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If k not equal to 0, 1 or -1 is 1/k>0? (1) 1/(k-1) >0

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If k not equal to 0, 1 or -1 is 1/k>0? (1) 1/(k-1) >0 [#permalink] New post 22 Oct 2006, 06:46
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If k not equal to 0, 1 or -1 is 1/k>0?

(1) 1/(k-1) >0
(2) 1/(k+1)>0

Any explanations will be appreciated.

Last edited by ishtmeet on 22 Oct 2006, 07:29, edited 1 time in total.
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 [#permalink] New post 22 Oct 2006, 07:10
I pick B
Since K>0 so 1/K>0
a. not suf, if K=- 0,5, equation holds true and K<0
b. suf, K has to be greater than 1
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 [#permalink] New post 22 Oct 2006, 07:24
For me (A),

k != -1,0,1

1/k>0 ?

NB: K could be any real numbers.

Stat1:
1/k-1 >0
<=> 1/k > 1 > 0

SUFF.

Stat2:
1/k+1 >0
<=> 1/k > -1

Pick numbers to be sure :)
o k = -2, -1/2 > -1 and -1/2 < 0
o k = 2, 1/2 > -1 and 1/2 > 0

INSUFF.
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 [#permalink] New post 22 Oct 2006, 07:28
Hey guys sorry for the confusion but it is 1/(k-1) and 1/(k+1). I have modified the question.
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 [#permalink] New post 22 Oct 2006, 07:35
ishtmeet wrote:
Hey guys sorry for the confusion but it is 1/(k-1) and 1/(k+1). I have modified the question.


The question changes but my answer remain the same ;) It's still (A) for me :)
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 [#permalink] New post 22 Oct 2006, 07:39
If k not equal to 0, 1 or -1 is 1/k>0?

(1) 1/(k-1) >0
(2) 1/(k+1)>0

question is

is k<0

from one

k-1<0thus k< 1 could be a +ve fraction OR -VE .......not suff

from two

k+1<0 thus k <-1 thus ure k<0.......suff

my answer is B
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 [#permalink] New post 22 Oct 2006, 07:45
yezz wrote:
If k not equal to 0, 1 or -1 is 1/k>0?

(1) 1/(k-1) >0
(2) 1/(k+1)>0

question is

is k<0

from one

k-1<0thus k< 1 could be a +ve fraction OR -VE .......not suff

from two

k+1<0 thus k <-1 thus ure k<0.......suff

my answer is B


Sorry Yezz, ... it should be for both bold > 0 :)
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 [#permalink] New post 22 Oct 2006, 07:56
Why fig ......can you ellaborate

i believe if we have

1/x<0 we flip to x>0......wuts wrong in this
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 [#permalink] New post 22 Oct 2006, 07:59
yezz wrote:
Why fig ......can you ellaborate

i believe if we have

1/x<0 we flip to x>0......wuts wrong in this


U cannot use the fonction f(x) = 1/x here.... we have 0 by right side :)

We have to use the properties of the operaiton "division" :)

Sign(1/x) < 0
<=> Sign(1)/Sign(x) <0
<=> Sign(x) < 0

or

1/x < 0
<=> x < 0 as 1 > 0 :)

Last edited by Fig on 22 Oct 2006, 08:00, edited 1 time in total.
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 [#permalink] New post 22 Oct 2006, 08:00
st 1 says 1/(k-1)>0 that means k-1>0 that means k>1 that means k is +ve and thus 1/k>0 ... suff

st2 says 1/(k+1) >0 that means k+1>0 i.e k>-1 that means k can -ve fraction or k is positive . if k is -neg fraction 1/k <0 and if k is +ve 1/k >0 so insuff

Thus the answer is A.

What is the OA?
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 [#permalink] New post 22 Oct 2006, 08:03
Hi Guys,
The OA is A. Can anybody explain how did thi answer come??

As per the solutions given by folks since one says 1/(k-1) >0 it implies k-1 <0 or k<1 therefore K maybe greater that 0 0r less than 0. SO how come the correct answer is A in GMAT PREP.
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 [#permalink] New post 22 Oct 2006, 08:10
ishtmeet wrote:
Hi Guys,
The OA is A. Can anybody explain how did thi answer come??

As per the solutions given by folks since one says 1/(k-1) >0 it implies k-1 <0 or k<1 therefore K maybe greater that 0 0r less than 0. SO how come the correct answer is A in GMAT PREP.


yogeshsheth did a great job on it :)

Stat1:
1/(k-1) >0
=> (k-1) > 0
<=> k > 1 > 0
=> 1/k > 0

SUFF

Stat2:
1/(k+1) >0
=> (k+1) > 0
<=> k > -1
=> 1/k < -1 or 1/k > 0

INSUFF
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 [#permalink] New post 22 Oct 2006, 08:12
Thanks Fig :)
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 [#permalink] New post 22 Oct 2006, 08:15
fig is this right or wrong why we did it your way??
eg: 1/x< 5 thus x>1/5

but if

1/x <0 we cant say x>1/0 because it is not defined ......right or wrong
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 [#permalink] New post 22 Oct 2006, 08:18
Correct yezz one can not take a reciprocal of inequality of type x<0

x<0 does not imply that 1/x >0 as 1/0 is not defined.

Hope that helps.
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 [#permalink] New post 22 Oct 2006, 08:20
yezz wrote:
fig is this right or wrong why we did it your way??
eg: 1/x< 5 thus x>1/5

but if

1/x <0 we cant say x>1/0 because it is not defined ......right or wrong


Correct :)

The fonction f(x) = 1/x decreases when x inscreases but it's not defined when x = 0. Thus,

5 > 1
<=> f(5) < f(1) : the fonction decreases
<=> 1/5 < 1


Another example:
g(x)= x+1

The fonction g increases when x increases.

5 > 1
<=> g(5) > g(1) : the fonction increases
<=> 6 > 2

We flip when the function decreases when x increases :)
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 [#permalink] New post 22 Oct 2006, 08:25
Thanks yougish and Fig ......u saved my life :-D

I have a problem with iniqualities i dont know how to get over it

well i will keep trying and solving as much as i can

do u suggest any thing else i can do??
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 [#permalink] New post 22 Oct 2006, 08:28
yezz wrote:
Thanks yougish and Fig ......u saved my life :-D

I have a problem with iniqualities i dont know how to get over it

well i will keep trying and solving as much as i can

do u suggest any thing else i can do??


U are welcome :D :)

Only thing, do not hesitate to come back to look some math basics :)... but the most important is as u said: the practice :)
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 [#permalink] New post 22 Oct 2006, 08:32
Fig wrote:
yezz wrote:
Thanks yougish and Fig ......u saved my life :-D

I have a problem with iniqualities i dont know how to get over it

well i will keep trying and solving as much as i can

do u suggest any thing else i can do??


U are welcome :D :)

Only thing, do not hesitate to come back to look some math basics :)... but the most important is as u said: the practice :)


Welcome yezz :)

Completely agree with you Fig practice helps in getting familiar in dealing with ineaualities
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 [#permalink] New post 22 Oct 2006, 10:28
yezz wrote:
fig is this right or wrong why we did it your way??
eg: 1/x< 5 thus x>1/5

but if

1/x <0 we cant say x>1/0 because it is not defined ......right or wrong


Yes don't use the flip sign method when you have zero on the right side. In stead, think this way. For a product or ratio of two numbers to be positive, the two numbers have to have the same sign; for the product or ratio to be negative, the two numbers must have different signs.
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  [#permalink] 22 Oct 2006, 10:28
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