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1/x^3,1/x^2,1/x,x,x^2,x^3
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28 Sep 2020, 02:45
28 Sep 2020, 02:45
Expert Reply
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Question Stats:
41% (02:30) correct
58% (01:55) wrong based on 97 sessions
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\(\frac{1}{x^3},\frac{1}{x^2},\frac{1}{x},x,x^2,x^3\)
If \(-1<x<0\) , what is the median for the six numbers in the list above ?
A. \(\frac{1}{x}\)
B. \(x^2\)
C. \(\frac{x^2(x+1)}{2}\)
D. \(\frac{x(x^2+1)}{2}\)
E. \(\frac{x^2+1}{2x}\)
If \(-1<x<0\) , what is the median for the six numbers in the list above ?
A. \(\frac{1}{x}\)
B. \(x^2\)
C. \(\frac{x^2(x+1)}{2}\)
D. \(\frac{x(x^2+1)}{2}\)
E. \(\frac{x^2+1}{2x}\)
Re: 1/x^3,1/x^2,1/x,x,x^2,x^3
[#permalink]
28 Sep 2020, 02:45
28 Sep 2020, 02:45
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Re: 1/x^3,1/x^2,1/x,x,x^2,x^3
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28 Sep 2020, 04:11
28 Sep 2020, 04:11
2
let x value be -0.1
given set will be -1000,100,-10,-0.1,0.01,-0.001
writing them in ascending order -1000,-10,-0.1,-0.001,0.01,100
median will be (-0.1-0.001)/2 =-0.0505
lets verify options
A) (1/x)=-10 not correct
B) \(x^2\)=0.01not correct
C) \(x^2(x+1)/2\)=0.0045not correct
D)\(x(x^2 +1)/2\)=-0.0505correct
E)\((x^2+1)/2x\)=5.05 not correct
so option D is correct
given set will be -1000,100,-10,-0.1,0.01,-0.001
writing them in ascending order -1000,-10,-0.1,-0.001,0.01,100
median will be (-0.1-0.001)/2 =-0.0505
lets verify options
A) (1/x)=-10 not correct
B) \(x^2\)=0.01not correct
C) \(x^2(x+1)/2\)=0.0045not correct
D)\(x(x^2 +1)/2\)=-0.0505correct
E)\((x^2+1)/2x\)=5.05 not correct
so option D is correct
