequality will hold if and only if x(-y)<0 or xy>0.

(1) Either x or y can be zero, and in addition, we don't know anything about their signs.
Not sufficient.

(2) Obviously sufficient, because we can state with certainty that the given inequality does not hold.

Answer B


There is another property worth remembering:

1. Always true: |x+y|\leq{|x|+|y|}, note that "=" sign holds for xy\geq{0} (or simply when x and y have the same sign);

2. Always true: |x-y|\geq{|x|-|y|}, note that "=" sign holds for xy>{0} (so when x and y have the same sign) and |x|>|y| (simultaneously).
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is |x – y| < |x| + |y|? [#permalink] New post 09 Aug 2012, 04:08
Bunuel wrote:
EvaJager wrote:
venmic wrote:
Is |x – y| < |x| + |y|?

(1) y < x

(2) xy < 0

Bunuel

I searched of this question on gmatclub culd not find a post please explain your method


Useful to remember the so-called "Triangle Inequality " for the absolute value:

For any real numbers x,y \,|x+y|\leq|x|+|y|.
Equality holds if and only if xy\geq{0}, which means, at least one of the two numbers is 0 or they both have the same sign (both positive or both negative).
Or, strict inequality holds if and only if xy<0.

In our case, we can see that always |x-y|=|x+(-y)|\leq|x|+|y|, because |y|=|-y|.
The strict inequality will hold if and only if x(-y)<0 or xy>0.

(1) Either x or y can be zero, and in addition, we don't know anything about their signs.
Not sufficient.

(2) Obviously sufficient, because we can state with certainty that the given inequality does not hold.

Answer B


There is another property worth remembering:

1. Always true: |x+y|\leq{|x|+|y|}, note that "=" sign holds for xy\geq{0} (or simply when x and y have the same sign);

2. Always true: |x-y|\geq{|x|-|y|}, note that "=" sign holds for xy>{0} (so when x and y have the same sign) and |x|>|y| (simultaneously).


Thanks for reminding, certainly worth remembering. Just for the practice, we can prove that the second inequality can be deduced from the first one quite easily:

|(x-y)+y|\leq|x-y|+|y|, or |x|\leq|x-y|+|y|, from which we get |x-y|\geq{|x|-|y|}.
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Re: Is |x y| < |x| + |y|? [#permalink] New post 06 Jun 2013, 02:56
Ans B

Ref to the attached pic

Stmt 1: has two possibilities

Stmt 2 : gives confirmed no

Therefore ans B
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Re: Is |x – y| < |x| + |y|? [#permalink] New post 10 Jun 2013, 10:29
Hi.

I am wondering how you got: x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0?

I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with -2xy while on the right side you are left with 2|xy|

Thanks!


Bunuel wrote:
Is |x – y| < |x| + |y|?

Since both sides of the inequality are non-negative, then we can safely square and we'll get: is x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0? Consider two cases:

If xy\leq{0}, then |xy|+xy=-xy+xy=0. So, when xy\leq{0}, then |xy|+xy>0 does NOT hold true.
If xy>0, then |xy|+xy=xy+xy=2xy>0 (because xy>0). So, when xy>{0}, then |xy|+xy>0 holds true.

As we can see the question basically asks whether xy>0.

(1) y < x. Not sufficient to say whether xy>0.

(2) xy < 0. Directly gives a NO answer to the question.

Answer: B.

Hope it's clear.
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Re: Is |x – y| < |x| + |y|? [#permalink] New post 10 Jun 2013, 10:57
Expert's post
WholeLottaLove wrote:
Hi.

I am wondering how you got: x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0?

I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with -2xy while on the right side you are left with 2|xy|

Thanks!


Bunuel wrote:
Is |x – y| < |x| + |y|?

Since both sides of the inequality are non-negative, then we can safely square and we'll get: is x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0? Consider two cases:

If xy\leq{0}, then |xy|+xy=-xy+xy=0. So, when xy\leq{0}, then |xy|+xy>0 does NOT hold true.
If xy>0, then |xy|+xy=xy+xy=2xy>0 (because xy>0). So, when xy>{0}, then |xy|+xy>0 holds true.

