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Is k^2/m < 0 (1) 3 < k < 5 (2) 2 < m < 4
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13 May 2012, 22:47
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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)
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13 May 2012, 22:51
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
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13 May 2012, 23:03



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Re: Is k^2/m < 0 (1) 3 < k < 5 (2) 2 < m < 4
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14 May 2012, 21:13
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
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14 May 2012, 22:13



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Re: Is k^2/m < 0 (1) 3 < k < 5 (2) 2 < m < 4
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14 May 2012, 22:45
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
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15 May 2012, 03:30
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
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15 May 2012, 10:18
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. Give me kudos if u like my post !!



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Re: Is k^2/m < 0 (1) 3 < k < 5 (2) 2 < m < 4
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15 May 2012, 10:19
That really helps! Thank you Bunuel!



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Re: Is k^2/m < 0 (1) 3 < k < 5 (2) 2 < m < 4
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15 May 2012, 10:41
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. Give me kudos if u like my post !! Please notice that the GMAT is dealing only with real numbers.
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Re: Need your help  DS (Inequalities)
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29 Oct 2012, 03:55
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 nonnegative. Is k^2/m < 0k^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. OA is given as C, can you please confirm the OA Cheers Manisha



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Re: Need your help  DS (Inequalities)
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29 Oct 2012, 04:02
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 nonnegative. Is k^2/m < 0k^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. 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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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Is k^2/m < 0 (1) 3 < k < 5 (2) 2 < m < 4
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06 Sep 2014, 19:34
Stem: "k" will always be a nonnegative 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. Posted from my mobile device
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Re: Is k^2/m < 0 (1) 3 < k < 5 (2) 2 < m < 4
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