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

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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, 05:06
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62% (02:57) correct 38% (01:56) wrong based on 66 sessions
(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)
[Reveal] Spoiler: OA

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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, 05:17
Expert's post
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 17 Mar 2015, 19:45
\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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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink] New post 17 Mar 2015, 21:08
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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] 17 Mar 2015, 21:08
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