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Data Sufficiency Pack 3, Question 2) If gh < 0 < gk...

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Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 07 Nov 2015, 16:23
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QUANT 4-PACK SERIES Data Sufficiency Pack 3 Question 2 If gh < 0 < gk...

If gh < 0 < gk, is g < 0?

1) k < h
2) 0 < h


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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 07 Nov 2015, 17:18
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EMPOWERgmatRichC wrote:
QUANT 4-PACK SERIES Data Sufficiency Pack 3 Question 2 If gh < 0 < gk...

If gh < 0 < gk, is g < 0?

1) k < h
2) 0 < h


48 Hour Window Answer & Explanation Window
Earn KUDOS! Post your answer and explanation.
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

Scroll Down For Official Explanation



Given gh<0<gk ---> gh<gk , is g<0?

Statement 1, k<h ---> h-k>0 but as gh<gk --> g(h-k)< 0 ---> as h-k>0 the only way g(h-k)< 0 is for g<0, "yes" . Sufficient.

Statement 2, h>0 ---> as gh<0 and h>0 ---> g<0, "yes". Sufficient.

D is the correct answer.
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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 08 Nov 2015, 18:49
1
IMO D is correct'


I] k<h but gh<gk so this only possible when g<0 .
Sufficient.

II] h>0 but gh<0 again this only possible when g<0

Sufficient.

So D is the correct answer.
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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 09 Nov 2015, 02:57
Please someone correct me. This is how I approached the problem.

gh<0 implies either g or h <0 ; (h can be >0 or h <0)
gk> 0 leads to scenarios : either bot g and k are > 0 or g and k are < 0

second statement was pretty simple to evaluate.

first statement : k< h ; this is possible when both k and h are positive --> g has to be positive ( hence g>0) since the equality gk>0 has to be true.
k<h is possible even when k< 0 , (and h< 0 or h> 0 both cases) in which case g has to be <0 for the equality gk>0 to be true. (g<0)

thus not sufficient.

where am i going wrong ?
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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 09 Nov 2015, 04:12
Rk91 wrote:
Please someone correct me. This is how I approached the problem.

gh<0 implies either g or h <0 ; (h can be >0 or h <0)
gk> 0 leads to scenarios : either bot g and k are > 0 or g and k are < 0

second statement was pretty simple to evaluate.

first statement : k< h ; this is possible when both k and h are positive --> g has to be positive ( hence g>0) since the equality gk>0 has to be true.
k<h is possible even when k< 0 , (and h< 0 or h> 0 both cases) in which case g has to be <0 for the equality gk>0 to be true. (g<0)

thus not sufficient.

where am i going wrong ?


hi,
if k<h and both are multiplied by same number...
if the number is positive, x*k<x*h..
only if we multiply both numbers by a negative number, there is a reversal of sign ie x*k>x*h
so x or g here has to be a negative number since gh < 0 < gk..

now why is your reasoning flawed?
why are you taking the two numbers to be positive..
both can be negative or one can be negative and other positive..
here the second is the case k is -ive and h is positive and therefore their signs change when multiplied by a -ive number..
hope it was helpful
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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 09 Nov 2015, 05:02
1
Hi,

Below was my thought process.

Question stem: gh<0<gk, g<0?

Below is what I inferred from the question stem:
gk>0 i.e. g,k<0 or g,k>0
gh<0 i.e. g<0, h>0 or g>0, h<0


1: k<h
For statement 1 to comply with question stem (gh<0<gk) , I have to multiply on both sides with a negative value i.e. (-ve mutliplier)*(k) > (-ve multiplier) * (h). Therefore, g will have to be that negative multiplier. i.e. g<0. I could not think of any other way the sign will get flipped.
Sufficient

2: 0< h
Based on the question stem, if h>0, g<0.
Sufficient

Hope this helps.

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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 09 Nov 2015, 05:29
1
Rk91 wrote:
Please someone correct me. This is how I approached the problem.

gh<0 implies either g or h <0 ; (h can be >0 or h <0)
gk> 0 leads to scenarios : either bot g and k are > 0 or g and k are < 0

second statement was pretty simple to evaluate.

first statement : k< h ; this is possible when both k and h are positive --> g has to be positive ( hence g>0) since the equality gk>0 has to be true.
k<h is possible even when k< 0 , (and h< 0 or h> 0 both cases) in which case g has to be <0 for the equality gk>0 to be true. (g<0)

thus not sufficient.

where am i going wrong ?


Whenever you are given variables to play with in inequalities, be very careful about your assumptions.

You are given, gh<0<gk ---> gh<0 , 2 cases either g<0 and h>0 or g>0 and h<0

Similarly, gk>0 ---> g>0 k>0 and g<0 and k<0

Combine the 4 cases by keeping the sign of g as constant, you get the following 2 cases,

g>0 h<0 k>0 and

g<0 h>0 k<0

Now look at statement 1, h>k ---> you can eliminate the case g<0 h>0 k<0, thus leaving you with only 1 case of g>0 h<0 k>0, making this statement sufficient to answer the question
"is g<0".

Hope this helps.

Statement in red above is not the correct interpretation of the given variables. Statement in green above is also not correct.
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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 09 Nov 2015, 07:12
Ohhhhk!! (lightbulb!) Now I feel so dumb!!!!

Thanks a lot guys for the explanations! Really appreciate it :)
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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 10 Nov 2015, 11:54
1
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If gh < 0 < gk, is g < 0?

1) k < h
2) 0 < h

We can get gh<gk, or gh-gk<0, or g(h-k)<0에서 g<0? --> h-k>0?, or h>k? when we modify the question.
Condition 1 answers the question 'yes' so it is sufficient,
Condition 2 also gives gh<0, and when h>0, g<0 so this is also sufficient; the answer becomes (D).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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New post 10 Nov 2015, 17:59
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EMPOWERgmatRichC wrote:
QUANT 4-PACK SERIES Data Sufficiency Pack 3 Question 2 If gh < 0 < gk...

If gh < 0 < gk, is g < 0?

1) k < h
2) 0 < h


Hi All,

This prompt is based heavily on NUMBER PROPERTY rules. As such, you can solve it by TESTing VALUES or by using Number Property Rules. Since the prompt itself starts us off with a Number Property rule, I'm going to start by breaking down those possibilities.

gh < 0 < gk

From this inequality, we know two things:
1) gh is NEGATIVE, so one of those variables is negative while the other is positive
2) gk is POSITIVE, so the two variables are either both negative OR both positive

When combining those pieces of information, we have a limited number of outcomes...

gh < 0 < gk
(-)(+) < 0 < (-)(-)
(+)(-) < 0 < (+)(+)

Thus, having information about one variable will impact the rest of the other variables. We're asked if g is less than 0. This is a YES/NO question.

1) k < h

From the above information, we know that k and h CANNOT be the same sign (one will be positive and the other will be negative, depending on g).

Since k is LESS than h, this means that k MUST be NEGATIVE and h MUST be POSITIVE. This fits the second 'option' (above) and proves that g MUST be POSITIVE. Thus, the answer to the question is ALWAYS NO.
Fact 1 is SUFFICIENT

2) 0 < h

This Fact proves that MUST be POSITIVE. This also fits just the second 'option' (above) and proves that g MUST be POSITIVE. Thus, the answer to the question is ALWAYS NO.
Fact 2 is SUFFICIENT

Final Answer:

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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk... [#permalink]

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Re: Data Sufficiency Pack 3, Question 2) If gh < 0 < gk...   [#permalink] 26 Mar 2017, 03:16
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