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16 Sep 2014, 00:11
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Is $$x$$ greater than 1?

(1) $$\frac{1}{x} \gt -1$$

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$

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16 Sep 2014, 00:11
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Official Solution:

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

B. $$x \lt 0$$ and $$1+x \lt 0$$, ($$x \lt -1$$). So, $$x \lt -1$$.

So, the given inequality holds true in two ranges: $$x \gt 0$$ and $$x \lt -1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$, two cases:

A. $$x \gt 0$$ (it's the same as $$x^5 \gt 0$$) and $$1-x^2 \gt 0$$, ($$-1 \lt x \lt 1$$). So, $$0 \lt x \lt 1$$;

B. $$x \lt 0$$ and $$1-x^2 \lt 0$$, ($$x \lt -1$$ or $$x \gt 1$$). So, $$x \lt -1$$;

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

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04 Dec 2014, 18:48
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Hi Bunuel,
I appreciate your answer, but the method looks bit clumsy to me.
I solved it in simpler way, please guide me whether my approach is correct.

1) 1/x > -1. It means 1 > -x
That is x > -1 .
So we are not sure that if the x is greater than 1. So it is insufficient.
2) 1/x^5 > 1/x^3. It means 1/x^2 > 1 .
That is x^2 < 1.
If x^2 is less than 1, that definitely implies x is less than 1. So it is sufficient.
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04 Dec 2014, 22:36
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Is x greater than 1?

(1) 1/x > -1

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

Alternate solution (Substituting numbers)
>> !!!

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05 Dec 2014, 05:01
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amitjathar wrote:
Hi Bunuel,
I appreciate your answer, but the method looks bit clumsy to me.
I solved it in simpler way, please guide me whether my approach is correct.

1) 1/x > -1. It means 1 > -x
That is x > -1 .
So we are not sure that if the x is greater than 1. So it is insufficient.
2) 1/x^5 > 1/x^3. It means 1/x^2 > 1 .
That is x^2 < 1.
If x^2 is less than 1, that definitely implies x is less than 1. So it is sufficient.

No, that's not correct.

You cannot multiply 1/x > -1 by x because you don't know the sign of x. If x is positive then yes, 1 > -x but if x is negative, then when multiplying by it you should flip the sign and you get 1 < -x.

Similarly you cannot reduce 1/x^5 > 1/x^3 by 1/x^3, because you don't know its sign.

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

Other tips on Inequalities are here: inequalities-tips-and-hints-175001.html
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06 Dec 2014, 01:33
Bunuel wrote:
Official Solution:

Is x greater than 1?

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

B. $$x \lt 0$$ and $$1+x \lt 0$$, ($$x \lt -1$$). So, $$x \lt -1$$.

So, the given inequality holds true in two ranges: $$x \gt 0$$ and $$x \lt -1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$, two cases:

A. $$x \gt 0$$ (it's the same as $$x^5 \gt 0$$) and $$1-x^2 \gt 0$$, ($$-1 \lt x \lt 1$$). So, $$0 \lt x \lt 1$$;

B. $$x \lt 0$$ and $$1-x^2 \lt 0$$, ($$x \lt -1$$ or $$x \gt 1$$). So, $$x \lt -1$$;

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

Hi Bunuel,
I did it like this:
Stmt1: $$\frac{1}{x} >-1$$
To make left side between -1 and 0, x must be negative.
To make left side greater than 0, x must be positive.
Not sufficient.

Stmt2: $$\frac{1}{x^5} > \frac{1}{x^3}$$
Multiply both sides by $$x^2$$ => $$\frac{1}{x^3} >\frac{1}{x}$$
Now: $$\frac{1}{x^3} >\frac{1}{x}$$is true only if x<1
Sufficient.

Tell me if I did something wrong.
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06 Dec 2014, 06:27
scofield1521 wrote:
Bunuel wrote:
Official Solution:

Is x greater than 1?

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

B. $$x \lt 0$$ and $$1+x \lt 0$$, ($$x \lt -1$$). So, $$x \lt -1$$.

So, the given inequality holds true in two ranges: $$x \gt 0$$ and $$x \lt -1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$, two cases:

A. $$x \gt 0$$ (it's the same as $$x^5 \gt 0$$) and $$1-x^2 \gt 0$$, ($$-1 \lt x \lt 1$$). So, $$0 \lt x \lt 1$$;

B. $$x \lt 0$$ and $$1-x^2 \lt 0$$, ($$x \lt -1$$ or $$x \gt 1$$). So, $$x \lt -1$$;

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

Hi Bunuel,
I did it like this:
Stmt1: $$\frac{1}{x} >-1$$
To make left side between -1 and 0, x must be negative.
To make left side greater than 0, x must be positive.
Not sufficient.

