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If n is a positive integer, is (1/2)^n > 0.125 ?

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If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 08:34
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post Updated on: 21 Aug 2019, 08:55
1
Quote:
If n is a positive integer, is \((\frac{1}{2})^n>0.125?\)

(1) n>3

(2) \((\frac{1}{2})^n−1<0.25\)


Given: n is a positive integer

Asked: Is \((\frac{1}{2})^n>0.125?\)

\(0.125 = (\frac{1}{2})^3\)
Is \((\frac{1}{2})^n>0.125\)
Mean Is n<3?

(1) n>3
If n>3 => \((\frac{1}{2})^n<0.125\)
SUFFICIENT

(2) \((\frac{1}{2})^{n−1}<0.25\)
n-1>2
n>3
If n>3 => \((\frac{1}{2})^n<0.125\)
SUFFICIENT

IMO D
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Originally posted by Kinshook on 21 Aug 2019, 08:46.
Last edited by Kinshook on 21 Aug 2019, 08:55, edited 3 times in total.
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 08:51
1
IMO-D

n is a positive integer
Check : 1/2 ^ n> 0.125 ?

0.125 = 1/2 ^ 3
So, Check (1/2)^n > (1/2)^3

(1) n>3
If n>3, (1/2)^n < (1/2)^3
Sufficient

(2) 1/2 ^ (n−1) <0.25
1/2 ^ (n−1) < (1/2)^2
n-1 > 2
n>3
If n>3, (1/2)^n < (1/2)^3
Sufficient
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If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post Updated on: 22 Aug 2019, 09:00
1
n = positive integer

We need to find whether \((\frac{1}{2})^n > 0.125\)?
=\((\frac{1}{2})^n > (\frac{125}{1000})\)?
=\((\frac{1}{2})^n > (\frac{1}{8})\)?
=\((\frac{1}{2})^n > (\frac{1}{2})^3\)?

indirectly asking whether n<3?

(1) n>3 Any number greater than 3 would always give NO.
Example:
when n=4 => \((\frac{1}{2})^4 > (\frac{1}{2})^3\)? NO
when n=5 => \((\frac{1}{2})^5 > (\frac{1}{2})^3\)? NO

Hence n always has to be less than 3, i.e, n<3

SUFFICIENT!

(2) \((\frac{1}{2})^{(n-1)} < 0.25\)
\((\frac{1}{2})^{(n-1)} < (\frac{25}{100})\)
\((\frac{1}{2})^{(n-1)} < (\frac{1}{2})^2\)
n-1 > 2 (sign changes here)
n>3.............same as statement 1, always no!
SUFFICIENT

IMO OPTION D
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Originally posted by EncounterGMAT on 21 Aug 2019, 09:03.
Last edited by EncounterGMAT on 22 Aug 2019, 09:00, edited 2 times in total.
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 09:04
n should be less than 3 to make sure \((1/2)^n > 0.125\) as (1/2)^n > (1/2)^3 => n < 3 (Fractional value decreases as power increases)

1) This is sufficient as n > 3, so answer is NO

2) With the same logic, n -1 > 2 => n > 3 , sufficient as it would result in answer NO

Answer is D
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 09:22
If n is a positive integer, is (12)n>0.125(12)n>0.125 ?

(1) n>3

(2) (12)n−1<0.25

simplify given expression
we get
(1/2)^n>(1/2)^3
only possible when n=1,2
#1
n>3
sufficient
#2
(12)n−1<0.25
solve we get (1/2)^n-1<(1/2)^2
possible when n>3
sufficient
IMO D
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 09:38
statement 1 gives

n>3 so 1/2^3 = .125

so if n >3 then it will be lesser than .125 which is clear no so sufficient



statement 2

1/2 ^(n-1)

1/2^n/1/2^1

multiply both side by 1/2 gives

1/2 ^n > .125 which is nothing but question so not sufficient

hence ans is A
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 09:48
If n is a positive integer, is \((\frac{1}{2})^n > 0.125\) ?

(1) n > 3

(2) \(\frac{1}{2}^(n−1)\) < 0.25

\((\frac{1}{2})^n > 0.125\)
 \((\frac{1}{2})^n > (\frac{1}{8})\) [Also \(0.125 = (0.5)^3\)]
 \((\frac{1}{2})^n > (\frac{1}{2})^3\)
 n > 3 [Since n >0]

Statement 1) n > 3

Yes n > 3

SUFFICIENT

Statement 2) \(\frac{1}{2}^(n−1)\) < 0.25
 \(\frac{1}{2}^(n−1)\) < \((0.5)^2\)
 n - 1 < 2 [Since n >0]
 n < 3
No, n > 3

SUFFICIENT

Answer (D).
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 11:07
If n is a positive integer, is \((\frac{1}{2})^{n}\) > 0.125 ?

\((\frac{1}{2})^{n}\) > \(\frac{1}{8}\)

\((\frac{1}{2})^{n}\) > \((\frac{1}{2})^{3}\)

is n < 3 ???

Statement1: n>3
--> Always NO.
Sufficient

Statement2: \((\frac{1}{2})^{n−1}\) < 0.25

\((\frac{1}{2})^{n-1}\) < \((\frac{1}{2})^2\)
n-1 > 2
n >3
Always YES
Sufficient

The answer is D.
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 11:38
1
Given: n >=1
Analysis: (1/2)^n > 0.125
(1/2)^n > 1/8 => n>3

St.1 : n>3 Sufficient
St.2 : (1/2)^n * 2 < 1/4
(1/2)^n < 1/8
n < 3 => Sufficient.

D ans.
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 11:43
1st Statement
n>3 , (1/2)^4 = 0.0625
Insuff. for the question being.

2nd Statement
(1/2)^n-1 <0.25
If n= 1,2,3 then statement false.
Insuff. for the question being

But (1) + (2)
n>3 and (1/2)^n-1 <0.25
n=4 , the answer comes to be 0.125<0.25
Hence together Sufficient.

IMO C
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 12:00
We know from the question stem that n is a positive integer. We are to find if (0.5)^n > (0.5)^3. This means we are simply asked if n>3.

Statement 1 says n>3. Clearly it is sufficient since we can answer that n>3 or that for all integers values n={4,5,5,...}, (0.5)^n is not greater than 0.125.

Statement 2: (0.5)^(n-1) < (0.5)^2
This means that n-1 < 2 hence n<3.
And for all positive values of n<3 (1,2), (0.5)^n > 0.125.
Hence we can answer yes to the question posed. Statement 2 is also sufficient on its own.

Answer is therefore D.

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Re: If n is a positive integer, is (1/2)^n > 0.125 ?  [#permalink]

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New post 21 Aug 2019, 12:38
Quote:
If n is a positive integer, is (1/2)^n>0.125 ?

(1) n>3
(2) (1/2)^(n−1)<0.25


\((1/2)^n>1/8…2^{(-n)}>2^{(-3)}…-n>-3…n<3\) so, find if \(n<3\)

(1) n>3: sufic.
(2) (1/2)^(n−1)<0.25: \((1/2)^{(n-1)}>1/4…2^{(-(n-1))}>2^{(-2)}…-(n-1)>-2…-n>-3…n<3\); sufic.

Answer (D)
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Re: If n is a positive integer, is (1/2)^n > 0.125 ?   [#permalink] 21 Aug 2019, 12:38
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