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Math Expert V
Joined: 02 Sep 2009
Posts: 59588
If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 51% (02:29) correct 49% (02:12) wrong based on 89 sessions

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If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?

(1) 1/k<0
(2) 1/(n+2)>0

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Math Expert V
Joined: 02 Aug 2009
Posts: 8284
Re: If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?  [#permalink]

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Bunuel wrote:
If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?

(1) 1/k<0
(2) 1/(n+2)>0

Bunuel , excellent Q..

lets see what the Q is asking..

$$\frac{1}{(n−k)}>0$$..
since 1 is positive, we have to find if n-k>0 OR n>k..

lets see the statements--

(1)$$\frac{1}{k}<0$$
we know k<0..
Insuff

(2)$$\frac{1}{(n+2)}>0$$..
but n+2>0 or n>-2..
Insuff

combined..
we know n>-2, so least value of n is -1..
Statement I tells us that k<0..
so k's largest possible value could be -1..
but n ≠ k..
so if n is -1, k will not be -1.

so if n=-1, the largest value of k will be -2.. so n>k..
if n>-1, k can be -1 at the maximum, so again n>k..

Suff
C
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Marshall & McDonough Moderator D
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Posts: 1682
Location: India
Re: If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?  [#permalink]

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St1: 1/k < 0 --> for the condition to hold true k must be negative
If k is negative, 1/(n−k) = 1/(n - (-k)) = 1/(n + k)
n can be positive or negative --> 1/(n + k) can be positive or negative
Not sufficient

St2: 1/(n + 2) > 0 --> n > -2
Not sufficient

Combining St1 and St2: k is negative and n > -2; Also n is not equal to k
If n = -1, k < -1 --> 1/(n - (k) > 0
If n is 0 or positive, --> 1/(n - (-k)) is always positive
Sufficient

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?  [#permalink]

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Bunuel wrote:
If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?

(1) 1/k<0
(2) 1/(n+2)>0

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

The quesiton 1/(n-k) > 0 is equivalent to n-k>0 or n>k.

Since we have 2 variables (n and k) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first.

Conditions 1) & 2):
1/k < 0 ⇔ k < 0 ⇔ k ≤ -1
1/(n+2) > 0 ⇔ n+2 > 0 ⇔ n > -2 ⇔ n ≥ -1
Since n≠k, we have n > k.
Both conditions 1) & 2) are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1):
This condition provides us with no information about the variable n, so it is not sufficient.

Condition 2):
This condition provides us with no information about the variable k, so it is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?  [#permalink]

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_________________ Re: If n ≠ k, k ≠ 0, n ≠ -2, and n and k are integers, is 1/(n−k)>0?   [#permalink] 08 Aug 2019, 11:35
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