Stmt2: $$\frac{1}{x^5} > \frac{1}{x^3}$$
Multiply both sides by $$x^2$$ => $$\frac{1}{x^3} >\frac{1}{x}$$
Now: $$\frac{1}{x^3} >\frac{1}{x}$$is true only if x<1
Sufficient.

Tell me if I did something wrong.

Do the ranges you got match the ranges from my solution?
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06 Dec 2014, 06:47
1
Bunuel wrote:
scofield1521 wrote:
Bunuel wrote:
Official Solution:

Is x greater than 1?

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

B. $$x \lt 0$$ and $$1+x \lt 0$$, ($$x \lt -1$$). So, $$x \lt -1$$.

So, the given inequality holds true in two ranges: $$x \gt 0$$ and $$x \lt -1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$, two cases:

A. $$x \gt 0$$ (it's the same as $$x^5 \gt 0$$) and $$1-x^2 \gt 0$$, ($$-1 \lt x \lt 1$$). So, $$0 \lt x \lt 1$$;

B. $$x \lt 0$$ and $$1-x^2 \lt 0$$, ($$x \lt -1$$ or $$x \gt 1$$). So, $$x \lt -1$$;

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

Hi Bunuel,
I did it like this:
Stmt1: $$\frac{1}{x} >-1$$
To make left side between -1 and 0, x must be negative.
To make left side greater than 0, x must be positive.
Not sufficient.

Stmt2: $$\frac{1}{x^5} > \frac{1}{x^3}$$
Multiply both sides by $$x^2$$ => $$\frac{1}{x^3} >\frac{1}{x}$$
Now: $$\frac{1}{x^3} >\frac{1}{x}$$is true only if x<1
Sufficient.

Tell me if I did something wrong.

Do the ranges you got match the ranges from my solution?

Yes I did try with some examples. I believe they are true.
For stmt 1: x= -.5, x= 1;
For stmt 2: x= 0.9, x= 1.1.
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06 Dec 2014, 07:07
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scofield1521 wrote:
Bunuel wrote:
scofield1521 wrote:
Hi Bunuel,
I did it like this:
Stmt1: $$\frac{1}{x} >-1$$
To make left side between -1 and 0, x must be negative.
To make left side greater than 0, x must be positive.
Not sufficient.

Stmt2: $$\frac{1}{x^5} > \frac{1}{x^3}$$
Multiply both sides by $$x^2$$ => $$\frac{1}{x^3} >\frac{1}{x}$$
Now: $$\frac{1}{x^3} >\frac{1}{x}$$is true only if x<1
Sufficient.

Tell me if I did something wrong.

Do the ranges you got match the ranges from my solution?

Yes I did try with some examples. I believe they are true.
For stmt 1: x= -.5, x= 1;
For stmt 2: x= 0.9, x= 1.1.

$$\frac{1}{x} \gt - 1$$ holds true for two ranges: $$x \gt 0$$ and $$x \lt -1$$. You got x>0 and x<0.

$$\frac{1}{x^5} \gt \frac{1}{x^3}$$ holds true for two ranges: $$0 \lt x \lt 1$$ and $$x \lt -1$$ you got x<1.

How do they match???
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19 May 2015, 13:00
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Hi Bunuel.

I am not able to understand how to arrive at the bold part:

B. x<0 and 1−x2<0, (x<−1 or x>1). So, x<−1;

Can you please explain
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20 May 2015, 02:43
Randude wrote:
Hi Bunuel.

I am not able to understand how to arrive at the bold part:

B. x<0 and 1−x2<0, (x<−1 or x>1). So, x<−1;

Can you please explain

1 - x^2 < 0 --> x^2 > 1 --> x < -1 or x > 1. Since we are also considering x < 0, then the intersection of these ranges gives x < -1.
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02 Oct 2015, 11:33
Bunuel wrote:
Official Solution:

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

B. $$x \lt 0$$ and $$1+x \lt 0$$, ($$x \lt -1$$). So, $$x \lt -1$$.

So, the given inequality holds true in two ranges: $$x \gt 0$$ and $$x \lt -1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$, two cases:

A. $$x \gt 0$$ (it's the same as $$x^5 \gt 0$$) and $$1-x^2 \gt 0$$, ($$-1 \lt x \lt 1$$). So, $$0 \lt x \lt 1$$;

B. $$x \lt 0$$ and $$1-x^2 \lt 0$$, ($$x \lt -1$$ or $$x \gt 1$$). So, $$x \lt -1$$;

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

Can you explain how did you arrive at (2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$,
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03 Oct 2015, 05:03
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harishbiyani8888 wrote:
Bunuel wrote:
Official Solution:

