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Is a < -1?

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Is a < -1? [#permalink]

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New post 02 Nov 2017, 08:51
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Question Stats:

54% (01:35) correct 46% (01:00) wrong based on 79 sessions

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Is \(a<-1\)?


(1) \(|a|>a+1\)

(2) \(a^2>1\)

Self made : tricky
[Reveal] Spoiler: OA

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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: Is a < -1? [#permalink]

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New post 02 Nov 2017, 10:14
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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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Is a < -1? [#permalink]

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New post 02 Nov 2017, 10:51
chetan2u wrote:
Is \(a<-1\)?


(1) \(|a|>a+1\)

(2) \(a^2>1\)

Self made : tricky


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

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Re: Is a < -1? [#permalink]

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New post 10 Dec 2017, 11:20
niks18 wrote:
chetan2u wrote:
Is \(a<-1\)?


(1) \(|a|>a+1\)

(2) \(a^2>1\)

Self made : tricky


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

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Re: Is a < -1? [#permalink]

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New post 11 Dec 2017, 03:17
hellosanthosh2k2 wrote:
niks18 wrote:
chetan2u wrote:
Is \(a<-1\)?


(1) \(|a|>a+1\)

(2) \(a^2>1\)

Self made : tricky


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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Re: Is a < -1? [#permalink]

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New post 11 Dec 2017, 04:44
chetan2u wrote:
Is \(a<-1\)?


(1) \(|a|>a+1\)

(2) \(a^2>1\)

Self made : tricky


(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.

Hence C answer

Kudos [?]: 184 [0], given: 289

Re: Is a < -1?   [#permalink] 11 Dec 2017, 04:44
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