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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
If n >2 and 2/n is substituted for all instances of n in the above
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21 May 2021, 08:16
21 May 2021, 08:16
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Question Stats:
50% (00:43) correct
50% (02:57) wrong based on 4 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. \((n – 2)^{(-1)}*(2 + n)\)
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. \((n – 2)^{(-1)}*(2 + n)\)
If n >2 and 2/n is substituted for all instances of n in the above
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24 May 2021, 08:00
24 May 2021, 08:00
1
Substituting 2/n in the above equation, we get
\(\frac{2}{n} + 2\)/\(\frac{2}{n} - 2\)
\(\frac{1}{n}+1\) / \(\frac{1}{n}-1\)
\(\frac{1+n}{1-n}\)
\(- (n+1)(n-1)^-1\)
Hence, B
\(\frac{2}{n} + 2\)/\(\frac{2}{n} - 2\)
\(\frac{1}{n}+1\) / \(\frac{1}{n}-1\)
\(\frac{1+n}{1-n}\)
\(- (n+1)(n-1)^-1\)
Hence, B
gmatclubot
If n >2 and 2/n is substituted for all instances of n in the above [#permalink]
24 May 2021, 08:00
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