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Re: In which of the following scenarios is p>q? [#permalink]
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Re: In which of the following scenarios is p>q? [#permalink]
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Sawant91 wrote:
In which of the following scenarios is p>q?

Indicate all possible scenarios.
A. \((0.11)^p>(0.11)^q\)
B. \(1^p>1^q\)
C. \((1.11)^p>(1.11)^q\)
D. \((1.01)^p>(1.01)^q\)
E. \((p+q)(p−q)>0\)
F. \(|p|>|q|\)


So where are the variables p and q - they are used as exponent or power..
some rules for the terms when the base is positive..
1) when the number, say x, is between 0 and 1 that is 0<x<1...
Higher the power, lower the value so x^3<x^2
2) when x=1
the values are always same irrespective of the power. 1^7 = 1^1
3) when x>1
Higher the power , higher the value so x^3>x^2

now let us see the choices..
A. \((0.11)^p>(0.11)^q\).....0<0.11<1 so case (1) p<q
B. \(1^p>1^q\).......... case (2).. cannot be determined
C. \((1.11)^p>(1.11)^q\).......1.11>1, so case (3).. p>q
D. \((1.01)^p>(1.01)^q\).......1.01>1, so case (3).. p>q
E. \((p+q)(p−q)>0.......p^2-q^2>0.....p^2>q^2\), we can just say |p|>|q|... say p is negative (-3)^2>2^2 but -3<2, so p<q and if both positive p>q
F. \(|p|>|q|\).... same as E above

thus only C and D

hope it helps
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Re: In which of the following scenarios is p>q? [#permalink]
chetan2u wrote:
Sawant91 wrote:
In which of the following scenarios is p>q?

Indicate all possible scenarios.
A. \((0.11)^p>(0.11)^q\)
B. \(1^p>1^q\)
C. \((1.11)^p>(1.11)^q\)
D. \((1.01)^p>(1.01)^q\)
E. \((p+q)(p−q)>0\)
F. \(|p|>|q|\)


So where are the variables p and q - they are used as exponent or power..
some rules for the terms when the base is positive..
1) when the number, say x, is between 0 and 1 that is 0<x<1...
Higher the power, lower the value so x^3<x^2
2) when x=1
the values are always same irrespective of the power. 1^7 = 1^1
3) when x>1
Higher the power , higher the value so x^3>x^2

now let us see the choices..
A. \((0.11)^p>(0.11)^q\).....0<0.11<1 so case (1) p<q
B. \(1^p>1^q\).......... case (2).. cannot be determined
C. \((1.11)^p>(1.11)^q\).......1.11>1, so case (3).. p>q
D. \((1.01)^p>(1.01)^q\).......1.01>1, so case (3).. p>q
E. \((p+q)(p−q)>0.......p^2-q^2>0.....p^2>q^2\), we can just say |p|>|q|... say p is negative (-3)^2>2^2 but -3<2, so p<q and if both positive p>q
F. \(|p|>|q|\).... same as E above

thus only C and D

hope it helps


I think E and F are also correct because the question stem is looking for possible scenarios and not must be scenarios so it is possible that p is greater than q in both cases when p and q are both +ve terms and P>Q for ex. P=4 and Q=3
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Re: In which of the following scenarios is p>q? [#permalink]
Expert Reply
amorphous wrote:
chetan2u wrote:
Sawant91 wrote:
In which of the following scenarios is p>q?

Indicate all possible scenarios.
A. \((0.11)^p>(0.11)^q\)
B. \(1^p>1^q\)
C. \((1.11)^p>(1.11)^q\)
D. \((1.01)^p>(1.01)^q\)
E. \((p+q)(p−q)>0\)
F. \(|p|>|q|\)


So where are the variables p and q - they are used as exponent or power..
some rules for the terms when the base is positive..
1) when the number, say x, is between 0 and 1 that is 0<x<1...
Higher the power, lower the value so x^3<x^2
2) when x=1
the values are always same irrespective of the power. 1^7 = 1^1
3) when x>1
Higher the power , higher the value so x^3>x^2

now let us see the choices..
A. \((0.11)^p>(0.11)^q\).....0<0.11<1 so case (1) p<q
B. \(1^p>1^q\).......... case (2).. cannot be determined
C. \((1.11)^p>(1.11)^q\).......1.11>1, so case (3).. p>q
D. \((1.01)^p>(1.01)^q\).......1.01>1, so case (3).. p>q
E. \((p+q)(p−q)>0.......p^2-q^2>0.....p^2>q^2\), we can just say |p|>|q|... say p is negative (-3)^2>2^2 but -3<2, so p<q and if both positive p>q
F. \(|p|>|q|\).... same as E above

thus only C and D

hope it helps


I think E and F are also correct because the question stem is looking for possible scenarios and not must be scenarios so it is possible that p is greater than q in both cases when p and q are both +ve terms and P>Q for ex. P=4 and Q=3



I believe the part - indicate all possible scenarios- is to tell you that you have to pick more than one choice.
The main question asks you "is p>q?".
However, maybe a word must can be added to make it more clear.
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