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Manager  B
Joined: 02 Nov 2009
Posts: 97
Is |x y| < |x| + |y|?  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 59% (01:48) correct 41% (01:53) wrong based on 239 sessions

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Is |x – y| < |x| + |y|?

(1) y < x
(2) xy < 0
Intern  Joined: 12 Apr 2012
Posts: 2
Re: Is |x y| < |x| + |y|?  [#permalink]

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1
1
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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58401
Re: Is |x y| < |x| + |y|?  [#permalink]

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1
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.

Hope it's clear.
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Director  Joined: 22 Mar 2011
Posts: 588
WE: Science (Education)
Re: Is |x y| < |x| + |y|?  [#permalink]

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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.

_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Math Expert V
Joined: 02 Sep 2009
Posts: 58401
Re: Is |x y| < |x| + |y|?  [#permalink]

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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.

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).
_________________
Director  Joined: 22 Mar 2011
Posts: 588
WE: Science (Education)
Re: Is |x y| < |x| + |y|?  [#permalink]

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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.

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 $$|x-y|\geq{|x|-|y|}.$$
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Intern  Joined: 24 May 2013
Posts: 18
Location: United Kingdom
WE: Project Management (Real Estate)
Re: Is |x y| < |x| + |y|?  [#permalink]

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1
Ans B

Ref to the attached pic

Stmt 1: has two possibilities

Stmt 2 : gives confirmed no

Therefore ans B
Attachments MOD.pdf [6.22 MiB]

_________________
Correct me If I'm wrong !! looking for valuable inputs
Senior Manager  Joined: 13 May 2013
Posts: 399
Re: Is |x y| < |x| + |y|?  [#permalink]

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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.

Hope it's clear.
Math Expert V
Joined: 02 Sep 2009
Posts: 58401
Re: Is |x y| < |x| + |y|?  [#permalink]

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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.

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.
_________________
Senior Manager  Joined: 13 May 2013
Posts: 399
Re: Is |x y| < |x| + |y|?  [#permalink]

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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.

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.
Senior Manager  Joined: 13 May 2013
Posts: 399
Re: Is |x y| < |x| + |y|?  [#permalink]

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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.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58401
Re: Is |x y| < |x| + |y|?  [#permalink]

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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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Senior Manager  Joined: 13 May 2013
Posts: 399
Re: Is |x y| < |x| + |y|?  [#permalink]

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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)
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Re: Is |x y| < |x| + |y|?  [#permalink]

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_________________ Re: Is |x y| < |x| + |y|?   [#permalink] 25 Mar 2019, 23:45
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