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 350,000 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:

Re: If n is a positive integer, is the value of b - a at least [#permalink]
18 Dec 2012, 08:03

Expert's post

2

This post was BOOKMARKED

If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?

Question: is \(b-a\geq{2(3^n - 2^n)}\)?

(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is \(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\)? --> is \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\)? --> is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.

Re: If n is a positive integer, is the value of b - a at least [#permalink]
29 Sep 2013, 01:59

Bunuel wrote:

If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?

Question: is \(b-a\geq{2(3^n - 2^n)}\)?

(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is \(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\)? --> is \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\)? --> is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.

(2) n = 3. Clearly insufficient.

Answer: A.

For statement (1), I understand how you arrived at \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\) but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?

For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value.

Re: If n is a positive integer, is the value of b - a at least [#permalink]
29 Sep 2013, 12:01

Expert's post

bulletpoint wrote:

Bunuel wrote:

If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?

Question: is \(b-a\geq{2(3^n - 2^n)}\)?

(1) a= 2^(n+1) and b= 3^(n+1). The question becomes: is \(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\)? --> is \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\)? --> is \(3^{n}\geq{0}\)? 3^n is always more than zero, so this statement is sufficient.

(2) n = 3. Clearly insufficient.

Answer: A.

For statement (1), I understand how you arrived at \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\) but how does this equation tell you whether b-a is at least twice the value of 3^n-2^n?

For example, if i sub in n=1 into the equation, it is true that b-a is at least twice the value (b-a= 5 and 3^n-2^n = 1 then 1*2). However, if i sub in n=2, the equation no longer is true as b-a then becomes 19 and the right side of the equation then becomes 10, which is not at least twice the value.

What am i missing here?

For the first statement after some manipulation the question becomes: is \(3^{n}\geq{0}\)? Irrespective of the actual value of n, 3^n will always be greater than zero, thus the answer to the question is YES. _________________

Re: If n is a positive integer, is the value of b - a at least [#permalink]
19 Jul 2014, 23:12

Do you think that it's better to approach this problem by plugging in numbers for n (as long as you're sure you can do it fast for cases n=1 and n=2), or to do the problem algebraically (which you know will take ~45-90seconds to crunch)?

Would be particularly interested in getting Bunuel's take...Thanks!

Re: If n is a positive integer, is the value of b - a at least [#permalink]
20 Jul 2014, 05:06

Expert's post

warriorsquared wrote:

Do you think that it's better to approach this problem by plugging in numbers for n (as long as you're sure you can do it fast for cases n=1 and n=2), or to do the problem algebraically (which you know will take ~45-90seconds to crunch)?

Would be particularly interested in getting Bunuel's take...Thanks!

You should take the approach which fits you the best. _________________

If n is a positive integer, is the value of b - a at least [#permalink]
03 Aug 2014, 09:09

1

This post received KUDOS

Bunuel wrote:

suhaschan wrote:

Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?

\(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\);

\(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\);

Cancel \(-2*2^{n}\): \(3*3^{n}\geq{2*3^n}\);

Subtract 2*3^n from both sides: \(3*3^{n}-2*3^n\geq{0}\);

\(3^{n}\geq{0}\)

Hope it's clear.

Hi Bunuel, In the end instead of subtracting \(2*3^n\) from both sides, can we cancel \(3^n\) from both sides (as it's always positive) and reach \(3>2\) making the statement sufficient? Is that also a correct approach? Thanks.

Re: If n is a positive integer, is the value of b - a at least [#permalink]
12 Aug 2014, 01:55

2

This post received KUDOS

Expert's post

Dienekes wrote:

Bunuel wrote:

suhaschan wrote:

Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?

\(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\);

\(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\);

Cancel \(-2*2^{n}\): \(3*3^{n}\geq{2*3^n}\);

Subtract 2*3^n from both sides: \(3*3^{n}-2*3^n\geq{0}\);

\(3^{n}\geq{0}\)

Hope it's clear.

Hi Bunuel, In the end instead of subtracting \(2*3^n\) from both sides, can we cancel \(3^n\) from both sides (as it's always positive) and reach \(3>2\) making the statement sufficient? Is that also a correct approach? Thanks.

Yes, this also would be a correct way of solving. _________________

If n is a positive integer, is the value of b - a at least [#permalink]
04 May 2015, 05:45

Bunuel wrote:

suhaschan wrote:

Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?

\(3^{n+1}-2^{n+1}\geq{2*(3^n - 2^n)}\);

\(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\);

Cancel \(-2*2^{n}\): \(3*3^{n}\geq{2*3^n}\);

Subtract 2*3^n from both sides: \(3*3^{n}-2*3^n\geq{0}\);

\(3^{n}\geq{0}\)

Hope it's clear.

You could stop at the part I have boldfaced to realize that the statement is sufficient, no? I mean \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}\); clearly shows that it it IS bigger. (Given that n is a positive integer)

gmatclubot

If n is a positive integer, is the value of b - a at least
[#permalink]
04 May 2015, 05:45

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

I have not posted in more than a month! It has been a super busy period, wrapping things up at Universal Music, completing most of the admin tasks in preparation for Stanford...