If n is a prime number greater than 2, is 1/x > 1? : GMAT Data Sufficiency (DS)
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 07 Dec 2016, 21:05

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If n is a prime number greater than 2, is 1/x > 1?

Author Message
TAGS:

### Hide Tags

Manager
Joined: 09 Feb 2013
Posts: 120
Followers: 1

Kudos [?]: 804 [2] , given: 17

If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

08 May 2013, 21:03
2
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

13% (03:08) correct 87% (01:56) wrong based on 352 sessions

### HideShow timer Statistics

If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$
[Reveal] Spoiler: OA

_________________

Kudos will encourage many others, like me.
Good Questions also deserve few KUDOS.

Last edited by Bunuel on 29 Oct 2014, 03:29, edited 2 times in total.
Formatted the question
Manager
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 123
Location: India
Concentration: General Management, Technology
GMAT 1: 640 Q42 V37
GPA: 3.5
WE: Web Development (Computer Software)
Followers: 17

Kudos [?]: 109 [1] , given: 14

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

08 May 2013, 22:28
1
KUDOS
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

In such a case, 1/x>1 or x belongs to {0,1} -ve values again will not satisfy the equation.

Statement 2: x^(n-1) > x^2(n-1), for n being odd, n-1 will be even. Hence for the above equation the solution set will be {-1,1}
As mentioned above, the negative values of x will not hold the 1/x>1 equation. Hence the range is again {0,1}.

Am I doing sumthing wrong?

Regards,
Arpan
_________________

Feed me some KUDOS! *always hungry*

Manager
Status: Pushing Hard
Affiliations: GNGO2, SSCRB
Joined: 30 Sep 2012
Posts: 89
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.33
WE: Analyst (Health Care)
Followers: 1

Kudos [?]: 82 [0], given: 11

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

08 May 2013, 22:56
emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a +ve proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X is a +ve proper fraction ?????

So, We have to find if X is a +ve Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a +ve proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a +ve proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

It can be written as ... .... $$x^{n+1} > x^{2n}$$

Therefore.. InSufficient ....

Hence, ...... A.......
_________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Last edited by manishuol on 09 May 2013, 00:42, edited 3 times in total.
Manager
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 123
Location: India
Concentration: General Management, Technology
GMAT 1: 640 Q42 V37
GPA: 3.5
WE: Web Development (Computer Software)
Followers: 17

Kudos [?]: 109 [0], given: 14

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

08 May 2013, 23:14
manishuol wrote:
emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X a proper fraction ?????

So, We have to find if X is a Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

In this one also if we plugin in the value of n as 3 or 5 ... we will get when n = 3 ...$$x^2$$ > $$x^4$$

& when n = 5 .. we will get $$x^4$$ > $$x^8$$ ....... Both of these states that x is a proper fraction .....

Therefore.. Sufficient ....

Hence, ...... D.......

I agree with Statement 1 being sufficient, but when it comes to Statement 2,

For example, x = -0.1, in such a case, assuming n is Odd and equal to 3, the inequality can be reduced to x^2>x^4. The value x = -0.1 satisfies.

Since the power term (n-1) will always be even, the inequality will hold good for even -ve numbers between {-1,0}. But when it comes to our parent equation:
1/-0.1 = -10 < 1. Hence the inequality doesn't satisfy. Therefore, I feel [A] should be the answer. Please look into my method and let me know if I am doing anything wrong! Also please refer to my post above for a general approach to the prob.

Regards,
Arpan
_________________

Feed me some KUDOS! *always hungry*

Last edited by arpanpatnaik on 08 May 2013, 23:35, edited 1 time in total.
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 631
Followers: 79

Kudos [?]: 1091 [3] , given: 136

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

08 May 2013, 23:41
3
KUDOS
4
This post was
BOOKMARKED
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The question is asking, is $$\frac{1}{x}$$>1 --> Is 0<x<1

From F.S 1, for n=3, we have$$x^3<x<x^{\frac{1}{3}}$$ . As $$x>x^3,$$ we have $$x(1-x^2)>0$$ --> x(x+1)(x-1)<0 .

