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If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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If n is a prime number greater than 2, is 1/x > 1? (1) \(x^n < x < x^{\frac{1}{n}}\) (2)\(x^{n1} > x^{2n2}\)
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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emmak wrote: If n is a prime number greater than 2, is 1/x > 1? (1) \(x^n < x < x^{\frac{1}{n}}\) (2)\(x^{n1} > x^{2n2}\) 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^(n1) > x^2(n1), for n being odd, n1 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
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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08 May 2013, 23:56
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)\(x^{n1} > x^{2n2}\) 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 ... \(x^{n1} > x^{2n2}\) ... \(x^{n1} > x^{2n2}\) It can be written as ... .... \(x^{n+1} > x^{2n}\) Therefore.. InSufficient .... Hence, ...... A.......
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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09 May 2013, 00:14
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)\(x^{n1} > x^{2n2}\) 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 ... \(x^{n1} > x^{2n2}\) ... \(x^{n1} > x^{2n2}\) 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 (n1) 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
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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09 May 2013, 00:41
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emmak wrote: If n is a prime number greater than 2, is 1/x > 1? (1) \(x^n < x < x^{\frac{1}{n}}\) (2)\(x^{n1} > x^{2n2}\) 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 \(x(1x^2)>0\) > x(x+1)(x1)<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.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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09 May 2013, 00:45
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)\(x^{n1} > x^{2n2}\) 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.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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09 May 2013, 00:59
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)\(x^{n1} > x^{2n2}\) 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 ... \(x^{n1} > x^{2n2}\) ... \(x^{n1} > x^{2n2}\) 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 (n1) 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 ...
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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09 May 2013, 01:06
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)\(x^{n1} > x^{2n2}\) 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}. Regards, Arpan
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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30 Oct 2014, 00:49
the question asks if 1/x > 1 or, 1/x  1 > 0 or, (1x) / x > 0 again for the above condition to be true; (1x) > 0 and x > 0 or (1x) < 0 and x<0 (for x<0, 1x 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^(n1) > x^(2n2) => (x^(n1))  (x^(2n2)) > 0 => (x^n/x)  (x^2n/x^2) > 0 => x^n/x (1x^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^(n1) < 1
when x^(n1) 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^(n1) 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.



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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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30 Oct 2014, 08:52
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emmak wrote: If n is a prime number greater than 2, is 1/x > 1?
(1) \(x^n < x < x^{\frac{1}{n}}\)
(2)\(x^{n1} > x^{2n2}\) 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. Answer > C



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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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01 Nov 2014, 08:12
emmak wrote: If n is a prime number greater than 2, is 1/x > 1?
(1) \(x^n < x < x^{\frac{1}{n}}\)
(2)\(x^{n1} > x^{2n2}\) 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^(n1) > x^(2n2) => x^(n1) > x^2(n1) n=odd, so n1=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.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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01 Nov 2014, 08:38
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)\(x^{n1} > x^{2n2}\) 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^(n1) > x^(2n2) => x^(n1) > x^2(n1) n=odd, so n1=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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If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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01 Nov 2014, 08:47
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)\(x^{n1} > x^{2n2}\) 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^(n1) > x^(2n2) => x^(n1) > x^2(n1) n=odd, so n1=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
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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11 May 2016, 13:45
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
Answer is B.



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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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11 May 2016, 15:40
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
Answer is B. The OA is given under the spoiler and it's C, not B. If x=1/2 and n=3, \(x^{n1} > x^{2n2}\) holds true: \([(\frac{1}{2})^{(31)} =\frac{1}{4}] > [\frac{1}{16}=(\frac{1}{2})^{(2*32)}]\)
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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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.



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