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

B. $$x \lt 0$$ and $$1+x \lt 0$$, ($$x \lt -1$$). So, $$x \lt -1$$.

So, the given inequality holds true in two ranges: $$x \gt 0$$ and $$x \lt -1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$, two cases:

A. $$x \gt 0$$ (it's the same as $$x^5 \gt 0$$) and $$1-x^2 \gt 0$$, ($$-1 \lt x \lt 1$$). So, $$0 \lt x \lt 1$$;

B. $$x \lt 0$$ and $$1-x^2 \lt 0$$, ($$x \lt -1$$ or $$x \gt 1$$). So, $$x \lt -1$$;

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

Can you explain how did you arrive at (2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$,

$$\frac{1}{x^5} \gt \frac{1}{x^3}$$;

$$\frac{1}{x^5} - \frac{1}{x^3} \gt 0$$;

$$\frac{1}{x^5} - \frac{x^2}{x^5} \gt 0$$;

$$\frac{1-x^2}{x^5} \gt 0$$.
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11 Dec 2016, 20:53
I think this is a high-quality question and I don't agree with the explanation. Can you please explain S1 (case 1). 1+x>0 and x>0. So it should be x>-1 and x>0. How have you written x>0 after that.
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12 Dec 2016, 01:25
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pratyushk1 wrote:
I think this is a high-quality question and I don't agree with the explanation. Can you please explain S1 (case 1). 1+x>0 and x>0. So it should be x>-1 and x>0. How have you written x>0 after that.

Can you please tell me what does it even mean x > -1 and x > 0?

Please check the links below to study how to derive the ranges for inequalities:

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Hope this helps.
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06 May 2017, 20:17
Bunuel wrote:
Official Solution:

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

I don't understand this. Is there a rule or something that says when a/b > 0 then a>0 and b>0?
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07 May 2017, 02:06
joondez wrote:
Bunuel wrote:
Official Solution:

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

I don't understand this. Is there a rule or something that says when a/b > 0 then a>0 and b>0?

Yes. a/b to be positive a and b must have the same sign, so both must be positive or both must be negative.

You should brush-up fundamentals before attempting questions, especially hard ones.

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

http://gmatclub.com/forum/inequalities-trick-91482.html
http://gmatclub.com/forum/data-suff-ine ... 09078.html
http://gmatclub.com/forum/range-for-var ... 09468.html
http://gmatclub.com/forum/everything-is ... 08884.html
http://gmatclub.com/forum/graphic-appro ... 68037.html

Hope this helps.
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03 Jun 2017, 22:45
Hi Bunuel,
How did you write 1/x-1>0 as x+1/x without knowing the sign of x?
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04 Jun 2017, 06:31
supra2411 wrote:
Hi Bunuel,
How did you write 1/x-1>0 as x+1/x without knowing the sign of x?

We are concerned about the sign when we are multiplying/dividing an inequality by a variable. Here we are simply re-arranging the inequality:

$$\frac{1}{x} \gt - 1$$.

$$\frac{1}{x} +1\gt 0$$.

$$\frac{1+x}{x} \gt 0$$.

Hope it helps.
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04 Jun 2017, 11:31
pratyushk1 wrote:
I think this is a high-quality question and I don't agree with the explanation. Can you please explain S1 (case 1). 1+x>0 and x>0. So it should be x>-1 and x>0. How have you written x>0 after that.

Hi

Let me try. You have written 1+x>0 and x>0 from Bunuel's first explanation to this question.
you are correct that 1+x > 0 translates to x > -1. And we already have x > 0.

So when we are given that x>-1 and x>0 (these are Two different ranges of x given) we have to figure out the common range - that will be our final range for x.

You see x>-1 includes all numbers greater than -1 till infinity. So this means -1<x<0 and 0<x<infinity
But when we say x>0 it includes all numbers from o to infinity. So this means 0<x<infinity

So if you look at both the ranges above.. which range is common between both? Its 0<x<infinity.. So when we say that x>-1 and x>0, then our conclusion is that x>0 (because that is the only range which is common to both the above ranges)..

I think thats why Bunuel wrote (in his very first explanation to this question) that x>0

(You can also think of it this way: Say your boss tells you that you will receive your salary after 10th June and your Super Boss tells you that you will receive your salary after 12th June.. and if you know that both these statements are true - then the only conclusion which can be drawn from these is that your salary will come only after 12th June..from 10th to 12th June is NOT possible.)
Re: D01-10   [#permalink] 04 Jun 2017, 11:31

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