If n is a positive integer, is the value of b - a at least twice the : Data Sufficiency (DS)

bulletpoint

Joined: 02 Jul 2013

Last visit: 13 May 2014

Posts: 16

Own Kudos:

1,274

Given Kudos: 16

Schools: LBS MIF '15

Posts: 16

Kudos: 1,274

29 Sep 2013, 02:59

Kudos

Bookmarks

Bunuel

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?

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 24 Mar 2025

Posts: 100,064

Own Kudos:

710,548

[1]

Given Kudos: 92,683

Products:

Expert

Expert reply

Posts: 100,064

Kudos: 710,548

[1]

29 Sep 2013, 13:01

1

Kudos

Bookmarks

bulletpoint

Bunuel

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.

suk1234

Joined: 29 Sep 2013

Last visit: 04 Jul 2016

Posts: 38

Own Kudos:

172

[9]

Given Kudos: 48

Posts: 38

Kudos: 172

[9]

30 Sep 2013, 18:49

7

Kudos

2

Bookmarks

Walkabout

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

(1) a= 2^(n+1) and b= 3^(n+1)
(2) n = 3

The Question asks: b - a > 2*(3^n - 2^n)

The answer should be a definitive yes/no

Statement 1:

a= 2^(n+1) and b= 3^(n+1)

Lets take the smallest possible value of a Positive Integer i.e 1 and put it in for "n"

3^(1+1) - 2^ (1+1) > 2*(3^1 - 2^1)
3^2 - 2^2 > 2* (3 - 2)
9 - 4 > 2 *1
5>2 (a definitive answer)

Lets test another number (just to be on the Safe Side), lets test n=2

3^(2+1) - 2^ (2+1) > 2*(3^2 - 2^2)
3^3 - 2^3 > 2* (9 - 4)
27 - 8 > 2* 5
19> 10 (a definitive answer)

Thus, Sufficient.

Statement 2:

n=3

Let's Put it in the in-equality b - a > 2*(3^n - 2^n)

b - a> 2*(3^3 - 2^3)
b - a> 2*(27 - 8)
b - a> 2*(19)
b - a> 38

If, b=100 and a=10, than definitive answer
If, b= 2 and a= 10, than definitive answer
But since it doesn't provide any value for either "a" or "b"

Thus, Not Sufficient

Therefore the answer is

Show Spoiler

A

suhaschan

Joined: 10 Dec 2013

Last visit: 20 Aug 2014

Posts: 11

Own Kudos:

8

Given Kudos: 61

Status:Preparing for GMAT

Location: India

Concentration: Marketing, Leadership

Schools: HEC Montreal '16 Rotman '16 Sauder '16 Schulich '16 Stanford '16 Wharton '15

GMAT 1: 530 Q46 V18 Verified score

WE:Other (Media/Entertainment)

Schools: HEC Montreal '16 Rotman '16 Sauder '16 Schulich '16 Stanford '16 Wharton '15

GMAT 1: 530 Q46 V18 Verified score

Posts: 11

Kudos: 8

20 Jul 2014, 09:17

Kudos

Bookmarks

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

suhaschan

Joined: 10 Dec 2013

Last visit: 20 Aug 2014

Posts: 11

Own Kudos:

8

Given Kudos: 61

Status:Preparing for GMAT

Location: India

Concentration: Marketing, Leadership

Schools: HEC Montreal '16 Rotman '16 Sauder '16 Schulich '16 Stanford '16 Wharton '15

GMAT 1: 530 Q46 V18 Verified score

WE:Other (Media/Entertainment)

Schools: HEC Montreal '16 Rotman '16 Sauder '16 Schulich '16 Stanford '16 Wharton '15

GMAT 1: 530 Q46 V18 Verified score

Posts: 11

Kudos: 8

21 Jul 2014, 05:50

Kudos

Bookmarks

Thanks a lot Bunuel. The explanation couldn't have been more lucid.

Dienekes

Joined: 13 Feb 2011

Last visit: 18 Jan 2020

Posts: 63

Own Kudos:

187

[4]

Given Kudos: 3,386

Schools: ISB '17 (A) Tuck '18 Insead Sept '17 (A)

GMAT 1: 730 Q49 V41 Verified score

Posts: 63

Kudos: 187

[4]

03 Aug 2014, 10:09

4

Kudos

Bookmarks

Bunuel

suhaschan

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.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 24 Mar 2025

Posts: 100,064

Own Kudos:

710,548

[4]

Given Kudos: 92,683

Products:

Expert

Expert reply

Posts: 100,064

Kudos: 710,548

[4]

12 Aug 2014, 02:55

4

Kudos

Bookmarks

Dienekes

Bunuel

suhaschan

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.

