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If x is an integer, is |x|>1 ? a:(1 - 2x)(1 + x) < 0

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If x is an integer, is |x|>1 ? a:(1 - 2x)(1 + x) < 0 [#permalink]

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New post 12 Oct 2008, 06:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks

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Re: Ineuality: How to solve [#permalink]

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New post 12 Oct 2008, 07:47
vishalgc wrote:
If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks

obviously each alone is not suff

both( multiply)

(1 - x^2)(1+4x^2) < 0

thus sure

x^2>1 hence

/x/>1 suff

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Re: Ineuality: How to solve [#permalink]

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New post 12 Oct 2008, 10:01
yezz wrote:
vishalgc wrote:
If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks

obviously each alone is not suff

both( multiply)

(1 - x^2)(1+4x^2) < 0

thus sure

x^2>1 hence

/x/>1 suff


Yezz .... I don't understand how you get (1 - x^2)(1+4x^2) < 0 .

Both statements give us 2 negative expressions. If we multiply them should we not get a positive number ??

Please explain.

Yezz can we solve it this way ?

|x| means x<-1 or x>1

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

a : from a we get x >1/2 (OR) x < -1

But a is still not sufficient.

b : x > 1 (OR) x < -1/2

But b is not sufficient.


Assume this is the number line ............(-1)......(-1/2)..........(0)..........(1/2)........(1)............

=====> x lies on the red portion below

from a we get x >1/2 (OR) x < -1

............(-1)......(-1/2)..........(0)..........(1/2)........(1)............

from b we get x > 1 (OR) x < -1/2

............(-1)......(-1/2)..........(0)..........(1/2)........(1)............

Together we can say x<-1 and x>1

............(-1)......(-1/2)..........(0)..........(1/2)........(1)............


So C

Is the approach right ??
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Re: Ineuality: How to solve [#permalink]

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New post 12 Oct 2008, 10:16
yezz wrote:
vishalgc wrote:
If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks

obviously each alone is not suff

both( multiply)

(1 - x^2)(1+4x^2) < 0

thus sure

x^2>1 hence

/x/>1 suff


if multiplied, (1 - x^2)(1+4x^2) < 0
if (1 - x^2) < 0, (1+4x^2) has to be > 0. possible
if (1 - x^2) > 0, (1+4x^2) has to be < 0 but (1+4x^2) < 0 is not possible because if so, (4x^2) < (-1), which unreal.

now only possibilities is if (1 - x^2) < 0, (1+4x^2) > 0.
then 1 < x^2, then lxl > 1.
but when (1+4x^2) > 0, x^2 > -1/4. in this case, x could be a fraction or 1 or > 1. so insuff.

shouls be E.
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Re: Ineuality: How to solve [#permalink]

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New post 12 Oct 2008, 10:20
amitdgr wrote:
yezz wrote:
vishalgc wrote:
If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks

obviously each alone is not suff

both( multiply)

(1 - x^2)(1+4x^2) < 0

thus sure

x^2>1 hence

/x/>1 suff


Yezz .... I don't understand how you get (1 - x^2)(1+4x^2) < 0 .

Both statements give us 2 negative expressions. If we multiply them should we not get a positive number ??

Please explain.

Yezz can we solve it this way ?

|x| means x<-1 or x>1

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

a : from a we get x >1/2 (OR) x < -1

But a is still not sufficient.

b : x > 1 (OR) x < -1/2

But b is not sufficient.


Assume this is the number line ............(-1)......(-1/2)..........(0)..........(1/2)........(1)............

=====> x lies on the red portion below

from a we get x >1/2 (OR) x < -1

............(-1)......(-1/2)..........(0)..........(1/2)........(1)............

from b we get x > 1 (OR) x < -1/2

............(-1)......(-1/2)..........(0)..........(1/2)........(1)............

Together we can say x<-1 and x>1

............(-1)......(-1/2)..........(0)..........(1/2)........(1)............


