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If n >2 and (2/n) is substituted for all instances of n in t

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Bunuel
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If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  04 Feb 2014, 06:06
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\((n – 2)^{-1}* (2 + n)\)
If n >2 and 2/n is substituted for all instances of n in the above expression, then the new expression will be equivalent to which of the following:

A. \((n + 1)(n – 1)^{(-1)}\)
B. \(–(n + 1)(n – 1)^{(-1)}\)
C. \(–(n – 1)(n + 1)^{(-1)}\)
D. \((2 + n)^{(-1)}*(n – 2)\)
E. \((n – 2)^{(-1)}*(2 + n)\)
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Bunuel
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  04 Feb 2014, 06:17
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Bunuel wrote:
(n – 2)^-1 (2 + n)
If n >2 and 2/n is substituted for all instances of n in the above expression, then the new expression will be equivalent to which of the following:

A. (n + 1)(n – 1)^(-1)
B. –(n + 1)(n – 1)^(-1)
C. –(n – 1)(n + 1)^(-1)
D. (2 + n)^(-1)*(n – 2)
E. (n – 2)^(-1)*(2 + n)


Number plugging:

Say \(n=3\), then substitute \(\frac{2}{n}=\frac{2}{3}\) into \((n - 2)^{-1} (2 + n)=-2\).

Now, substitute \(n=3\) into the answer choices and see which of them gives -2. Only B fits.

Answer: B.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Alebra:

Substitute \(\frac{2}{n}\) into \((n - 2)^{-1} (2 + n)\):

\((\frac{2}{n} - 2)^{-1} (2 + \frac{2}{n})=(\frac{2-2n}{n})^{-1}(\frac{2n+2}{n})=(\frac{n}{2-2n})(\frac{2n+2}{n})=\frac{n+1}{1-n}=-\frac{n+1}{n-1}=-(n+1)(n-1)^{-1}\).

Answer: B.
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  20 May 2021, 05:20
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  17 Mar 2015, 20:45
\(\frac{(n+1)}{(1-n)}=-\frac{(n+1)}{(n-1)}\)

How do you do this kind of operation? Can you share any study materials on this topic? Thanks
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KarishmaB
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  17 Mar 2015, 22:08
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evdo wrote:
\(\frac{(n+1)}{(1-n)}=-\frac{(n+1)}{(n-1)}\)

How do you do this kind of operation? Can you share any study materials on this topic? Thanks


This is basic algebra and fraction.

You take our "-" common from the denominator to get

\(\frac{(n+1)}{(1-n)}=\frac{(n+1)}{-(n-1)}\)

And from theory of fractions, you know that

\(\frac{(n+1)}{-(n-1)}=-\frac{(n+1)}{(n-1)} = \frac{-(n+1)}{(n-1)}\)

For these basics, check purplemath.com or khanacademy.org
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  12 Jun 2015, 15:37
Moderators please move this question into PS section.This is not DS question!
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  02 Apr 2016, 19:06
Bunuel wrote:
(n – 2)^-1 (2 + n)
If n >2 and 2/n is substituted for all instances of n in the above expression, then the new expression will be equivalent to which of the following:

A. (n + 1)(n – 1)^(-1)
B. –(n + 1)(n – 1)^(-1)
C. –(n – 1)(n + 1)^(-1)
D. (2 + n)^(-1)*(n – 2)
E. (n – 2)^(-1)*(2 + n)


PLEASE FORMAT THE QUESTION AS IT IS NOT CLEAR WHAT THE ROLE OF (2+n) IS
IS THE EXPONENT -1(2+n) OR IS EVERYTHING DIVIDED BY 2+n???????????
OR MULTIPLIED? OR WHAT?
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Bunuel
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  03 Apr 2016, 11:32
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mvictor wrote:
Bunuel wrote:
(n – 2)^-1 (2 + n)
If n >2 and 2/n is substituted for all instances of n in the above expression, then the new expression will be equivalent to which of the following:

A. (n + 1)(n – 1)^(-1)
B. –(n + 1)(n – 1)^(-1)
C. –(n – 1)(n + 1)^(-1)
D. (2 + n)^(-1)*(n – 2)
E. (n – 2)^(-1)*(2 + n)


PLEASE FORMAT THE QUESTION AS IT IS NOT CLEAR WHAT THE ROLE OF (2+n) IS
IS THE EXPONENT -1(2+n) OR IS EVERYTHING DIVIDED BY 2+n???????????
OR MULTIPLIED? OR WHAT?

_____________
Done. Thank you.
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  21 May 2017, 17:29
Bunuel wrote:
\((n – 2)^{-1}* (2 + n)\)
If n >2 and 2/n is substituted for all instances of n in the above expression, then the new expression will be equivalent to which of the following:

A. \((n + 1)(n – 1)^{(-1)}\)
B. \(–(n + 1)(n – 1)^{(-1)}\)
C. \(–(n – 1)(n + 1)^{(-1)}\)
D. \((2 + n)^{(-1)}*(n – 2)\)
E. \((n – 2)^{(-1)}*(2 + n)\)


Picking n = 3 is the way to go for this problem. If n=3, then the formula simplifies to -2 when 2/n replaces n.

Then it's about neatness and working carefully. Start with answer choice A: the target value = 2, not -2. Eliminate A.

Move on to B. -(3+1)(3-1)^-1 = -4/(2)^-1 = -4/2 = -2. At this point, you should test the rest of the choices to ensure that there are no other answer choices that work for n = 3.
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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post  20 May 2021, 05:36
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sm1510 wrote:
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