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# Is |ab| > -ab?

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Intern
Joined: 01 Apr 2013
Posts: 12

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27 Oct 2013, 14:46
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Difficulty:

75% (hard)

Question Stats:

54% (02:18) correct 46% (02:07) wrong based on 297 sessions

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Is |ab| > -ab?

1) a > 1/b

2) a + b > 0
Math Expert
Joined: 02 Sep 2009
Posts: 52390
Re: Is |ab| > -ab?  [#permalink]

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31 Oct 2013, 09:29
4
1
TheKraken wrote:
SleeB wrote:
Is |ab| > -ab?

1) a > 1/b

2) a + b > 0

but if b < 0 then that inequality won't hold as the inequality sign will flip

The sign flip is only upon multiplying and dividing by a negative number.

That's the point. Do you know whether b is negative or positive? Do you know whether you should flip the sign of the inequality when multiplying by b?

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign.

So you CANNOT multiply a>1/b by b since you don't know the sign of b: if b>0, then you'll have ab>1 but if b<0, then you'll have ab<1 (flip the sign when multiplying by negative value).

Hope it's clear.
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Re: Is |ab| > -ab?  [#permalink]

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Updated on: 27 Oct 2013, 15:54
5
1
Is |ab| > -ab?

--> Is |ab| + ab > 0 ?

Notice that this is true only when the product ab is positive. (If ab is negative, |ab| + ab = 0, which is not greater than 0. If ab is equal to zero, |ab| + ab = 0, which is not greater than 0.)

Thus, what we really want to know...

Is ab > 0 ?

1) a > 1/b
If a = 2 and b = 1, the answer is YES.
If a = 2 and b = -1, the answer is NO. --> Insufficient.

2) a + b > 0
This can be shown to be insufficient in the same manner; indeed, with the very same numbers.
If a = 2 and b = 1, the answer is YES.
If a = 2 and b = -1, the answer is NO. --> Insufficient.

Thus, even combined the statements are insufficient. Answer E.

Originally posted by Reinfrank2011 on 27 Oct 2013, 15:36.
Last edited by Reinfrank2011 on 27 Oct 2013, 15:54, edited 1 time in total.
##### General Discussion
Intern
Joined: 01 Apr 2013
Posts: 12
Re: Is |ab| > -ab?  [#permalink]

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27 Oct 2013, 15:44
Awesome, thanks for that explanation, it makes a lot more sense now
Manager
Joined: 24 Sep 2013
Posts: 75
Location: United States
Concentration: Technology, Strategy
WE: Analyst (Transportation)
Re: Is |ab| > -ab?  [#permalink]

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Updated on: 31 Oct 2013, 09:20
Reinfrank2011 wrote:
Is |ab| > -ab?

--> Is |ab| + ab > 0 ?

Notice that this is true only when the product ab is positive. (If ab is negative, |ab| + ab = 0, which is not greater than 0. If ab is equal to zero, |ab| + ab = 0, which is not greater than 0.)

Thus, what we really want to know...

Is ab > 0 ?

1) a > 1/b
If a = 2 and b = 1, the answer is YES.
If a = 2 and b = -1, the answer is NO. --> Insufficient.

2) a + b > 0
This can be shown to be insufficient in the same manner; indeed, with the very same numbers.
If a = 2 and b = 1, the answer is YES.
If a = 2 and b = -1, the answer is NO. --> Insufficient.

Thus, even combined the statements are insufficient. Answer E.

Based on statement 1 however, can't you convert the equation a > 1/b to a*b>1?

Originally posted by TheKraken on 31 Oct 2013, 08:33.
Last edited by TheKraken on 31 Oct 2013, 09:20, edited 1 time in total.
Intern
Joined: 01 Apr 2013
Posts: 12
Re: Is |ab| > -ab?  [#permalink]

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31 Oct 2013, 09:13
but if b < 0 then that inequality won't hold as the inequality sign will flip
Manager
Joined: 24 Sep 2013
Posts: 75
Location: United States
Concentration: Technology, Strategy
WE: Analyst (Transportation)
Re: Is |ab| > -ab?  [#permalink]

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Updated on: 31 Oct 2013, 09:31
1
SleeB wrote:
but if b < 0 then that inequality won't hold as the inequality sign will flip

The sign flip is only upon multiplying and dividing by a negative number. Therefore I think OA should be A.

Is |ab| > -ab?

1) a > 1/b

** ab>1 both AB has to positive or AB is two negative (3 negatives = Negative). Therefore, AB

2) a + b > 0

Not sufficient as we don't know what values A or B are.

Originally posted by TheKraken on 31 Oct 2013, 09:21.
Last edited by TheKraken on 31 Oct 2013, 09:31, edited 1 time in total.
Manager
Joined: 24 Sep 2013
Posts: 75
Location: United States
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WE: Analyst (Transportation)
Re: Is |ab| > -ab?  [#permalink]

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31 Oct 2013, 09:33
Bunuel wrote:
TheKraken wrote:
SleeB wrote:
Is |ab| > -ab?

