If n and k are integers and (-2) n^5 > 0, is k^37 < 0?
[1] (n k)^z > 0, where z is an integer that is not divisible by 2.
[2] k < n.
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(-2) n^5 > 0, is k^37 < 0?
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sanju09
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Since (-2) n^5 > 0, n^5 is is negative => n is negative.sanju09 wrote:If n and k are integers and (-2) n^5 > 0, is k^37 < 0?
[1] (n k)^z > 0, where z is an integer that is not divisible by 2.
[2] k < n.
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1) Z is odd.(as it is integer and not divisible by 2)
(nk)^Z >0 =>nk is positive => K is negative -- Sifficient.
2)k<n and n is negative => k is definitely negative -- Sifficient.
D.
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Night reader
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given: (-2)n^5>0 only possible when n^5>0 or n>0 is k^37<0 OR k<0 (since 37 is the power of odd number and will keep sign of the base number)sanju09 wrote:If n and k are integers and (-2) n^5 > 0, is k^37 < 0?
[1] (n k)^z > 0, where z is an integer that is not divisible by 2.
[2] k < n.
https://www.platinumgmat.com
st(1) (nk)^z>0, z is odd --> nk>0 only possible when k<0 Sufficient
st(2) k<n OR k<0 Sufficient
D
