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# Is a < -1?

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Math Expert
Joined: 02 Aug 2009
Posts: 6216

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02 Nov 2017, 09:51
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Difficulty:

55% (hard)

Question Stats:

55% (01:25) correct 45% (01:10) wrong based on 110 sessions

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Is $$a<-1$$?

(1) $$|a|>a+1$$

(2) $$a^2>1$$

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Joined: 22 May 2017
Posts: 117
Re: Is a < -1? [#permalink]

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02 Nov 2017, 11:14
1
My try on this.. not sure approach is optimal

(1) |a|>a+1

Checking with examples for validity

Positive:
a=1

1>1+1
1>2 -----> No

Negative
a=-1
1>-1+1
1>0 ------> Yes

So above eqtn is satisfied when
a<=-1

Hence A cannot be the answer

(2) a^2>1

a cannot take value 1,-1,0

So we can conclude a can take value
1<a<-1

Hence B cannot be the answer

Combining both I and II
C gives a<-1

Posted from my mobile device
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Joined: 25 Feb 2013
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Location: India
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02 Nov 2017, 11:51
chetan2u wrote:
Is $$a<-1$$?

(1) $$|a|>a+1$$

(2) $$a^2>1$$

Statement 1: implies that $$a>a+1 =>0>1$$. Not Possible

or $$a<-(a+1) => a<-0.5$$ but $$a$$ can be less than $$-1$$ or greater than $$-1$$. Hence insufficient

Another way to analyse this statement -

$$|a|>a+1$$, square both sides to get

$$a^2>a^2+2a+1 => a<-0.5$$

Statement 2: $$a^2>1 => a>1$$ or $$a<-1$$. Hence Insufficient

Combining 1 & 2: $$a<-0.5$$ & $$a<-1$$, the overlapping region is $$a<-1$$. Sufficient

Option C
Senior Manager
Joined: 02 Apr 2014
Posts: 486
GMAT 1: 700 Q50 V34
Re: Is a < -1? [#permalink]

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10 Dec 2017, 12:20
niks18 wrote:
chetan2u wrote:
Is $$a<-1$$?

(1) $$|a|>a+1$$

(2) $$a^2>1$$

Statement 1: implies that $$a>a+1 =>0>1$$. Not Possible

or $$a<-(a+1) => a<-0.5$$ but $$a$$ can be less than $$-1$$ or greater than $$-1$$. Hence insufficient

Another way to analyse this statement -

$$|a|>a+1$$, square both sides to get

$$a^2>a^2+2a+1 => a<-0.5$$

Statement 2: $$a^2>1 => a>1$$ or $$a<-1$$. Hence Insufficient

Combining 1 & 2: $$a<-0.5$$ & $$a<-1$$, the overlapping region is $$a<-1$$. Sufficient

Option C

Hi niks18,
I am not sure, if we can square on both sides of this expression without knowing sign of a, |a| > a + 1
if a is negative, after squaring the equality sign stays, but if a is positive, the equality sign is flipped

Please correct me if i am wrong.

Thanks
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1181
Location: India
GPA: 3.82
Re: Is a < -1? [#permalink]

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11 Dec 2017, 04:17
hellosanthosh2k2 wrote:
niks18 wrote:
chetan2u wrote:
Is $$a<-1$$?

(1) $$|a|>a+1$$

(2) $$a^2>1$$

Statement 1: implies that $$a>a+1 =>0>1$$. Not Possible

or $$a<-(a+1) => a<-0.5$$ but $$a$$ can be less than $$-1$$ or greater than $$-1$$. Hence insufficient

Another way to analyse this statement -

$$|a|>a+1$$, square both sides to get

$$a^2>a^2+2a+1 => a<-0.5$$

Statement 2: $$a^2>1 => a>1$$ or $$a<-1$$. Hence Insufficient

Combining 1 & 2: $$a<-0.5$$ & $$a<-1$$, the overlapping region is $$a<-1$$. Sufficient

Option C

Hi niks18,
I am not sure, if we can square on both sides of this expression without knowing sign of a, |a| > a + 1
if a is negative, after squaring the equality sign stays, but if a is positive, the equality sign is flipped

Please correct me if i am wrong.

Thanks

Hi hellosanthosh2k2

$$a$$ can be negative but $$|a|$$ is always positive

so we know that $$|a| > a + 1$$ implies that some positive number > some number for e.g if $$a=-5$$, then $$|a|=|-5|=5$$

$$5>-5+1 =>5>-4$$, clearly you can square both sides of the inequality without changing the sign.

In this case $$a$$ cannot be positive because in that case the equation will be

$$a>a+1 => 0>1$$ which is impossible.
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Joined: 22 Aug 2013
Posts: 1273
Location: India
Re: Is a < -1? [#permalink]

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11 Dec 2017, 05:44
chetan2u wrote:
Is $$a<-1$$?

(1) $$|a|>a+1$$

(2) $$a^2>1$$

(1) |a| - a > 1
Now if a is positive or 0, then |a| = a, so the above becomes a-a > 1 or 0 > 1, which is NOT possible. So a cannot be positive or 0.
If a is negative, |a| = -a, so the above becomes -a-a > 1 or a < -1/2. So we know that a is < -1/2, but we dont know whether a is < -1 or not (because a also might lie between -1/2 and -1). So Insufficient.

(2) a^2 > 1
This means a > 1 if a is positive, and a < -1 if a is negative. But we dont know whether a is positive or negative. So Insufficient.

Combining the two statements, first statement rules out that a can be positive or zero. So a can only be negative, and so according to statement 2, it must be < -1. Sufficient.

Re: Is a < -1?   [#permalink] 11 Dec 2017, 05:44
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