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Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4

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Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 13 May 2012, 23:47
2
5
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

65% (00:39) correct 35% (01:02) wrong based on 122 sessions

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Is k^2/m < 0

(1) -3 < k < 5
(2) 2 < m < 4
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Re: Need your help - DS (Inequalities)  [#permalink]

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New post 13 May 2012, 23:51
2
I believe the answer should be B.

As K^2 is always positive, we only need to check for value of m. Statement 1 says m can be -ve or +ve. Statement 2(2<m<4) confirms that m is always positive, hence answer is B. K^2/m is not less than 0.
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Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 14 May 2012, 00:03
2
1
gmihir wrote:
I believe the answer should be B.

As K^2 is always positive, we only need to check for value of m. Statement 1 says m can be -ve or +ve. Statement 2(2<m<4) confirms that m is always positive, hence answer is B. K^2/m is not less than 0.


The square of a number is not always positive, it's non-negative.

Is k^2/m < 0

k^2/m<0 to hold true k must not equal to zero AND m must be negative.

(1) -3 < k < 5. Not sufficient.
(2) 2 < m < 4. Since the second condition is already violated then the answer to the question is NO: \(\frac{k^2}{m}=\frac{nonnegative}{positive}=nonnegative\) . Sufficient.

Answer: B.
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 14 May 2012, 22:13
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i also have chosen B, but then thought what if k=0, we don't know from the statements, so then answer should be E.
Did i miss anything?
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 14 May 2012, 23:13
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 14 May 2012, 23:45
Excellent, i always forget that in such DS problems yes or no is sufficient.
Thanks!
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 15 May 2012, 04:30
k^2 must be equal to m, then that is possible.
but here is not, in case of inequality + or - matters
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 15 May 2012, 11:18
Bunuel wrote:
ziko wrote:
i also have chosen B, but then thought what if k=0, we don't know from the statements, so then answer should be E.
Did i miss anything?


Notice that if k=0 then k^2/m=0 and the answer to the question whether k^2/m<0 is still no.

Hope it's clear.

No the answer to this one could be C.

The stem asks is k^2/m<0
or u can write :
Is k^2*m < 0 ??
now u will wonder k^2 will always be positive.so it depends on m only whether the product is <0 or not.
But it is not the case.
As k could be an imaginary number say k = (-1)^0.5 or 'iota' as we call it, for which it fails.Since it is not mentioned here that k is a real number we cant guess on the nature on k. Thus we also need to determine the nature of k(real or imaginary).
And hence answer is C.



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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 15 May 2012, 11:19
That really helps! Thank you Bunuel! :-D
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 15 May 2012, 11:41
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piyushksharma wrote:
Bunuel wrote:
ziko wrote:
i also have chosen B, but then thought what if k=0, we don't know from the statements, so then answer should be E.
Did i miss anything?


Notice that if k=0 then k^2/m=0 and the answer to the question whether k^2/m<0 is still no.

Hope it's clear.

No the answer to this one could be C.

The stem asks is k^2/m<0
or u can write :
Is k^2*m < 0 ??
now u will wonder k^2 will always be positive.so it depends on m only whether the product is <0 or not.
But it is not the case.
As k could be an imaginary number say k = (-1)^0.5 or 'iota' as we call it, for which it fails.Since it is not mentioned here that k is a real number we cant guess on the nature on k. Thus we also need to determine the nature of k(real or imaginary).
And hence answer is C.



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Please notice that the GMAT is dealing only with real numbers.
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Re: Need your help - DS (Inequalities)  [#permalink]

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New post 29 Oct 2012, 04:55
Bunuel wrote:
gmihir wrote:
I believe the answer should be B.

As K^2 is always positive, we only need to check for value of m. Statement 1 says m can be -ve or +ve. Statement 2(2<m<4) confirms that m is always positive, hence answer is B. K^2/m is not less than 0.


The square of a number is not always positive, it's non-negative.

Is k^2/m < 0

k^2/m<0 to hold true k must not equal to zero AND m must be negative.

(1) -3 < k < 5. Not sufficient.
(2) 2 < m < 4. Since the second condition is already violated then the answer to the question is NO: \(\frac{k^2}{m}=\frac{non-negative}{positive}=non-negative\) . Sufficient.

Answer: B.



OA is given as C, can you please confirm the OA

Cheers
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Re: Need your help - DS (Inequalities)  [#permalink]

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New post 29 Oct 2012, 05:02
magicmanisha wrote:
Bunuel wrote:
gmihir wrote:
I believe the answer should be B.

As K^2 is always positive, we only need to check for value of m. Statement 1 says m can be -ve or +ve. Statement 2(2<m<4) confirms that m is always positive, hence answer is B. K^2/m is not less than 0.


The square of a number is not always positive, it's non-negative.

Is k^2/m < 0

k^2/m<0 to hold true k must not equal to zero AND m must be negative.

(1) -3 < k < 5. Not sufficient.
(2) 2 < m < 4. Since the second condition is already violated then the answer to the question is NO: \(\frac{k^2}{m}=\frac{non-negative}{positive}=non-negative\) . Sufficient.

Answer: B.



OA is given as C, can you please confirm the OA

Cheers
Manisha


OA for this question is wrong. It should be B, not C.
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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New post 06 Sep 2014, 20:34
Stem: "k" will always be a non-negative number (or zero if k=0). To get the answer we need the values for "m".

i) "k" lies between -2 and 4. We do not know the values of "m". Insufficient.

ii) "m" is positve. We have two cases
Case a) When "k"= any number (any non zero number including decimals) and "m" is postive, then k^2/m>0
Eg: k=3 and m=3 --> 9/3=3 --> 3 > 0
Eg: k=-3 and m=3 --> 9/3=3 --> 3 > 0

Case b) When k=0 and m=3
k^2/m=0. We are asked to find k^2/m<0. Therefore 0<0 is not possible. Case b is invalid.

Therefore using case a, B is sufficient to answer.

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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4  [#permalink]

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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4   [#permalink] 31 Jul 2019, 02:28
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