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Is n negative?

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Is n negative? [#permalink] New post 19 Apr 2011, 04:54
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Is n negative?

(1) n^5(1 - n^4) < 0
(2) n^4 - 1 < 0
[Reveal] Spoiler: OA

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Re: Inequalities DS [#permalink] New post 19 Apr 2011, 05:35
(1)
n^5 (1-n^4) < 0

So either n^5 < 0 and (1-n^4) > 0

Or n^5 > 0 and 1 - n^4 < 0

So insufficient (n^4 is always > 0)


(2)

n^4 - 1 < 0

But we're not sure about sign of n

(1) and (2) say :

n^5 < 0, so n < 0

Answer - C
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Re: Inequalities DS [#permalink] New post 19 Apr 2011, 05:49
Just to be sure does the statement 1 reads n^ (5^(1-n^4)) or n^5*(1-n^4)
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Re: Inequalities DS [#permalink] New post 19 Apr 2011, 07:30
gmatpapa wrote:
Is n negative?

1. n^5(1-n^4) < 0
2. n^4-1 < 0


Is n<0

1. n^5(1-n^4)< 0

n^5(n^4-1)> 0
is similar to saying:
n(n^2-1)>0
(n-1)n(n+1)>0

Three roots: -1, 0, 1.

The above expression would be true for;
n>1 \hspace{2} OR \hspace{2} -1<n<0
Not Sufficient.

2. n^4-1 < 0

Similar to:
n^2-1<0
(n+1)(n-1)<0

Two roots: -1,1.
The above expression would be true for;
-1<n<1
Not Sufficient.

Combining both;
-1<n<0
Sufficient.

Ans: "C"
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Re: Inequalities DS [#permalink] New post 23 Jun 2013, 12:40
For the first statement:
isn't n^5*(1-n^4) the same as n^5/n^5n^4??

In this case would A be sufficient. What am I missing?
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Re: Inequalities DS [#permalink] New post 23 Jun 2013, 12:41
For the first statement:
isn't n^5*(1-n^4) the same as n^5/n^5n^4??

In this case would A be sufficient. What am I missing?
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Re: Inequalities DS [#permalink] New post 23 Jun 2013, 23:59
BankerRUS wrote:
For the first statement:
isn't n^5*(1-n^4) the same as n^5/n^5n^4??

In this case would A be sufficient. What am I missing?


Hi BankerRUS

You mean n^5*(1-n^4) the same as
(1) [n^5/n^5n]^4
--OR--
(2) n^5/[(n^5n)^4].

Both of them are incorrect.

First, n^5*(1-n^4) = n^5 - n^9
But:
(1) [n^5/n^5n]^4 = n^20/n^20n
(2) n^5/[(n^5n)^4] = n^5/n^20n

So (1) and (2) differ from n^5 - n^9

Hope it's clear.
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Re: Inequalities DS [#permalink] New post 24 Jun 2013, 00:14
pqhai wrote:
BankerRUS wrote:
For the first statement:
isn't n^5*(1-n^4) the same as n^5/n^5n^4??

In this case would A be sufficient. What am I missing?


Hi BankerRUS

You mean n^5*(1-n^4) the same as
(1) [n^5/n^5n]^4
--OR--
(2) n^5/[(n^5n)^4].

Both of them are incorrect.

First, n^5*(1-n^4) = n^5 - n^9
But:
(1) [n^5/n^5n]^4 = n^20/n^20n
(2) n^5/[(n^5n)^4] = n^5/n^20n

So (1) and (2) differ from n^5 - n^9

Hope it's clear.


I think the question here is whether (1-n^4) is in the exponent or not...this is the misunderstanding I suppose.....if it in the exponent then your statement is not true..correct?
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Re: Inequalities DS [#permalink] New post 24 Jun 2013, 00:21
BankerRUS wrote:
pqhai wrote:
BankerRUS wrote:
For the first statement:
isn't n^5*(1-n^4) the same as n^5/n^5n^4??

In this case would A be sufficient. What am I missing?


Hi BankerRUS

You mean n^5*(1-n^4) the same as
(1) [n^5/n^5n]^4
--OR--
(2) n^5/[(n^5n)^4].

Both of them are incorrect.

First, n^5*(1-n^4) = n^5 - n^9
But:
(1) [n^5/n^5n]^4 = n^20/n^20n
(2) n^5/[(n^5n)^4] = n^5/n^20n

So (1) and (2) differ from n^5 - n^9

Hope it's clear.


I think the question here is whether (1-n^4) is in the exponent or not...this is the misunderstanding I suppose.....if it in the exponent then your statement is not true..correct?


The question says: n^5*(1-n^4) ==> (1 - n^4) cannot be in the exponent.

If (1 - n^4) is in the exponent ==> the question will be n^[5*(1-n^4)]


Hope it helps.
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Re: Is n negative? [#permalink] New post 24 Jun 2013, 00:30
Expert's post
Is n negative?

(1) n^5(1 - n^4) < 0. For convenience rewrite this as n^5(n^4-1) > 0 --> n can be positive as well as negative: consider n=2 and n=-1/2. Not sufficient.

(2) n^4 - 1 < 0 --> n^4<1 --> -1<n<1. Not sufficient.

(1)+(2) Since from (2) n^4 - 1 < 0, then from (1) n^5 must also be negative (in order n^5(n^4-1) to be positive). n^5<0 means that n is negative. Sufficient.

Answer: C.

Hope it's clear.
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Re: Is n negative?   [#permalink] 24 Jun 2013, 00:30
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