Hence 0<x<1 OR x<-1. Again, from $$x<x^{\frac{1}{3}}$$, we can cube on both sides and get, $$x^3<x$$, which is the same as above. Thus, from F.S 1, we get :

0<x<1 OR x<-1. Insufficient.

From F.S 2, we know that for n=3, $$x^2 > x^4 --> x^2<1$$ --> -1<x<1. Clearly Insufficient.

Taking both together, we get 0<x<1 as the common range and a YES for the question stem. Sufficient.

C.
_________________
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 631
Followers: 79

Kudos [?]: 1091 [0], given: 136

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

08 May 2013, 23:45
arpanpatnaik wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

Not true. As x>2 and is a prime number, x will always be an odd integer. For x = -8,$$x^{1/3}$$ = -2.
_________________
Manager
Status: Pushing Hard
Affiliations: GNGO2, SSCRB
Joined: 30 Sep 2012
Posts: 89
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.33
WE: Analyst (Health Care)
Followers: 1

Kudos [?]: 82 [0], given: 11

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

08 May 2013, 23:59
arpanpatnaik wrote:
manishuol wrote:
emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X a proper fraction ?????

So, We have to find if X is a Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

In this one also if we plugin in the value of n as 3 or 5 ... we will get when n = 3 ...$$x^2$$ > $$x^4$$

& when n = 5 .. we will get $$x^4$$ > $$x^8$$ ....... Both of these states that x is a proper fraction .....

Therefore.. Sufficient ....

Hence, ...... D.......

I agree with Statement 1 being sufficient, but when it comes to Statement 2,

For example, x = -0.1, in such a case, assuming n is Odd and equal to 3, the inequality can be reduced to x^2>x^4. The value x = -0.1 satisfies.

Since the power term (n-1) will always be even, the inequality will hold good for even -ve numbers between {-1,0}. But when it comes to our parent equation:
1/-0.1 = -10 < 1. Hence the inequality doesn't satisfy. Therefore, I feel [A] should be the answer. Please look into my method and let me know if I am doing anything wrong! Also please refer to my post above for a general approach to the prob.

Regards,
Arpan

My answer is still A ...
_________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Last edited by manishuol on 09 May 2013, 00:34, edited 2 times in total.
Manager
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 123
Location: India
Concentration: General Management, Technology
GMAT 1: 640 Q42 V37
GPA: 3.5
WE: Web Development (Computer Software)
Followers: 17

Kudos [?]: 109 [0], given: 14

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

09 May 2013, 00:06
vinaymimani wrote:
arpanpatnaik wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

Not true. As n>2 and is a prime number, will n always be an odd integer. For x = -8,$$x^{1/3}$$ = -2.

My Bad!! I missed that... Thanks for correcting me Vinay! I got the answer now! [C] it is
Can't believe I used the odd principle in Statement 2 and failed to use it in Statement 1. The range x < -1 will be valid and the solution set needs to be an intersection of Statement 1 and 2 i.e. x belongs to {0,1}.

Regards,
Arpan
_________________

Feed me some KUDOS! *always hungry*

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12891
Followers: 561

Kudos [?]: 158 [0], given: 0

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

28 Oct 2014, 10:26
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 06 Aug 2013
Posts: 92
Followers: 0

Kudos [?]: 3 [0], given: 17

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

29 Oct 2014, 23:49
the question asks if 1/x > 1
or, 1/x - 1 > 0
or, (1-x) / x > 0
again for the above condition to be true;
(1-x) > 0 and x > 0 or (1-x) < 0 and x<0 (for x<0, 1-x can never be less than 0, hence ignore this)
basically the question asks if 0 < x < 1 <----------- this is what we need to find, does x lie between 0 and 1??

statement 1 is clearly insufficient

i've got a doubt here with statement 2.

