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

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Math Expert
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If n >2 and (2/n) is substituted for all instances of n in t [#permalink]

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04 Feb 2014, 06:06
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Difficulty:

65% (hard)

Question Stats:

63% (02:41) correct 37% (01:51) wrong based on 120 sessions

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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. $$(n – 2)^{(-1)}*(2 + n)$$
[Reveal] Spoiler: OA

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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink]

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04 Feb 2014, 06:17
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.

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 $$(n - 2)^{-1} (2 + n)$$:

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

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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink]

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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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Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink]

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17 Mar 2015, 22:08
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]

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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]

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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?
Math Expert
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Posts: 34514
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Kudos [?]: 80033 [0], given: 10024

Re: If n >2 and (2/n) is substituted for all instances of n in t [#permalink]

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03 Apr 2016, 11:32
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] 03 Apr 2016, 11:32
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