1) a > 1/b

2) a + b > 0

but if b < 0 then that inequality won't hold as the inequality sign will flip

The sign flip is only upon multiplying and dividing by a negative number.

That's the point. Do you know whether b is negative or positive? Do you know whether you should flip the sign of the inequality when multiplying by b?

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign.

So you CANNOT multiply a>1/b by b since you don't know the sign of b: if b>0, then you'll have ab>1 but if b<0, then you'll have ab<1 (flip the sign when multiplying by negative value).

Hope it's clear.

Thanks, Bunuel! The helps a lot.
SVP
Joined: 06 Sep 2013
Posts: 1705
Concentration: Finance
Re: Is |ab| > -ab?  [#permalink]

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28 May 2014, 11:51
I think that tough part is to be able to rephrase the question asked

Is |ab| > -ab = |ab|+ab > 0

Is ab<0?

We need to know if a,b have different signs

1) a>1/b, Insufficient

2) a>-b, Again could be either positive or negative

Both together, Still insufficient

Intern
Joined: 08 Aug 2011
Posts: 21
Re: Is |ab| > -ab?  [#permalink]

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29 May 2014, 04:04
jlgdr wrote:
I think that tough part is to be able to rephrase the question asked

It is almost always easier to evaluate the behavior of an inequality or equation when everything is moved to one side of the arrow or equal sign. When in doubt, get it all on one side, especially when you're dealing with inequalities in DS. There are exceptions, of course, but as a rule of thumb it is useful.

Posted from my mobile device
Intern
Joined: 18 Mar 2014
Posts: 30

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27 Aug 2014, 10:17
SleeB wrote:
Is |ab| > -ab?

1) a > 1/b

2) a + b > 0

What is the flaw in my approach ? Experts kindly assist.
The question is transformed to (ab)^2+ab>0 => ab ( ab+1) > 0

Hence i need to find if ab is +ve or ab > 0

case 1:
Now with a> 1/b ...By cross multiplying, this is transformed to ab > 1 (Cross Multiply) Hence this is sufficient
case 2:
a+b > 0 means ab is necessarily > 0.Hence this is sufficient

Hence i selected option D, which is not the answer.

_________________

[color=#007ec6]Would love to get kudos whenever it is due.. Though the mountain is steep, nevertheless once i reach the finishing point, i would enjoy the view [/color]....

Math Expert
Joined: 02 Sep 2009
Posts: 52390
Re: Is |ab| > -ab?  [#permalink]

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27 Aug 2014, 10:20
dshuvendu wrote:
SleeB wrote:
Is |ab| > -ab?

1) a > 1/b

2) a + b > 0

What is the flaw in my approach ? Experts kindly assist.
The question is transformed to (ab)^2+ab>0 => ab ( ab+1) > 0

Hence i need to find if ab is +ve or ab > 0

case 1:
Now with a> 1/b ...By cross multiplying, this is transformed to ab > 1 (Cross Multiply) Hence this is sufficient
case 2:
a+b > 0 means ab is necessarily > 0.Hence this is sufficient

Hence i selected option D, which is not the answer.

How did you get (ab)^2+ab>0 ?
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Joined: 18 Mar 2014
Posts: 30

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27 Aug 2014, 13:16
Bunuel wrote:
dshuvendu wrote:
SleeB wrote:
Is |ab| > -ab?

1) a > 1/b

2) a + b > 0

What is the flaw in my approach ? Experts kindly assist.
The question is transformed to (ab)^2+ab>0 => ab ( ab+1) > 0

Hence i need to find if ab is +ve or ab > 0

case 1:
Now with a> 1/b ...By cross multiplying, this is transformed to ab > 1 (Cross Multiply) Hence this is sufficient
case 2:
a+b > 0 means ab is necessarily > 0.Hence this is sufficient

Hence i selected option D, which is not the answer.

How did you get (ab)^2+ab>0 ?

I have committed a crime now.I missed the square root. I am in a deadlock after the correction, cannot proceed with the transformation As always thanks for the poke .
_________________

[color=#007ec6]Would love to get kudos whenever it is due.. Though the mountain is steep, nevertheless once i reach the finishing point, i would enjoy the view [/color]....

Manager
Joined: 16 Oct 2015
Posts: 72
Location: India
Concentration: Technology, General Management
GMAT 1: 520 Q44 V17
GMAT 2: 530 Q44 V20
GMAT 3: 710 Q48 V40
GPA: 3.45
WE: Research (Energy and Utilities)

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11 Apr 2017, 07:00
When we combine both equations :

a>(1/b) ------> a-(1/b) > 0 (1)

a+b>0 (2)

subracting 1 and 2

a-(1/b)>0
- a+ b>0
___________
-b-(1/b)>0
-(1+b^2)/(b) >0

(1+b^2)/(b)<0 (3)

From equation 3 we can say that b is negative since numerator is +ve

hence b<0

in order to equation 1 and 2 to be valid ; a should be positive and greater than b

hence we can say a>0

so a>0 and b<0 ; now we know a and b have opposite signs. Hence C should be sufficient.

Please correct where I am going wrong....
Is |ab| > -ab? &nbs [#permalink] 11 Apr 2017, 07:00
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