Is k + k - 2 > 0 (1) k < 1 (2) k > -1

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Is k + k - 2 > 0 (1) k < 1 (2) k > -1 [#permalink]

09 Nov 2005, 21:24

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Is kÂ² + k - 2 > 0
(1) k < 1
(2) k > -1

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09 Nov 2005, 21:33

(1) Insufficient.

If k =0 then k^2 + K -2 < 0
If k = -10 then k^2 + K -2 > 0

(2) Insufficient.

Same reasoning above

(1/2) The boundary is now -1 < k < 1

Which implies that k^2 + K -2 will always be less than 0.

Therefore the answer to the question is NO.

I pick C.

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Re: DS: Inequality [#permalink]

09 Nov 2005, 21:35

sudhagar wrote:

Is kÂ² + k - 2 > 0
(1) k < 1
(2) k > -1

C
Is (k-1)(k+2)>0
A) Insuff.
k=-2 -> Expression is False
k=-3 -> Expression is True
B Insuff.
k=1 -> Expression is False
k=2 -> Expression is True

Together, -1<k<1
This keeps (k-1) always -ve and (k+2) always +ve. Hence Sufficient.

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Re: DS: Inequality [#permalink]

09 Nov 2005, 22:54

gsr wrote:

sudhagar wrote:

Is kÂ² + k - 2 > 0
(1) k < 1
(2) k > -1

C. Is (k-1)(k+2)>0
A) Insuff.
k=-2 -> Expression is False
k=-3 -> Expression is True
B Insuff.
k=1 -> Expression is False
k=2 -> Expression is True

Together, -1<k<1. This keeps (k-1) always -ve and (k+2) always +ve. Hence Sufficient.

Is kÂ² + k - 2 > 0?
or, is (k-1)(k+2)>0?
or, is k>1, and/or k>-2?

from i, k is smaller than 1 but greater than -2. so not suff.
from ii, k is greater than -1 but could be greater than -2. so not suff.

from i and ii, 1>k>-1.

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09 Nov 2005, 23:27

C.

A and B are both insufficient ...
and equation is always < 0 when -1 <k<1 ....hence C

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Re: DS: Inequality [#permalink]

09 Nov 2005, 23:39

C
Stmt1: Let K = 0 then -2 > 0 false.
Let K = -5 then 18> 0 true. So stmt 1 is insufficient.

Stmt2: Let k = 0 then -2>0 false
Let k = 5 then 28 > 0 true. So stmt 2 is insufficient.

Combing both -1<K<1. Let K = 0 -2 > 0 false. Let k = .5 then -1.25 > 0 false. Let k = -.5 then -2.25 > 0 false. Hence the combined statments are sufficient.

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11 Nov 2005, 00:50

Thats correct guys, OA is C

[#permalink] 11 Nov 2005, 00:50

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Is k + k - 2 > 0 (1) k < 1 (2) k > -1

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Is k + k - 2 > 0 (1) k < 1 (2) k > -1

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