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If |x|>1, is x^(1/a)>x^(1/b)? 1) a/b>1 2) a+b<0

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If |x|>1, is x^(1/a)>x^(1/b)? 1) a/b>1 2) a+b<0  [#permalink]

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New post Updated on: 13 Mar 2019, 21:20
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

38% (01:38) correct 62% (01:55) wrong based on 13 sessions

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If \(|x|>1\), is \(x^{\frac{1}{a}}>x^{\frac{1}{b}}\)?


(1) \(\frac{a}{b}>1\)

(2) \(a+b<0\)

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Originally posted by AsadAbu on 13 Mar 2019, 13:05.
Last edited by Bunuel on 13 Mar 2019, 21:20, edited 1 time in total.
Formatted.
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Re: If |x|>1, is x^(1/a)>x^(1/b)? 1) a/b>1 2) a+b<0  [#permalink]

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New post 16 Mar 2019, 20:47
Combining:
a=-3 & b=-1 suffice both statements.
If x=8, x^(a/b)>x^(a/b)
If x=-8, x^(a/b)<x^(a/b)

My answer is E

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Re: If |x|>1, is x^(1/a)>x^(1/b)? 1) a/b>1 2) a+b<0  [#permalink]

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New post 16 Mar 2019, 22:40
I will go with A. As question asks whether 1/a>1/b ? (As same base of exponents) Or b/a >1 ?

Stmnt 1 itself says a/b >1. So this is sufficient.
Stmnt 2 doesn’t give info to find value of b/a.

Experts answers please...

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Re: If |x|>1, is x^(1/a)>x^(1/b)? 1) a/b>1 2) a+b<0  [#permalink]

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New post 16 Mar 2019, 23:09
AN12 wrote:
I will go with A. As question asks whether 1/a>1/b ? (As same base of exponents) Or b/a >1 ?

Stmnt 1 itself says a/b >1. So this is sufficient.
Stmnt 2 doesn’t give info to find value of b/a.

Experts answers please...

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b/a>1 is true if the base is positive
What would be the case if base is negative.
According to the question, base can be 3,4,-3,-4 etc.
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Re: If |x|>1, is x^(1/a)>x^(1/b)? 1) a/b>1 2) a+b<0  [#permalink]

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New post 17 Mar 2019, 11:15
AN12 wrote:
I will go with A. As question asks whether 1/a>1/b ? (As same base of exponents) Or b/a >1 ?

Stmnt 1 itself says a/b >1. So this is sufficient.
Stmnt 2 doesn’t give info to find value of b/a.

Experts answers please...

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Assume, two combinations for a, b & x

1. a=0.5 & b=0.25 , X=+ve ----> then eqn becomes, x^(1/a) < x^(1/b)
2. a=-4 & b=-3 , X=-ve ----> now eqn becomes, x^(1/a) > x^(1/b)

Hence insufficient
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Re: If |x|>1, is x^(1/a)>x^(1/b)? 1) a/b>1 2) a+b<0   [#permalink] 17 Mar 2019, 11:15
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