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If n >2 and 2/n is substituted

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If n >2 and 2/n is substituted

by Vincen » Wed Oct 11, 2017 6:28 pm
$$(n-2)^{-1}\cdot(2+n)$$

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:

$$A.\ (n+1)(n-1)^{\left(-1\right)}$$
$$B.\ -(n+1)(n-1)^{\left(-1\right)}$$
$$C.\ -(n-1)(n+1)^{\left(-1\right)}$$
$$D.\ (2+n)^{\left(-1\right)}\cdot(n-2)$$
$$E.\ (n-2)^{\left(-1\right)}\cdot(2+n)$$

The OA is B.

I need some help making the calculus. Experts, can you help me?
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by Jay@ManhattanReview » Thu Oct 12, 2017 1:57 am
Vincen wrote:$$(n-2)^{-1}\cdot(2+n)$$

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:

$$A.\ (n+1)(n-1)^{\left(-1\right)}$$
$$B.\ -(n+1)(n-1)^{\left(-1\right)}$$
$$C.\ -(n-1)(n+1)^{\left(-1\right)}$$
$$D.\ (2+n)^{\left(-1\right)}\cdot(n-2)$$
$$E.\ (n-2)^{\left(-1\right)}\cdot(2+n)$$

The OA is B.

I need some help making the calculus. Experts, can you help me?
We have the expression [(n - 2)^(-1)]*(2 + n).

=> [(n - 2)^(-1)]*(2 + n) => (2 + n) / (n - 2)

Taking n common from the numerator and the denominator, we get,

(2 + n) / (n - 2) = [n(2/n + 1)] / [n(1 - 2/n)] = (2/n + 1) / (1 - 2/n )

Now replacing 2/n with n, we get

(2/n + 1) / (1 - 2/n ) => (n + 1) / (1 - n)

We see that no option matches (n + 1) / (1 - n). Let's manipulate (n + 1) / (1 - n).

(n + 1) / (1 - n) = (n + 1)* [(1 - n)^(-1)].

We see that no option matches (n + 1)* [(1 - n)^(-1)]. Let's manipulate (n + 1)* [(1 - n)^(-1)].

(n + 1)* [(1 - n)^(-1)] = (n + 1)* (-1)^(-1)*(n - 1)^(-1) = -(n + 1)* (n - 1)^(-1).

This one matches option B -- the correct answer.

The correct answer: B

Hope this helps!

-Jay

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by GMATGuruNY » Thu Oct 12, 2017 2:36 am
Vincen wrote:$$(n-2)^{-1}\cdot(2+n)$$

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:

$$A.\ (n+1)(n-1)^{\left(-1\right)}$$
$$B.\ -(n+1)(n-1)^{\left(-1\right)}$$
$$C.\ -(n-1)(n+1)^{\left(-1\right)}$$
$$D.\ (2+n)^{\left(-1\right)}\cdot(n-2)$$
$$E.\ (n-2)^{\left(-1\right)}\cdot(2+n)$$
Let n=4, with the result that 2/n = 2/4 = 1/2.
Substituting 2/n=1/2 for all instances of n in (n-2)¯¹(2+n), we get:
(1/2 - 2)¯¹(2 + 1/2)

= (-3/2)¯¹(5/2)

= (-2/3)(5/2)

= -5/3.

The target value is -5/3.
Now plug n=4 into the answer choices to see which yields the target value of -5/3.
Of the five answer choices, only B and C will yield negative values.
Test B:
-(n+1)(n-1)¯¹ = -(4+1)(4-1)¯¹ = -(5)(3)¯¹ = -(5)(1/3) = -5/3.

The correct answer is B.
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