BrentGMATPrepNow

Major Poster

Joined: 12 Sep 2015

Last visit: 13 May 2024

Posts: 6,769

Own Kudos:

33,132

[1]

Given Kudos: 799

Location: Canada

Expert

Expert reply

Posts: 6,769

Kudos: 33,132

[1]

01 Dec 2020, 12:03

1

Kudos

Bookmarks

tonebeeze

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

(1) \(a= 2^{n+1}\) and \(b= 3^{n+1}\)

(2) n = 3

Target question: Is \(b - a \geq 3^n - 2^n\)?

Statement 1: \(a= 2^{n+1}\) and \(b= 3^{n+1}\)

Substitute values into target question to get: Is \(b= 3^{n+1}- 2^{n+1} \geq 3^n - 2^n\)?

Rearrange terms: Is \(3^{n+1}- 3^n \geq 2^{n+1}-2^n \)?

Factor both sides: Is \(3^n(3- 1) \geq 2^n(2- 1)\)?

Simplify: Is \(3^n(2) \geq 2^n\)?

Divide both sides of the inequality by 2 to get: Is \(3^n \geq 2^{n-1}\)?

At this point, it's clear that the left side is greater than the right side (since the left side of the inequality is the product of n 3's, while the right side of the inequality is the product of n-1 2's)

As such we can be certain that \(3^n \geq 2^{n-1}\)
Statement 1 is SUFFICIENT

Statement 2: n = 3
Since we have no information about the variables a and b, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

Anthony99

Joined: 08 Jul 2020

Last visit: 23 Jan 2024

Posts: 6

Given Kudos: 8

Posts: 6

Kudos: 0

16 Feb 2021, 09:52

Kudos

Bookmarks

Bunuel

suhaschan

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.

Hey - can someone explain the 'cancel' step? I am lost on how the cancelling is done here, thx.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 24 Mar 2025

Posts: 100,064

Own Kudos:

710,548

Given Kudos: 92,683

Products:

Expert

Expert reply

Posts: 100,064

Kudos: 710,548

16 Feb 2021, 10:12

Kudos

Bookmarks

Anthony99

Bunuel

suhaschan

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.

Hey - can someone explain the 'cancel' step? I am lost on how the cancelling is done here, thx.

Notice that we have like terms, \(-2*2^{n}\), in both sides of \(3*3^{n}-2*2^{n}\geq{2*3^n-2*2^{n}}\), so we can cancel them. Add \(2*2^{n}\) to both sides to get \(3*3^{n}\geq{2*3^n}\).

Basshead

Joined: 09 Jan 2020

Last visit: 07 Feb 2024

Posts: 930

Own Kudos:

273

Given Kudos: 432

Location: United States

Posts: 930

Kudos: 273

28 Oct 2021, 02:03

Kudos

Bookmarks

Bunuel

I'm probably missing something really obvious here, but if we plug in n = 2 for statement 1:

\(3^3 - 2^3 = 19\)

\(2 (3^2 - 2^2) = 10\)

19 is not twice the value of 10. Don't we get yes and no answers?

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 24 Mar 2025

Posts: 100,064

Own Kudos:

710,548

[1]

Given Kudos: 92,683

Products:

Expert

Expert reply

Posts: 100,064

Kudos: 710,548

[1]

28 Oct 2021, 02:14

1

Kudos

Bookmarks

Basshead

Bunuel

I'm probably missing something really obvious here, but if we plug in n = 2 for statement 1:

\(3^3 - 2^3 = 19\)

\(2 (3^2 - 2^2) = 10\)

19 is not twice the value of 10. Don't we get yes and no answers?

19 is at least twice the value of 5 (3^2 - 2^2).

Is the value of b - a at least twice the value of 3^n - 2^n?

If n = 2, then:

b - a = 19
3^n - 2^n = 5. Twice the value of 5 is 10.

19 > 10.

