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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]  04 Feb 2014, 05:06
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Question Stats:

64% (02:51) correct 36% (01:54) wrong based on 76 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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Math Expert
Joined: 02 Sep 2009
Posts: 27215
Followers: 4228

Kudos [?]: 41022 [0], given: 5654

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

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