x^(n-1) > x^(2n-2)
=> (x^(n-1)) - (x^(2n-2)) > 0
=> (x^n/x) - (x^2n/x^2) > 0
=> x^n/x (1-x^n/x) > 0 taking x^n/x common outside the bracket
=> for the above condition to be true
either or
x^n/x > 0 and (1- x^n/x) > 0 x^n/x < 0 and (1 - x^n/x) < 0
when x^n/x > 0, when x^n/x < 0, (1 - x^n/x) can never be less than 0
it means for (1 - x^n/x) > 0 hence, we can ignore this option
x^n/x has to be a fraction between 0 and 1
i.e. 0< x^n/x < 1
or, 0 < x^(n-1) < 1

when x^(n-1) is between 0 and 1, and n is a prime number greater than 2,
it implies that x is a fraction between 0 and 1 and x^(n-1) will always be positive and true for any value between 0 and 1.
in that case Statement 2 is satisfying what we are asked to find out. statement 2 should be sufficient.
correct me if i am wrong.
Intern
Joined: 04 Jul 2014
Posts: 48
Schools: Smeal
Followers: 0

Kudos [?]: 27 [0], given: 40

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

30 Oct 2014, 07:52
2
This post was
BOOKMARKED
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$

Statement 1 -> will be true for negative integers and positive fractions. So X can either be a negative integer or positive fraction. If X is a negative integer 1/X will not be greater than 1 but if X is a positive fraction 1/X will be greater than 1. So we have two possibilities.

Hence Statement 1 alone is insufficient.

Statement 2 -> Will be true for negative and positive fractions. If X is a negative fraction 1/X will not be greater than 1 but if X is a positive fraction 1/X will be greater than 1. So we have two possibilities.

Hence Statement 2 alone is insufficient.

But for both statement 1 and 2 to be true, X can only be a positive fraction. Here 1/X will be greater than 1.

Hence both statements together are sufficient.

Manager
Joined: 22 Jan 2014
Posts: 138
WE: Project Management (Computer Hardware)
Followers: 0

Kudos [?]: 51 [0], given: 135

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

01 Nov 2014, 07:12
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$

I got A for this one.

1) x^n < x < x^(1/n)
n = odd (since it is prime and > 2)
so the inequality will hold true only for 0 < x < 1
it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number.
so this is sufficient.

2) x^(n-1) > x^(2n-2)
=> x^(n-1) > x^2(n-1)
n=odd, so n-1=even=2k(say)
x^2k > x^4k
this inequality holds true for -1<x<1
so insufficient.

Is the OA correct? Bunuel can you please have a look at this one.
_________________

Illegitimi non carborundum.

Math Expert
Joined: 02 Sep 2009
Posts: 35912
Followers: 6851

Kudos [?]: 90032 [0], given: 10402

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

01 Nov 2014, 07:38
thefibonacci wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$

I got A for this one.

1) x^n < x < x^(1/n)
n = odd (since it is prime and > 2)
so the inequality will hold true only for 0 < x < 1
it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number.
so this is sufficient.

2) x^(n-1) > x^(2n-2)
=> x^(n-1) > x^2(n-1)
n=odd, so n-1=even=2k(say)
x^2k > x^4k
this inequality holds true for -1<x<1
so insufficient.

Is the OA correct? Bunuel can you please have a look at this one.

Odd root from negative number is negative, not imaginary. For, example, $$\sqrt[3]{-8}=-2$$.

The question asks whether 0 < x < 1. For (1) check x = -8 and n = 3 prime.
_________________
Manager
Joined: 22 Jan 2014
Posts: 138
WE: Project Management (Computer Hardware)
Followers: 0

Kudos [?]: 51 [0], given: 135

If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

01 Nov 2014, 07:47
Bunuel wrote:
thefibonacci wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) $$x^n < x < x^{\frac{1}{n}}$$

(2)$$x^{n-1} > x^{2n-2}$$

I got A for this one.

1) x^n < x < x^(1/n)
n = odd (since it is prime and > 2)
so the inequality will hold true only for 0 < x < 1
it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number.
so this is sufficient.

