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# D01-10

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Math Expert
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15 Sep 2014, 23:11
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Is $$x$$ greater than 1?

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

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$
[Reveal] Spoiler: OA

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15 Sep 2014, 23:11
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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, 17:48
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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, 21:36
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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, 04:01
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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, 00:33
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, 05:27
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, 05:47
1
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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, 06:07
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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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07 Dec 2014, 02:55
Bunuel wrote:

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

After taking more examples into consideration:

Stmt 1: Range: $$x \gt 0$$ and $$x \lt -1$$.
Eg:
If x=-0.5 Stmt is true and x is not greater than 1.
If x=2 Stmt still holds true but is greater than 1.
Not sufficient.

Stmt 2: Range: $$0 \lt x \lt 1$$ and $$x \lt -1$$
Eg:
If x=0.5 Stmt is true and x is not greater than 1.
If x= 1.1 Stmt is false and x must not be greater than 1.
If x=-2 Stmt is true and x must be less than -1.
If x=-0.5 Statement is false and x must be less than -1.

These satisfy the range you have mentioned. I hope these are correct.
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19 May 2015, 12:00
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;

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20 May 2015, 01:43
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;

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, 10:33
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, 04:03
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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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30 Jul 2016, 06:31
arunspanda wrote:
Is x greater than 1?

(1) 1/x > -1

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

Alternate solution (Substituting numbers)

Solving this way consumes more time. We need to pick right numbers. And if we miss any of it, we dont get correct values.

Using inequality wavy curve method you can solve much early in a generic way [ w/o picking the numbers ]

Please search for wavy curve method in this forum or in google.
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02 Aug 2016, 10:11
it is obvious for many people, but important to remember

x>0 and x<0, since x≠0

cannot be divided by 0
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19 Aug 2016, 04:39
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.

Hi Bunuel,

I don’t understand why after getting x > -1 we are writing so x > o
Even in 2nd statement
-1< x <1 is re written as 0< x <1

Please explain this in both cases; A of statement 1 and in A of Statement 2

Thanks
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11 Dec 2016, 19:53
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, 00:25
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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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09 Jan 2017, 11:49
I think this is a high-quality question and I agree with explanation.
Re D01-10   [#permalink] 09 Jan 2017, 11:49

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# D01-10

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