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# Is x greater than 1?

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Is x greater than 1? [#permalink]  06 Oct 2011, 07:11
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Is x greater than 1?

(1) 1/x>-1
(2) 1/x^5 > 1/x^3

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-greater-than-134052.html
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Re: DS- Is X greater than 1? [#permalink]  06 Oct 2011, 07:53
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kkalyan wrote:
Is X greater than 1?

1. 1/X>-1
2. 1/X^5 > 1/X^3

1. 1/x>-1
1/x+1>0
(1+x)/x>0
If x>0; 1+x>0 OR x>-1; Combining both; x>0
If x<0; 1+x<0 OR x<-1; Combining both; x<-1
So,
x can be 2 OR -2.
Not Sufficient.

2. 1/X^5 > 1/X^3
Multiplying these with x^2
1/x^3 > 1/x
1/x^3 - 1/x > 0
(1-x^2)/x^3 > 0
(x+1)(x-1)/x^3<0
If x<0; (x+1)(x-1)>0 OR x>1 OR x<-1; Combing both; x<-1
If x>0; (x+1)(x-1)<0 OR -1<x<+1; Combing both; 0<x<1
Either case; x<1 OR NOT greater than 1.
Sufficient.

Ans: "B"
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Re: DS- Is X greater than 1? [#permalink]  06 Oct 2011, 08:48
1.
1/x>-1
x can be any positive number or any negative number lesser than -1 such as -4,-5 (-1/4>-1; -1/5>-1)
Insufficient

2.
1/X^5 > 1/X^3
Multiply both sides by x^6 (x^6 is positive)
x>x^3
This is possible only when x is a positive fraction
Sufficient

B
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Re: DS- Is X greater than 1? [#permalink]  06 Oct 2011, 12:32
Statement 1

1/X > -1

so x can be greater than 1(if x = 2 then 1/x will be greater than -1)
Also x can be less than 1(if x = -2 then 1/x will be greater than -1)

So A not sufficient

Statement 2

1/x^5 > 1/X^3

If x = 2 will not satisfy the statement as it will become 1/32 > 1/8 which is not correct.

If x=-2 then it will satisfy as it will become -1/32 > -1/8 which is true

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Re: DS- Is X greater than 1? [#permalink]  06 Oct 2011, 22:37
1.
1/x>-1
x can be any positive number or any negative number lesser than -1 such as -4,-5 (-1/4>-1; -1/5>-1)
Insufficient

2.
1/X^5 > 1/X^3
Multiply both sides by x^6 (x^6 is positive)
x>x^3
This is possible only when x is a positive fraction
Sufficient

B

Hi, cau you plz tell me whether my way is working correct or not........
1.1/X > -1

So X<-1(as we did reciprocla the sing changes...)
nd X<-1 does not hold true for all say for X=2;So it is not suff.

2.1/X^5 > 1/X^3
After reciprocal and cancelling X^3 on both sides we will get X^2 < 1 rgt?
So X^2 < 1 will be possible only when 0<X<1. As we got X<1 i consider this is sufficent. hence B

Please correct me if iam worng
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Re: DS- Is X greater than 1? [#permalink]  07 Oct 2011, 01:22
kkalyan wrote:
1.1/X > -1
So X<-1(as we did reciprocla the sing changes...)

Please correct me if iam worng

1/5>-1/4
kkalyan wrote:
as we did reciprocla the sing changes

5<-4
Now is this expression correct?

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Re: Is X greater than 1? 1. 1/X>-1 2. 1/X^5 > 1/X^3 [#permalink]  28 Jul 2012, 09:10
Please clarify, how it'll work for x=-1/2?
In this case statement modifies to -32>-8, which is not true. Thus X is not always <1

What am I missing here?
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Re: Is X greater than 1? 1. 1/X>-1 2. 1/X^5 > 1/X^3 [#permalink]  28 Jul 2012, 12:54
Since in both statements x appears in the denominator, it is clear that x is non-zero.
Inequalities can be multiplied by non-zero positive expressions and the direction of the inequality will be preserved.

(1) Multiplying both sides by x^2, we obtain x>-x^2 or x(1+x)>0. If x > 1, the inequality holds, but it also holds for x < -1.
Not sufficient.

(2) Multiplying both sides by x^6, we obtain x>x^3 or x(1-x^2)>0. If x > 1, the inequality certainly does not hold. So, the answer to the question is x > 1 is a definite NO. Therefore (2) is sufficient.

Note: I think I saw this question somewhere on the forum before.
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Re: Is x greater than 1? [#permalink]  29 Jul 2012, 01:15
Is x greater than 1?

Question is x>1?

(1) \frac{1}{x}>- 1 --> \frac{1+x}{x}>0, two cases:

A. x>0 and 1+x>0, x>-1 --> x>0;

B. x<0 and 1+x<0, x<-1 --> x<-1.

We got that given inequality holds true in two ranges: x>0 and x<-1, thus x may or may not be greater than one. Not sufficient.

(2) \frac{1}{x^5}> \frac{1}{x^3} --> \frac{1-x^2}{x^5}>0, two cases:

A. x>0 (it's the same as x^5>0) and 1-x^2>0, -1<x<1 --> 0<x<1;

B. x<0 and 1-x^2<0, x<-1 or x>1 --> x<-1;

We got that given inequality holds true in two ranges: 0<x<1 and x<-1, ANY x from this ranges will be less than 1. So the answer to our original question is NO. Sufficient.

Hope it's clear.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-greater-than-134052.html
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Re: Is x greater than 1?   [#permalink] 29 Jul 2012, 01:15
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