Basshead

Joined: 09 Jan 2020

Last visit: 07 Feb 2024

Posts: 930

Own Kudos:

273

Given Kudos: 432

Location: United States

Posts: 930

Kudos: 273

28 Oct 2021, 02:32

Kudos

Bookmarks

Bunuel

Basshead

Bunuel

I'm probably missing something really obvious here, but if we plug in n = 2 for statement 1:

\(3^3 - 2^3 = 19\)

\(2 (3^2 - 2^2) = 10\)

19 is not twice the value of 10. Don't we get yes and no answers?

19 is at least twice the value of 5 (3^2 - 2^2).

Is the value of b - a at least twice the value of 3^n - 2^n?

If n = 2, then:

b - a = 19
3^n - 2^n = 5. Twice the value of 5 is 10.

19 > 10.

Ah, yes. Silly mistake indeed.

Thank you Bunuel.

MasteringGMAT

Joined: 14 Feb 2022

Last visit: 10 Mar 2025

Posts: 79

Own Kudos:

2,026

Given Kudos: 69

Concentration: General Management, Strategy

Posts: 79

Kudos: 2,026

19 Oct 2022, 06:27

Kudos

Bookmarks

Hi Bunuel,

Can you please explain the step highlighted in red

Bunuel

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.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 24 Mar 2025

Posts: 100,064

Own Kudos:

710,548

[1]

Given Kudos: 92,683

Products:

Expert

Expert reply

Posts: 100,064

Kudos: 710,548

[1]

19 Oct 2022, 09:45

1

Kudos

Bookmarks

Traj201090

Hi Bunuel,

Can you please explain the step highlighted in red

Bunuel

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.

\(3^{n+1}=3*3^n\);

\(2^{n+1}=2*2^n\);

\(2*(3^n - 2^n)=2*3^n - 2*2^n\).

Or you meant something else?

MasteringGMAT

Joined: 14 Feb 2022

Last visit: 10 Mar 2025

Posts: 79

Own Kudos:

2,026

Given Kudos: 69

Concentration: General Management, Strategy

Posts: 79

Kudos: 2,026

19 Oct 2022, 10:13

Kudos

Bookmarks

Bunuel

Traj201090

Hi Bunuel,

Can you please explain the step highlighted in red

Bunuel

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.

\(3^{n+1}=3*3^n\);

\(2^{n+1}=2*2^n\);

\(2*(3^n - 2^n)=2*3^n - 2*2^n\).

Or you meant something else?

I was trying to understand how LHS = RHS in this particular step \(3^{n+1}=3*3^n\), which was quite confusing for me.

we can basically remove +1 from the exponent, and simply multiply the coefficient with the same value.

for example
\(7^{n+1}=7*7^n\)

or

\(7^{n+2}= 7*7*7^n\)

or

\(7^{n+3}= 49*7*7^n\)

is my understanding correct?

and thanks for the previous explanation!

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 24 Mar 2025

Posts: 100,064

Own Kudos:

710,548

[1]

Given Kudos: 92,683

Products:

Expert

Expert reply

Posts: 100,064

Kudos: 710,548

[1]

19 Oct 2022, 10:41

1

Kudos

Bookmarks

Traj201090

Bunuel

I was trying to understand how LHS = RHS in this particular step \(3^{n+1}=3*3^n\), which was quite confusing for me.

we can basically remove +1 from the exponent, and simply multiply the coefficient with the same value.

for example
\(7^{n+1}=7*7^n\)

or

\(7^{n+2}= 7*7*7^n\)

or

\(7^{n+3}= 49*7*7^n\)

is my understanding correct?

and thanks for the previous explanation!

Yes, generally \(a^n*a^m=a^{n+m}\).

8. Exponents and Roots of Numbers

Theory
Exponents and Roots on the GMAT: Tips and hints
Cyclicity and the remainders on the GMAT

Questions
Units digits, exponents, remainders problems
Tough and tricky DS exponents and roots questions with detailed solutions
Tough and tricky PS exponents and roots questions with detailed solutions
Exponents DS Questions
Exponents PS Questions
Roots DS Questions
Roots PS Questions

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.

If n is a positive integer, is the value of b - a at least twice the

Prep Toolkit

My Rewards

GMAT Data Sufficiency (DS) Questions

If n is a positive integer, is the value of b - a at least twice the

Prep Toolkit

My Rewards