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# If –8 < k < 8, is k <2> 0

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Manager
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If –8 < k < 8, is k <2> 0 [#permalink]  10 Jun 2007, 12:04
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If –8 < k < 8, is k <2> 0
(2) 1/k > 1/2

Manager
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[#permalink]  10 Jun 2007, 12:06
Not sure why the questions did post correctly, retrying

If –8 < k < 8, is k <2> 0
2. 1/k > 1/2
SVP
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[#permalink]  10 Jun 2007, 12:08
U should try to check the checkbox "Disable HTML in this post" just under the area to tape your question
Manager
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[#permalink]  10 Jun 2007, 12:09
OK, final attempt

If -8<k<8, is k<2> 0
2. 1/k > 1/2
Manager
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[#permalink]  10 Jun 2007, 12:11
Thanks Fig, final final...

If –8 < k < 8, is k < –2 ?

(1) k^2 – 7k – 18 > 0
(2) 1/k > 1/2
SVP
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[#permalink]  10 Jun 2007, 12:47
(A) for me

We have : |k| < 8

k < -2 ?

From 1
k^2 – 7k – 18 > 0

If k = -2, then (-2)^2 -7*(-2) - 18 = 18 - 18 = 0. -2 is a root.

k^2 – 7k – 18 > 0
<=> (k+2)*(k-9) > 0

Outside of the roots, a*x^2 + b*x + c is of the signt of a.

Here, we have so:
(k+2)*(k-9) > 0
<=> k > 9 or k < -2.

As |k| < 8, we are sure that -8 < k < -2.

SUFF.

From 2
1/k > 1/2 > 0... So, k > 0.

Thus,
k < 2.

INSUFF.
Senior Manager
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[#permalink]  11 Jun 2007, 12:28
I feel it is B.

1/K > 1/2 => k is between 0 and 2 (both excluded)

In stmt 1 K > 9 or k < -2 so insufficient
Manager
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[#permalink]  11 Jun 2007, 12:50
Looks like D to me.
1 is explained by Fig

2. 1/k>1/2 --->>> k must be positive so it has to be more than -2 in any case.

What is the OA?
VP
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The answer is B [#permalink]  11 Jun 2007, 17:25
If –8 < k < 8, is k < –2 ?

(1) k^2 – 7k – 18 > 0
(2) 1/k > 1/2
________________________________________

Rephase the question: The question is really asking if -8<k<-2?

(1) k^2 - 7k -18 >0 is the same as (k-9)*(k+2) > 0
From this, we know that the critical points are k=9 and k=-2. The range of k where (k-9)*(k+2) > 0 are k<-2 and k>9. INSUFFICIENT.

(2) 1/k > 1/2
From this, we know that 0<k<2. This is SUFFICIENT since k will never be between -8 and -2.
VP
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[#permalink]  11 Jun 2007, 17:33
Fig wrote:
Here, we have so:
(k+2)*(k-9) > 0
<k> 9 or k < -2.

As |k| < 8, we are sure that -8 < k < -2.

No, this is incorrect. |k|<8 is the same as -8<k<8. We don't know for sure that k is between -8 and 8. The answer only say k<-2. If k=-13, then k will NOT be between -8 and 8.
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[#permalink]  11 Jun 2007, 17:36
St1:
(k-9)(k+2) > 0
valid range: k <2> 9

Insufficient.

St2:
k<2.
Sufficient. Because k must be at positive but not 0.

Ans B
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[#permalink]  11 Jun 2007, 22:38
Hayabusa wrote:
Looks like D to me.
1 is explained by Fig

2. 1/k>1/2 --->>> k must be positive so it has to be more than -2 in any case.

What is the OA?

U are right... (D) it is ... The second statment is 0 < k < 2
SVP
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[#permalink]  11 Jun 2007, 22:44
bkk145 wrote:
Fig wrote:
Here, we have so:
(k+2)*(k-9) > 0
<k> 9 or k < -2.

As |k| < 8, we are sure that -8 < k < -2.

No, this is incorrect. |k|<8 is the same as -8<k<8. We don't know for sure that k is between -8 and 8. The answer only say k<-2. If k=-13, then k will NOT be between -8 and 8.

We analyse the inequation without constraint and then we add |k| < 8 to conclude.... So, no, we do not say k = -13
Manager
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[#permalink]  11 Jun 2007, 23:49
[u]Statement 1[/u]
k^2-7k-18>0
(k-9)(k+2)>0
it implies that either k-9 and k+2 both are positive or negative. If they are positive then, k>9 (which is not possible according to the stem), and if they are negative, then k<2.
Since, k<2 does not clearly tells us whether k<2>1/2
This statement tell us two things. 1) that K is positive and 2) that k<2. Since K is positive and smaller than 2, so this statement definitely tells us that K cannot be <-2. So it is SUFFICIENT.
Therefore, the answer is B
VP
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[#permalink]  12 Jun 2007, 01:13
Tricky one...

D for me:

1) The inequality gives us as already stated:
(K-9)*(K+2)>0
So K<2>9; As -8<k<8 only the "left side" of the inequality counts giving us -8<k<2>1/2

The result is k<2, k must be positive, , so 0<k<2; SUF
Director
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[#permalink]  12 Jun 2007, 07:59
If (k+2)*(k-9) > 0
does not that means
either k+2>0 or k-9>0
either k>-2 OR k>9

How did you get k < -2?
VP
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[#permalink]  12 Jun 2007, 08:09
[quote="asaf"]If (k+2)*(k-9) > 0
does not that means
either k+2>0 or k-9>0
either k>-2 OR k>9

How did you get k <2>0...

So for K<2>9, both are positive....
VP
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[#permalink]  12 Jun 2007, 08:10
asaf wrote:
If (k+2)*(k-9) > 0
does not that means
either k+2>0 or k-9>0
either k>-2 OR k>9

How did you get k < -2?

Take care, is not "or", but "and"

(K+2) "and" (k-9)>0...

So for K<-2 both terms are negative, and the product is positive, also for k>9, both are positive....
Manager
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[#permalink]  12 Jun 2007, 20:29
OA is D

2) 1/k > 1/2 means K must be positive and less than 2 ==> Suff

1) (k-9)*(k+2) > 0 implies
either (K-9) > 0 AND (k+2) > 0 -----> Both positive -- (a)
or (K-9) < 0 AND (k+2) < 0 -----> Both negative -- (b)

But since |k| < 8, (k-9) can never be > 0. This rules out (a)

from (b) we can conclude k < -2

Hence 2) alone is sufficient
Senior Manager
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[#permalink]  12 Jun 2007, 21:47
D

a tells us that K is either -2 or 9. 9 is out since we already know that it is less than 8. Sufficient

B tells us that K is positive. It could be a fraction, but in all cases it will be larger then -2 also sufficient
[#permalink] 12 Jun 2007, 21:47
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# If –8 < k < 8, is k <2> 0

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