2) x^(n-1) > x^(2n-2)
=> x^(n-1) > x^2(n-1)
n=odd, so n-1=even=2k(say)
x^2k > x^4k
this inequality holds true for -1<x<1
so insufficient.

Is the OA correct? Bunuel can you please have a look at this one.

Odd root from negative number is negative, not imaginary. For, example, $$\sqrt[3]{-8}=-2$$.

The question asks whether 0 < x < 1. For (1) check x = -8 and n = 3 prime.

Is this case GMAT specific?

Please have a look at this:

http://www.wolframalpha.com/input/?i=%2 ... 281%2F3%29
_________________

Illegitimi non carborundum.

Math Expert
Joined: 02 Sep 2009
Posts: 35912
Followers: 6851

Kudos [?]: 90032 [0], given: 10402

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

01 Nov 2014, 07:51
thefibonacci wrote:
Is this case GMAT specific?

Please have a look at this:

http://www.wolframalpha.com/input/?i=%2 ... 281%2F3%29

No, that's generally true.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12891
Followers: 561

Kudos [?]: 158 [0], given: 0

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

20 Nov 2015, 05:29
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 01 Mar 2016
Posts: 6
Followers: 0

Kudos [?]: 1 [0], given: 2

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

11 May 2016, 12:45
I believe the answer should be B

Statement 1: Insufficient [0<x<1] or [x<-1] (refer to others)

Statement 2: Fractions get closer to zero when raised to a power greater than 1. I think of it like this - when fractions are raised to a power greater than 1, the fraction gets closer to zero. A positive fraction would decrease, while a negative fraction would increase.
When plugging in x= (1/2) and n=3, statement 2 is correct ([1/4] is greater than [1/16])
When plugging in x=(-1/2) and n=3, statement 2 is incorrect ([-1/4] is NOT greater than [-1/16])
Therefore: [0<x<1] only ; [1/x]>1 in all cases ; Statement 2 Sufficient

Math Expert
Joined: 02 Sep 2009
Posts: 35912
Followers: 6851

Kudos [?]: 90032 [0], given: 10402

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

11 May 2016, 14:40
mattyahn wrote:
I believe the answer should be B

Statement 1: Insufficient [0<x<1] or [x<-1] (refer to others)

Statement 2: Fractions get closer to zero when raised to a power greater than 1. I think of it like this - when fractions are raised to a power greater than 1, the fraction gets closer to zero. A positive fraction would decrease, while a negative fraction would increase.
When plugging in x= (1/2) and n=3, statement 2 is correct ([1/4] is greater than [1/16])
When plugging in x=(-1/2) and n=3, statement 2 is incorrect ([-1/4] is NOT greater than [-1/16])
Therefore: [0<x<1] only ; [1/x]>1 in all cases ; Statement 2 Sufficient

The OA is given under the spoiler and it's C, not B.

If x=-1/2 and n=3, $$x^{n-1} > x^{2n-2}$$ holds true: $$[(-\frac{1}{2})^{(3-1)} =\frac{1}{4}] > [\frac{1}{16}=(-\frac{1}{2})^{(2*3-2)}]$$
_________________
Intern
Joined: 01 Mar 2016
Posts: 6
Followers: 0

Kudos [?]: 1 [0], given: 2

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]

### Show Tags

11 May 2016, 15:22
Bunuel, thanks - I overlooked the fact that the exponent will always be even given that n is prime greater than 2.
Man, this a tricky one.
Re: If n is a prime number greater than 2, is 1/x > 1?   [#permalink] 11 May 2016, 15:22
Similar topics Replies Last post
Similar
Topics:
4 If x = n!+1, is x a prime number? 5 06 May 2016, 22:26
Is x^2>x ? (1) x is a prime number (2) -1≤ x ≤1 3 27 Feb 2016, 23:29
4 Is x^2>x (1) x^2 is greater than 1 (2) x is greater than -1 6 14 May 2012, 09:39
10 Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t 10 04 Feb 2012, 08:23
Is y greater than x? (1) |y|<1 and |x| <2 (2) y>0 2 22 Jul 2010, 12:07
Display posts from previous: Sort by