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# Is n negative?

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22 Aug 2011, 14:47
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Difficulty:

45% (medium)

Question Stats:

67% (01:57) correct 33% (01:09) wrong based on 27 sessions

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Is n negative?

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

Last edited by Bunuel on 10 Jul 2013, 13:49, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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22 Aug 2011, 15:24
shashankp27 wrote:
Is n negative?
1. n5(1 – n4) < 0
2. n4 – 1 < 0

Exponents are denoted by "^"

$$n^5$$ should be written as n^5

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22 Aug 2011, 21:59

Statemet 1: to get a negative result we must have one of the two variables < 0 and the other > 0

in our case n^5 < 0 and 1-n^4 > 0 gives a negative result.

the same out come is arrrived at when n^5 is > 0 and 1 - n^4 is < 0 so not sifficient

statment 2 is also not sifficient; n could be - 2 or + 2

combining we will get the result
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22 Aug 2011, 23:03
shashankp27 wrote:
Is n negative?
1. n^5(1 – n^4) < 0
2. n^4 – 1 < 0

Go step by step.

Question: Is n negative?

Statement 1: $$n^5(1 - n^4) < 0$$
Re-write as: $$n^5(n^4 - 1) > 0$$ (I do this only for convenience. It is easier to handle >0 because either both are positive or both negative so less confusion.)
This tells me that either both $$n^5$$ and $$(n^4 - 1)$$ are positive (which means n is positive) or both are negative (which means n is negative). n can be positive or negative so not sufficient.

Statement 2: $$n^4 - 1 < 0$$
This implies that n is between -1 and 1.

How?
$$n^4 - 1 = (n^2 + 1)(n^2 - 1) = (n^2 + 1)(n + 1)(n - 1)$$
$$n^2 + 1$$ is always positive so we can ignore it.
We get, $$(n+1)(n - 1) < 0$$
-1 < n < 1 (check out: inequalities-trick-91482.html#p804990)

Since n can be positive or negative, not sufficient alone.

Using both statements together,
$$n^4 - 1$$ is negative, so from the analysis of statement 1, $$n^5$$ must be negative too. This means n must be negative. Sufficient.
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23 Aug 2011, 00:06
is n < 0?

1) n^5(1 – n^4) < 0
n^5 - n^9 < 0
n^5 < n^9
n < n^5

n > 1 or -1 < n < 0. NS.

2) n^4 - 1 < 0

-1 < n < 1. NS.

1+2)
n > 1 or -1 < n < 0.
n < 1, therefore, -1 < n < 0

C.
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23 Aug 2011, 00:35
1. n^5 (1- n^4) <0
for product of two numbers to be negative, they ahve to have different signs.
If n^ 5 is -ve, (1- n^4) has to be +. that is n has to be a negatve number, and also n has to be a fraction (only then (1- n^4) will be positive. Thus this gives n is a negative, proper fraction
If n^5 is +, (1- n^4) has to be negative. n has to be positive (for n^5 to be positive) , and n has to be a number greaters than - for (1- n^4) to be negative - any valuse less than 1 will always give (1- n^4) as +.
Thus from one n can be a positive number more than 1 or a negative fraction. Insufficient to say whether n is negative or not

2. (n^4 - 1) < 0
(n^4 - 1) is negative that means n^ 4 is less than 1, can be true for both negative and positive fratcions. thus not sufficient.

Together: if (n^4 - 1) < 0, (1- n^4) > 0. From 1, n^5 (1- n^4) <0
That means n^5 has to be negative (only then the product can be negative)
for n ^5 to be negative, n has to be negative
Hence Both togetehr are sufficient.
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Re: Is n negative? 1. n^5(1 n^4) < 0 2. n^4 1 < 0 [#permalink]

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04 Jan 2012, 00:56
Nice question. C is the answer.
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Re: Is n negative? 1. n^5(1 n^4) < 0 2. n^4 1 < 0   [#permalink] 04 Jan 2012, 00:56
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