So C

Is the approach right ??



a: if (1 - 2x)(1 + x) < 0, you cannot say that (1 - 2x) < 0 and (1 + x) < 0 because (1 - 2x)(1 + x) is -ve. so if (1 - 2x) is +ve, (1 + x) is -ve and vice versa.
b: apply same to b as well.
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Re: Ineuality: How to solve [#permalink]

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New post 12 Oct 2008, 10:32
vishalgc wrote:
If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks

IMO A

a) answers saying : -1<x<1/2 => always |x|<1
b)does not answer since x <-1/2 or x>1
|x| >1 or |x|<1 hence INSUFFI

IMO A
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Re: Ineuality: How to solve [#permalink]

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New post 12 Oct 2008, 14:03
yezz wrote:
vishalgc wrote:
If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks

obviously each alone is not suff

both( multiply)
(1-x)(1+x)(1-2x)(1+2x)
(1 - x^2)(1 - 4x^2) < 0 ( i wrote the sign as +ve by mistake before editing)

thus

negativivty source is unknown
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Re: Ineuality: How to solve [#permalink]

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New post 12 Oct 2008, 14:27
E

a:(1 - 2x)(1 + x) < 0
Either : 1-2x <0 and 1+x >0 or 1-2x>0 and 1+x <0
x>1/2, x>-1 or x<1/2, x<-1
Not Sufficient

b:(1 - x)(1 + 2x) < 0
Either : 1+2x <0 and 1-x >0 or 1+2x>0 and 1-x <0
x<-1/2, x<1 or x>-1/2, x>1
Not Sufficient

We can't really multiply and solve this equation :

a= -2, b= -3 -- > a<0, b<0 but that does not mean that ab(6)<0

I am kind of stuck here, so any math gurus out there ?

In fact I think : a<0, b<0 --> ab >0

so (1-x^2)(1-4x^2) >0

x^2<1 and 4x^2<1

Not Sufficient

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Re: Ineuality: How to solve [#permalink]

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New post 13 Oct 2008, 00:17
LiveStronger wrote:
E

a:(1 - 2x)(1 + x) < 0
Either : 1-2x <0 and 1+x >0 or 1-2x>0 and 1+x <0
x>1/2, x>-1 or x<1/2, x<-1
Not Sufficient

b:(1 - x)(1 + 2x) < 0
Either : 1+2x <0 and 1-x >0 or 1+2x>0 and 1-x <0
x<-1/2, x<1 or x>-1/2, x>1
Not Sufficient

We can't really multiply and solve this equation :

a= -2, b= -3 -- > a<0, b<0 but that does not mean that ab(6)<0

I am kind of stuck here, so any math gurus out there ?

In fact I think : a<0, b<0 --> ab >0

so (1-x^2)(1-4x^2) >0

x^2<1 and 4x^2<1

Not Sufficient

Agreed on this !!i did a silly mistake in solving inequality !!
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Re: Ineuality: How to solve [#permalink]

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New post 13 Oct 2008, 05:31
C for me.

From stmt1: either x > 1/2 or x < -1
From stmt2: either x > 1 or x < -1/2

Combining two, either x > 1 or x < -1. Hence, sufficient.

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Re: Ineuality: How to solve [#permalink]

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New post 14 Oct 2008, 08:39
vishalgc wrote:
If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks


Do you have OA for this one?
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Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


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Re: Ineuality: How to solve [#permalink]

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New post 14 Oct 2008, 11:17
vishalgc wrote:
If x is an integer, is |x|>1 ?

a:(1 - 2x)(1 + x) < 0
b:(1 - x)(1 + 2x) < 0

How to solve this kind of question very fast? or which is proper and authenticate way to solve this kind of question?

Thanks


a: (1-2x)*(1+x) < 0
(x-1/2)*(x+1) > 0 -> x < -1 or x > 1/2 -> since x is an integer x < -1 or x>=1 -> insufficient

b: (x-1)*(x+1/2)> 0 -> x < -1/2 or x > 1 -> since x is an integer x<=-1 or x>1 -> insufficient

a&b: x < -1 or x> 1: sufficient -> C

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Re: Ineuality: How to solve   [#permalink] 14 Oct 2008, 11:17
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