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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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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.

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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4 [#permalink]
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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]
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]
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]
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 [#permalink]
That really helps! Thank you Bunuel!
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4 [#permalink]
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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.

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) [#permalink]
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.

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

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

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]
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]
a) k^2 will always be positive. No information on m. Not sufficient. (cancel A,D)

b) 2<m<4; non negative/positive>0
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Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4 [#permalink]
Is K^2/M < 0?

(1) -3 < K < 5
(2) 2 < M < 4

Being the data sufficiency question, I think the answer is E but the correct answer is B?

This is how I solved it. Help me understand where I am going wrong.

The solution to this problem is based on the sign of the denominator (M)

1) The first statement has no information on M so this option is insufficient.
However, since K lies between K -3 and 5, what if the K is 0? Then (K^2/M) will be equal to 0.

So this option is not sufficient.

2) The second statement is positive since m lies between 2 and 4 but again what if K is 0? T hen (K^2/M) will be equal to 0 irrespective of what M is.

Hence my answer is E. Both statements are not sufficient. (C combining both the statements is of no help either).

Help me understand why B is the answer! m

Originally posted by Rucha96 on 03 Jun 2022, 04:12.
Last edited by Bunuel on 03 Jun 2022, 09:36, edited 2 times in total.
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Re: Is K^2/M < 0? (1) -3 < K < 5 (2) 2 < M < 4 [#permalink]
k^2 can also be 0, that situation needs to be considered. IMO E
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4 [#permalink]
Yes, I think it is clear. It is a yes or a now question. Even if it is Zero, it is still not less than 0.
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Re: Is K^2/M < 0? (1) -3 < K < 5 (2) 2 < M < 4 [#permalink]
target is k^2/m<0
only possible when M is -ve

#1
-3 < K < 5
insufficient as M is not given
#2
2 < M < 4
M is +ve so target value is +ve always
sufficient
option B

Rucha96 wrote:
Is K^2/M < 0?

(1) -3 < K < 5
(2) 2 < M < 4

Being the data sufficiency question, I think the answer is E but the correct answer is B?

This is how I solved it. Help me understand where I am going wrong.

The solution to this problem is based on the sign of the denominator (M)

1) The first statement has no information on M so this option is insufficient.
However, since K lies between K -3 and 5, what if the K is 0? Then (K^2/M) will be equal to 0.

So this option is not sufficient.

2) The second statement is positive since m lies between 2 and 4 but again what if K is 0? T hen (K^2/M) will be equal to 0 irrespective of what M is.

Hence my answer is E. Both statements are not sufficient. (C combining both the statements is of no help either).

Help me understand why B is the answer! m
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Re: Is K^2/M < 0? (1) -3 < K < 5 (2) 2 < M < 4 [#permalink]
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ArpanKanjilal wrote:
k^2 can also be 0, that situation needs to be considered. IMO E

Target question: Is K^2/M < 0?

If k^2 = 0, and M is positive, then the answer to the target question is "NO, K^2/M is not less than zero"
If k^2 > 0, and M is positive, then the answer to the target question is "NO, K^2/M is not less than zero"

Since the answer to the target question is the same in both cases (NO), statement 2 is sufficient.
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Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4 [#permalink]
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: k2m=nonnegativepositive=nonnegativek2m=nonnegativepositive=nonnegative . Sufficient

@bunnuel in the above which you is it pre fixed that k^2/m<0 to hold true k must not equal to zero AND m must be negative or will question missed to mention it out?
Re: Is k^2/m < 0 (1) -3 < k < 5 (2) 2 < m < 4 [#permalink]
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