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# Is 5^n < 0.04?

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Math Expert
Joined: 02 Sep 2009
Posts: 53066

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20 Oct 2014, 05:59
1
3
00:00

Difficulty:

15% (low)

Question Stats:

79% (02:01) correct 21% (02:10) wrong based on 144 sessions

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Tough and Tricky questions: Absolute Values.

Is 5^n < 0.04?

(1) (1/5)^n > 25
(2) n^3 < n^2

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Joined: 25 Apr 2012
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GPA: 3.21
Re: Is 5^n < 0.04?  [#permalink]

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20 Oct 2014, 21:15
Is 5^n < 0.04?

(1) (1/5)^n > 25
(2) n^3 < n^2[/quote]

Notice that for $$n\geq{0}$$ so RHS of the Q stem is always greater than zero so we need to confirm whether n<-2 because$$5^{-2}$$ is 1/25 =0.04

St 1 says (1/5)^n > 25 or n>2 so st 1 is sufficient as n is positive

St 2 says n^3<n^2 or n(n^2-1)<0 or n(n-1)(n+1) <0
Key points are n=0,-1 and 1

So we have 4 range

Case 1: n<-1 The expression n(n-1)(n+1) <0 is true
Case 2: -1<n<0, The expression is is false as it will be >0
Case3 : 0<n<1, The expression will be true
and Case 4: n>1 The expression will be true for all values

Since, we have 2 ranges in which expression holds true and for those range n <-2 or n>-2..

Ans is A
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Joined: 03 May 2014
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Concentration: Operations, Marketing
GMAT 1: 680 Q48 V34
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28 Aug 2015, 13:38
WoundedTiger wrote:
Is 5^n < 0.04?

(1) (1/5)^n > 25
(2) n^3 < n^2

Notice that for $$n\geq{0}$$ so RHS of the Q stem is always greater than zero so we need to confirm whether n<-2 because$$5^{-2}$$ is 1/25 =0.04

St 1 says (1/5)^n > 25 or n>2 so st 1 is sufficient as n is positive

St 2 says n^3<n^2 or n(n^2-1)<0 or n(n-1)(n+1) <0
Key points are n=0,-1 and 1

So we have 4 range

Case 1: n<-1 The expression n(n-1)(n+1) <0 is true
Case 2: -1<n<0, The expression is is false as it will be >0
Case3 : 0<n<1, The expression will be true
and Case 4: n>1 The expression will be true for all values

Since, we have 2 ranges in which expression holds true and for those range n <-2 or n>-2..

Ans is A[/quote]
I think red part is wrong as It should be n<-2 for statement 1
For statement 2 it comes down to n^3-n^2<0----> n^2(n-1)<0 SO for sure n-1<0 as n^2 can't be less than 0 which becomes n<1 excluding n=0, with this we can't be sure is n<-2 always.
Correct me if I am Wrong
Manager
Joined: 07 Apr 2015
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29 Aug 2015, 04:12
Here is my approach:

Statement 1:
$$\frac{1}{5}^n > 25$$
$$5^-n > 5^2$$
$$-n > 2$$
$$n < -2$$

plug in n=-3 into the equation will yield that 1/125 is < than 0,04

Sufficient

Statement 2:
per statement n can be either a negative number or a proper fraction. 5^(1/2) would be > 0,04 while to a negative power would be <. not sufficient.

--> A
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Joined: 09 Mar 2018
Posts: 1003
Location: India

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27 Jan 2019, 06:35
Bunuel wrote:

Tough and Tricky questions: Absolute Values.

Is 5^n < 0.04?

(1) (1/5)^n > 25
(2) n^3 < n^2

Is this a validate approach ??

is 5^n < 5^-2
becomes is n < -2 ??

1) 5^(-n) > 5^2
=> -n > 2
=> n < -2
Sufficient

2) Can't tell anything from this.
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Re: Is 5^n < 0.04?  [#permalink]

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27 Jan 2019, 18:49
1
Hii

Your approach to use first statement is correct.

For second statement.
n^3 < n^2
n^2(n-1)<0
n < 1
But since now n can be < or > -2, statement 2 is not sufficient

KanishkM wrote:
Bunuel wrote:

Tough and Tricky questions: Absolute Values.

Is 5^n < 0.04?

(1) (1/5)^n > 25
(2) n^3 < n^2

Is this a validate approach ??

is 5^n < 5^-2
becomes is n < -2 ??

1) 5^(-n) > 5^2
=> -n > 2
=> n < -2
Sufficient

2) Can't tell anything from this.

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Re: Is 5^n < 0.04?   [#permalink] 27 Jan 2019, 18:49
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