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If p and q are integers, is p^q >1 ? (1) pq < 0 (2) q^p < 1

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If p and q are integers, is p^q >1 ? (1) pq < 0 (2) q^p < 1  [#permalink]

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New post Updated on: 02 Oct 2018, 06:07
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If p and q are integers, is p^q >1


(1) pq < 0

(2) q^p < 1

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Originally posted by push12345 on 02 Oct 2018, 05:28.
Last edited by Bunuel on 02 Oct 2018, 06:07, edited 1 time in total.
Renamed the topic and edited the question.
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If p and q are integers, is p^q > 1?  [#permalink]

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New post 30 Dec 2018, 06:43
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If p and q are integers, is p^q > 1?

(1) pq < 0

(2) q^p < 1
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Re: If p and q are integers, is p^q > 1?  [#permalink]

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New post 30 Dec 2018, 07:47
1
The answer is E.

1) This could be true in two tays:
- p>0 and q<0, in which case 0<p^q<1 (ex: 3^(-1) = 1/3) - answer is yes!
- p<0 and q>0, in which case p^q can be greater than 1 (ex = -2^2=4), or smaller than it (ex: -2^-1=-1/2) - inconclusive!
insufficient.

2) This could be true in the exact same two ways as in (1). This means this as well is inconclusive.
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Re: If p and q are integers, is p^q >1 ? (1) pq < 0 (2) q^p < 1  [#permalink]

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New post 02 Oct 2018, 05:45
If p and q are integers, is p^q >1?

When will this happen :- when p>1 and q>0 or even when q is even >0 and p^2>1

i)pq<0
p and q are of OPPOSITE signs..
p is 3 and q<0...ans will be NO as 3^(-1)=1/3<1
If p -2 and q is 2... Ans will be yes
Insufficient

ii)q^p<1
Same example as in statement I
Insuff

Combined
Same examples remain
Insufficient

E
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Re: If p and q are integers, is p^q >1 ? (1) pq < 0 (2) q^p < 1  [#permalink]

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New post 31 Dec 2018, 01:41
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Re: If p and q are integers, is p^q >1 ? (1) pq < 0 (2) q^p < 1   [#permalink] 31 Dec 2018, 01:41
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