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m < 0 < n

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Sonalika42
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m < 0 < n [#permalink] New post   Updated on: 14 Feb 2023, 10:27
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\(m < 0 < n\)

Quantity A
Quantity B
\(m^{-1}\) minus \(n^{-1}\)
0






A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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B

Originally posted by Sonalika42 on 29 Aug 2016, 23:52.
Last edited by Carcass on 14 Feb 2023, 10:27, edited 4 times in total.
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Saurabh03121992
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Re: m < 0 < n [#permalink] New post   30 Aug 2016, 00:21
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B

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Re: m < 0 < n [#permalink] New post   30 Aug 2016, 04:54
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Sonalika42 wrote:
m < 0 < n

Quantity A
Quantity B
m^(-1) - n^(-1)
0




A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Quantity A: m^(-1) - n^(-1)
Quantity B: 0

NOTE: If m is NEGATIVE, then m^(-1) = 1/m = 1/NEGATIVE = NEGATIVE
If n is POSITIVE, then n^(-1) = 1/n = 1/POSITIVE = POSITIVE

So, we get:
Quantity A: NEGATIVE - POSITIVE
Quantity B: 0

Since a negative value minus a positive value will always be negative, we get:
Quantity A: NEGATIVE
Quantity B: 0

Answer:
Show: ::
B


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Kimberly99
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Re: m < 0 < n [#permalink] New post   14 Feb 2023, 10:11
Hi Brent GreenlightTestPrep, I got D here. Not sure where did it go wrong?
Thanks Brent

m^(-1) - n^(-1)
1/m - 1/n
n-m/mn
multiply with mn and left with n-m
n=1, m=-1 > n-m = 0 = with B :0
n=2, m=-3 > n-m = 5 > B :0
So is D.
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Carcass
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m < 0 < n [#permalink] New post   14 Feb 2023, 10:36
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I am not sure how you did manipulate

we do have \(\frac{1}{m} -\frac{1}{n}\)

\(\frac{n-m}{mn}\)

Now, iff you pick numbers such as m=-2 and n =1

we have

\(\frac{1+2}{-2*1}=\frac{2}{-2}=-1\) A will be always negative

B is the answer
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Kimberly99
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Re: m < 0 < n [#permalink] New post   16 Feb 2023, 05:19
Carcass wrote:
I am not sure how you did manipulate

we do have \(\frac{1}{m} -\frac{1}{n}\)

\(\frac{n-m}{mn}\)

Now, iff you pick numbers such as m=-2 and n =1

we have

\(\frac{1+2}{-2*1}=\frac{2}{-2}=-1\) A will be always negative

B is the answer


Thanks Carcass for your reply.
m < 0 < n , think I probably made the mistake here by taking m as positive here too of doing multiplication of mn to both side to elimiate it. That's why it has gone wrong.
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Carcass
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Re: m < 0 < n [#permalink] New post   16 Feb 2023, 06:37
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You are doing careless mistakes that a student cannot afford when he/she takes the GRE test.

Be careful
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Re: m < 0 < n [#permalink]
16 Feb 2023, 06:37
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