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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
08 May 2013, 22:28

emmak wrote:

If n is a prime number greater than 2, is 1/x > 1? (1) \(x^n < x < x^{\frac{1}{n}}\) (2)\(x^{n-1} > x^{2n-2}\)

Couldn't the answer be [A],

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}.

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.

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
08 May 2013, 23:41

2

This post received KUDOS

Expert's post

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.

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
08 May 2013, 23:45

Expert's post

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

Couldn't the answer be [A],

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. _________________

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.

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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)\(x^{n-1} > x^{2n-2}\)

Couldn't the answer be [A],

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}.

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
28 Oct 2014, 10:26

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

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
30 Oct 2014, 07:52

1

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.

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

Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
01 Nov 2014, 07:38

Expert's post

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. _________________

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

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