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

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If n is a positive integer, is the value of b - a at least [#permalink] New post 18 Dec 2012, 08:00
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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
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 18 Dec 2012, 08:03
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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.
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 29 Sep 2013, 01:59
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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?
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 29 Sep 2013, 12:01
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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.
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 30 Sep 2013, 17:49
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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
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 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!
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 20 Jul 2014, 05:06
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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.
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 20 Jul 2014, 08:17
Can someone please show me the simplification as to how the equation becomes whether 3^n > 0?
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 20 Jul 2014, 08:43
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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.
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 21 Jul 2014, 04:50
Thanks a lot Bunuel. The explanation couldn't have been more lucid.
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If n is a positive integer, is the value of b - a at least [#permalink] New post 03 Aug 2014, 09:09
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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.
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Re: If n is a positive integer, is the value of b - a at least [#permalink] New post 12 Aug 2014, 01:55
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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.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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