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

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Is |x y| < |x| + |y|? [#permalink] New post 08 Aug 2012, 22:08
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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|? (1) y < x (2) xy < 0 [#permalink] New post 08 Aug 2012, 23:21
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 09 Aug 2012, 00:57
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, 04:55
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, 05:00
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).
_________________

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

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. NEW!!!

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. NEW!!!


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Re: Is |x – y| < |x| + |y|? [#permalink] New post 09 Aug 2012, 05:08
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] 09 Aug 2012, 05:08
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