Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If k is not equal to 0, 1, or -1 is 1/k > 0? [#permalink]

Show Tags

10 Aug 2010, 12:05

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

56% (00:47) correct
44% (01:06) wrong based on 286 sessions

HideShow timer Statistics

If k is not equal to 0, 1, or -1 is 1/k > 0?

(1) 1/(k-1) > 0 (2) 1/(k+1) > 0

My method for this sort of questions is way underdeveloped. I will appreciate any hints on how to tackle this sort of problems. Many thanks in advance.

(1) \(\frac{1}{k-1}> 0\) --> the denominator must be positive: \(k-1>0\)--> \(k>1\), hence \(\frac{1}{k}>0\). Sufficient.

(2) \(\frac{1}{k+1}> 0\)--> the denominator must be positive: \(k+1>0\) --> \(k>-1\), hence \(k\) can be negative as well as positive: \(\frac{1}{k}\) may or may not be \(>0\). Not sufficient.

Merging similar topics. Please refer to the solutions above.

tonebeeze wrote:

*should I always assume that the variable can be a real number, unless otherwise stated?

The GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. So all numbers on the GMAT are real numbers and otherwise never will be stated.
_________________

(2) \(\frac{1}{k+1}> 0\)--> denominator is positive: \(k+1>0\) --> \(k>-1\), hence \(k\) can be negative as well as positive: \(\frac{1}{k}\) may or may not be \(>0\). Not sufficient.

Re: If k <> 0, 1, or -1, is 1/k > 0 ? [#permalink]

Show Tags

25 Aug 2013, 05:32

It must be A.

Again it is not mentioned that k is an integer.

1) 1/k-1>0 => k can be any value greater than 1 to satisfy this equation hence 1/k >0; sufficient. 2) 1/1+k>0 => if k=0.5 then 1/k >0 it holds true, but if k=-0.5 this equation still holds whereas, 1/k<0; therefore not sufficient.

-- Ramandeep
_________________

--It's one thing to get defeated, but another to accept it.

(2) \(\frac{1}{k+1}> 0\)--> \(k+1>0\) --> \(k>-1\), hence \(k\) can be negative as well as positive: \(\frac{1}{k}\) may or may not be \(>0\). Not sufficient.

Answer: A.

Hello Bunuel,

I am not getting this. why is is k>0?

Thanks

Turk

Which statement are you referring to?

In your explanation you said " Basically the question asks: is k>0? " I am not getting how the question is asking whether k>0?

In your explanation you said " Basically the question asks: is k>0? " I am not getting how the question is asking whether k>0?

The question asks whether \(\frac{1}{k}\) is positive. The numerator, which is 1, is positive, thus in order the fraction to be positive, the denominator must also be positive.

(2) \(\frac{1}{k+1}> 0\)--> denominator is positive: \(k+1>0\) --> \(k>-1\), hence \(k\) can be negative as well as positive: \(\frac{1}{k}\) may or may not be \(>0\). Not sufficient.

Answer: A.

Hope it's clear.

A very fundamental or maybe silly question regarding such questions: when we are deducing from (1/K)>0 -> K>0.. how are we doing that, as assuming 0/0 on the RHS would be undefined.

(2) \(\frac{1}{k+1}> 0\)--> denominator is positive: \(k+1>0\) --> \(k>-1\), hence \(k\) can be negative as well as positive: \(\frac{1}{k}\) may or may not be \(>0\). Not sufficient.

Answer: A.

Hope it's clear.

A very fundamental or maybe silly question regarding such questions: when we are deducing from (1/K)>0 -> K>0.. how are we doing that, as assuming 0/0 on the RHS would be undefined.

Thanks

No, we are not assuming 0/0. When you are given a/b > 0 this means that the fraction a/b > 0 ---> 2 cases possible,

Case 1, either both a,b > 0 or

Case 2, both a,b < 0.

You can try the following examples to see the above 2 cases:

Case 1: a=1, b = 2, a/b = 0.5 > 0

Case 2, a=-1, b = -5 , a/b = 0.2 > 0

But if lets say you have a=-1 and b = 2 or a=2 and b= -3 , a/b <0 and NOT >0.

You can even remember this that in order for a fraction a/b to be >0 ---> both a,b MUST have the same signs. Either both of them are >0 or both of them are negative.

Coming back to the question,

When you are given 1/k > 0 ---> 1 is already >0 so based on the 'rule' above, k must also be positive in order for 1 and k to have the same 'sign'. Thus k>0.

Similarly, if 1/(k-1) > 0 ---> k-1 > 0 ---> k >1 etc.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If k does not equal 0, 1 or -1, is 1/k >0

(1) 1/(k-1) > 0 (2) 1/(k+1) > 0

When it comes to inequality DS questions, 2 things are important at all times. First is square. Secondly, when range of que includes range of con, the con is sufficient. Modify the original condition and the question. Multiply k^2 on the both equations and the sign of inequality doesn't change as k^2 is still a positive integer even when it's multiplied. There is 1 variable(k), which should match with the number of equations. So you need 1 equation, for 1) 1 equation, for 2) q equation, which is likely to make D the answer. For 1), multiply (k-1)^2 on the both equations, they become k-1>0, k>1. The range of que includes the range of con, which is sufficient. For 2), multiply (k-1)^2 on the both equations, they become k-1>0, k>1. The range of que doesn't include the range of con, which is not sufficient Therefore, the answer is A.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

Re: If k is not equal to 0, 1, or -1 is 1/k > 0? [#permalink]

Show Tags

07 Mar 2016, 21:58

Bull78 wrote:

If k is not equal to 0, 1, or -1 is 1/k > 0?

(1) 1/(k-1) > 0 (2) 1/(k+1) > 0

My method for this sort of questions is way underdeveloped. I will appreciate any hints on how to tackle this sort of problems. Many thanks in advance.

Fun Fact => if K was an integer than then D would be the choice.
_________________

Re: If k does not equal 0, 1 or -1, is 1/k >0 (1) 1/(k-1) > 0 [#permalink]

Show Tags

07 Feb 2017, 07:08

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...