As we can see the question basically asks whether xy>0.

(1) y < x. Not sufficient to say whether xy>0.

(2) xy < 0. Directly gives a NO answer to the question.

Answer: B.

Hope it's clear.


x^2-2xy+y^2<x^2+2|xy|+y^2 --> cancel x^2+y^2 in both sides: -2xy<2|xy| --> reduce by 2: -xy<|xy| --> |xy|+xy>0.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is |x – y| < |x| + |y|? [#permalink] New post 10 Jun 2013, 11:04
I understand how to reduce on both sides, I'm just wondering why one has absolute value signs (after simplifying) and why one doesn't.

Thanks!

Bunuel wrote:
WholeLottaLove wrote:
Hi.

I am wondering how you got: x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0?

I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with -2xy while on the right side you are left with 2|xy|

Thanks!


Bunuel wrote:
Is |x – y| < |x| + |y|?

Since both sides of the inequality are non-negative, then we can safely square and we'll get: is x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0? Consider two cases:

If xy\leq{0}, then |xy|+xy=-xy+xy=0. So, when xy\leq{0}, then |xy|+xy>0 does NOT hold true.
If xy>0, then |xy|+xy=xy+xy=2xy>0 (because xy>0). So, when xy>{0}, then |xy|+xy>0 holds true.

As we can see the question basically asks whether xy>0.

(1) y < x. Not sufficient to say whether xy>0.

(2) xy < 0. Directly gives a NO answer to the question.

Answer: B.

Hope it's clear.


x^2-2xy+y^2<x^2+2|xy|+y^2 --> cancel x^2+y^2 in both sides: -2xy<2|xy| --> reduce by 2: -xy<|xy| --> |xy|+xy>0.

Hope it's clear.
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Re: Is |x – y| < |x| + |y|? [#permalink] New post 10 Jun 2013, 11:05
I understand how to reduce on both sides, I'm just wondering why one has absolute value signs (after simplifying) and why one doesn't.

Thanks!

Bunuel wrote:
WholeLottaLove wrote:
Hi.

I am wondering how you got: x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0?

I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with -2xy while on the right side you are left with 2|xy|

Thanks!


Bunuel wrote:
Is |x – y| < |x| + |y|?

Since both sides of the inequality are non-negative, then we can safely square and we'll get: is x^2-2xy+y^2<x^2+2|xy|+y^2? --> is |xy|+xy>0? Consider two cases:

If xy\leq{0}, then |xy|+xy=-xy+xy=0. So, when xy\leq{0}, then |xy|+xy>0 does NOT hold true.
If xy>0, then |xy|+xy=xy+xy=2xy>0 (because xy>0). So, when xy>{0}, then |xy|+xy>0 holds true.

As we can see the question basically asks whether xy>0.

(1) y < x. Not sufficient to say whether xy>0.

(2) xy < 0. Directly gives a NO answer to the question.

Answer: B.

Hope it's clear.


x^2-2xy+y^2<x^2+2|xy|+y^2 --> cancel x^2+y^2 in both sides: -2xy<2|xy| --> reduce by 2: -xy<|xy| --> |xy|+xy>0.

Hope it's clear.
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Re: Is |x y| < |x| + |y|? [#permalink] New post 29 Jun 2013, 15:16
Is |x – y| < |x| + |y|?

(1) y < x

If y < x then |x - y| will always be positive. However, we don't know the signs for x and y so we don't know what |x| and |y| is.
INSUFFICIENT

(2) xy < 0

This means that either x or y is negative and neither x or y are zero.

If we test for positive and negative values of x and y we will find that regardless of what x and y are (as long as one is positive and one is negative) the inequality will always fail (LHS and RHS will always equal one another)
SUFFICIENT

(B)
Re: Is |x y| < |x| + |y|?   [#permalink] 29 Jun 2013, 15:16
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