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Bunuel
Is 3^n > 2^k ?

(1) k = n + 1
(2) n is a positive integer.

(1) k=n+1, k=2 and n=1 NO, k=3 and n=2 Yes, NOT SUFFICIENT
(2) n > 0, no info about k, which could be 1 or 100, NOT SUFFICIENT
(1)+(2) Still not sufficient considering the cases from (1)
Asnwer E


Hi! Can you elaborate on your answer?

I get 3 cases for statement 1

1. n=2 , k=3 9>8 - yes
2. n =3, k=4 27>16 - yes
3. n=-1, k=0 1/3 > 1 ? - no

So statement 2 tells us n is positive -- so why not c?

Thanks

hi,
another option when n is positive and answer is NO..
n=1, k=2..3>4... no
so dont miss out few values
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Hi! Can you elaborate on your answer?

I get 3 cases for statement 1

1. n=2 , k=3 9>8 - yes
2. n =3, k=4 27>16 - yes
3. n=-1, k=0 1/3 > 1 ? - no

So statement 2 tells us n is positive -- so why not c?

Thanks[/quote]

hi,
another option when n is positive and answer is NO..
n=1, k=2..3>4... no
so dont miss out few values[/quote]


Hi, I thought the same case too. Here is my reasoning.

St 1: k = n + 1. Clearly insufficient, cause we don't have any case (positive and negative would have different outcomes)

St 2: n is positive integer. Means more than zero and positive ; hence cannot be 0 but is greater than one.

So one plus two statement means: k = n +1 and n is positive integer.
3^n > 2^n+1 (replacing k).
We get 3 to power 1 NOT GREATER than 2 to power 2. Then if n = 2; k = n +1 => k =2. So 3 to power is always less than 2 to power n +1.

Thus it is sufficient.
Request experts to help out on why E is correct, and point out the issue

Thank you
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Hi! Can you elaborate on your answer?

I get 3 cases for statement 1

1. n=2 , k=3 9>8 - yes
2. n =3, k=4 27>16 - yes
3. n=-1, k=0 1/3 > 1 ? - no

So statement 2 tells us n is positive -- so why not c?

Thanks

hi,
another option when n is positive and answer is NO..
n=1, k=2..3>4... no
so dont miss out few values


Hi, I thought the same case too. Here is my reasoning.

St 1: k = n + 1. Clearly insufficient, cause we don't have any case (positive and negative would have different outcomes)

St 2: n is positive integer. Means more than zero and positive ; hence cannot be 0 but is greater than one.

So one plus two statement means: k = n +1 and n is positive integer.
3^n > 2^n+1 (replacing k).
We get 3 to power 1 NOT GREATER than 2 to power 2. Then if n = 2; k = n +1 => k =2. So 3 to power is always less than 2 to power n +1.

Thus it is sufficient.
Request experts to help out on why E is correct, and point out the issue

Thank you


Hi...
You are wrong at one point above..
When n=1, k =1+1=2.....3^1>2^2....NO
But when n=2 k=2+1=3...3^2>2^3... YES
Rather for all values of n above 2, Ans will be YES..

So ans is E as different answers possible
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Bunuel
Is \(3^n > 2^k \)?

(1) \(k = n + 1\)

(2) n is a positive integer.

is 3^n > 2^k?

(1) \(k = n + 1\)

if n=1, k=2
3^n < 2^k

if n=2, k=3 (or for all other values where n is greater than 2)
3^n > 2^k
Therefore, statement (1) alone is not sufficient.

(2) n is a positive integer.
Still above mentioned 2 cases can be considered.
Therefore, statement (2) alone is not sufficient.

Statements (1) and (2) together, still no definite answer.
Therefore, answer is E.
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Bunuel
Is \(3^n > 2^k \)?

(1) \(k = n + 1\)

(2) n is a positive integer.


(1) If \(n=1, k=1+1=2;3^1>2^2;3^n < 2^k\)

If \(n=2, k=2+1=3;3^2>2^3;3^n > 2^k\)

Insufficient.

(2) n is a positive integer; The above examples are sufficient to consider Yes/No.

Considering both:
Even considering both the options we can't answer.

if n=1, k=2

\(3^1, 2^2\)

Insufficient.

The answer is E.
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Is 3^n > 2^k

(1) k=n+1
n = 1, k = 2 ---> 3^1 < 2^2 NO
n = 2, k = 3 ---> 3^2 > 2^2 YES
Insufficient

(2) n is a positive integer.
Clearly insufficient b/c we don't know k.

Combo pack: Still insufficient...all we need to do is look at the sample numbers above.

Answer is E.
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