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Is x > 0?
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Updated on: 12 Apr 2013, 02:24
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66% (01:21) correct 34% (01:26) wrong based on 175 sessions
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Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8
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Originally posted by aDAMS on 11 Apr 2013, 18:16.
Last edited by Bunuel on 12 Apr 2013, 02:24, edited 1 time in total.
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Re: DS question on Exponents
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11 Apr 2013, 18:49
Stmt 1 is Not Sufficientx^6 > x^7 means x < 1, so x > 0 when 0< x < 1 and when x < 0 so x < 0 Stmt 2 is Sufficientx^7 > x ^8 means 0<x<1 so x > 0 A variable with smaller power is more than a variable with larger power iff the variable is in the range 0 < x < 1 Answer is B//kudos please, if the above explanation is good.
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Re: DS question on Exponents
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11 Apr 2013, 23:09
aDAMS wrote: Q. Is x > 0?
1. x^6>x^7 2. x^7>x^8 Any number with an even exponent is \(>0\), but I wanna give you an alternative solution: 1. \(x^6>x^7\) so \(0>x^7x^6, 0>x^6(x1)\) at this point we can divide by x^6 because we know that does not equal 0(otherwise x^6=x^7 and not >) \(0>x1\) and finally \(1>x\). Is x positive? it could be(0,5) or not (1) 2.\(x^7>x^8\) becomes \(0>x^6(x^2x)\) divide \(0>x^2x\) so \(x(x1)<0\) and \(0<x<1\). Is x positive? yes B
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Re: Is x > 0?
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12 Apr 2013, 02:38



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Re: Is x > 0?
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12 Apr 2013, 04:35
Felt the following would be a simpler explanation The two option would go as follows 1. x^6 > x^7 This option is only possible either when 1>x>0 or x<0 Like if x is <1 but >0 then the values will keep on decreasing with every increasing exponent and if x<0 then the even powers will be positive and odd ones would be ve therefore irrespective of the value of x the even exponents will always be greater 2. x^7 > x^8 This is only possible when 1>x>0, as the situation would be as in case 1 but when X<0 the even exponents will always be greater Only 2 is sufficient but 1 is not B
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Re: Is x > 0?
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12 Apr 2013, 06:59
Sorry this is probably as silly as it gets but i need help in understanding the below In x^7 > x^8, say we divide by x^6, x> x^2 0>x(x1) Now 0>x and 0>(x1) for 0>x1 ==> x <1, till here i am fine but how do we get x>0 to make 0<x<1 ??
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Re: Is x > 0?
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12 Apr 2013, 07:04
summer101 wrote: Sorry this is probably as silly as it gets but i need help in understanding the below
In x^7 > x^8, say we divide by x^6, x> x^2 0>x(x1) Now 0>x and 0>(x1) for 0>x1 ==> x <1, till here i am fine but how do we get x>0 to make 0<x<1 ?? x*(x1)<0 implies either x<0 and x1 > 0 or x>0 and x1<0 first is impossible and hence from second we have 0<x<1



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Re: Is x > 0?
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12 Apr 2013, 07:56
summer101 wrote: Sorry this is probably as silly as it gets but i need help in understanding the below
In x^7 > x^8, say we divide by x^6, x> x^2 0>x(x1) Now 0>x and 0>(x1) for 0>x1 ==> x <1, till here i am fine but how do we get x>0 to make 0<x<1 ?? Till here you are fine x> x^2 so \(x^2x<0\) we have to solve this, and to solve let me use an old trick. Lets solve \(x^2x=0,x(x1)=0\) so x=1 or x=0. Now because the sign of x^2 is + and the operator is < we take the INTERNAL values: \(0<x<1\). Remember: to solve inequalities like this (x^2) treat them like equations (replace <,> with = ) then, once you have the results take a look at the sign of x^2 an the operator. (<,) or (>,+) take ESTERNAL values. (if they are the "same") (>,) or (<,+) take INTERNAL values(like this case). Let me know if it's clear now
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Re: Is x>0 ?
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10 Jun 2016, 12:53
AbdurRakib wrote: Is x>0 ?
1. \(x^6\)>\(x^7\)
2. \(x^7\) >\(x^8\) 1. \(x^6\)>\(x^7\), either 0 < x< 1 or x < 1 (the result of even power of a negative number is positive). Not Sufficient. 2. \(x^7\) >\(x^8\), 0 < x< 1. x cannot be negative, If x were negative, \(x^7\) will be negative and \(x^8\) will be positive and this statement won't hold true. Sufficient. Answer is B.



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Re: Is x > 0?
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14 Jun 2016, 01:06
aDAMS wrote: Is x > 0?
(1) x^6 > x^7 (2) x^7 > x^8 \(x^6 > x^7\) \(x^6  x^7 > 0\) \(x^6\)*(1  x) > 0 \(x^6\) is always a positive value. so (1x)> 0 or x < 1 x can be (1/2) a positive value or x can be 2 , a negative value, so not sufficient. Statement (2)\(x^7 > x^8\) \(x^7  x^8 > 0\) \(x^7\)*(1  x) > 0 \(x^6.x.(1x)\) > 0 Now \(x^6\) is always a positive value. so x(1x)> 0 or x(x1) < 0 0< x <1Thus x is always positive. B is sufficient.



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Re: Is x > 0?
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17 Jun 2019, 18:31
Simply assume values. Let's say x = 0.5 or 0.5
From statement 1, X^6>X^7 Case 1, if X is 0.5, X^6 > X^7 is true Case 2, if X is 0.5, X^>X^7 is true We cannot prove is X is >0. Insufficient.
From Statement 2, X^7>X^8 Case 1, if X is 0.5, X^7>X^8 is true Case 2, if X is 0.5, X^7>X^8 is false Therefore, sufficient. X>0
Answer is B.










