Is \(d\) greater than or equal to 0?

1) \(d\) is the median of \(d, \frac{1}{d}, -d\)
2) \(d^3\) is the median of \(d, d^2, d^3\)

Not sure if i got it right...

In II), other than negative fractions like -1/2, d could also be -1, 0 or 1 that will make d^3 a median of d, d^3, d^2. So, II) is insuff and B is not answer.

I) is insuff as d could be both -1 or 1 making d a median of d^2, d, d^3.


And because -1 and 1 satisfy both I) and II), combining both statements is still insuff to say whether d>=0, so answer is E.

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Is d greater than or equal to 0?

1) d is the median of d,1/d,−d
Therefore either a) 1/d<=d<=-d or b) -d<=d<=1/d

a) from d<=-d we get d<=0 and from 1/d<=d we get d not equal 0 (as 1/d then will be undefined) and d>=-1 (as if d-in denominator is higher in magnitude with negative sign the fraction will be (-1,0) while d will be smaller). we write it as (-1,0].
b) from -d<=d we get d>0 and from d<=1/d we get d<=1 but not 0. we write it as [0,1)

From a and be we get d is (-1,0] or [0,1), Hence, from I alone we cant say definitely that d>=0

2) d^3 is the median of d,d^2,d^3
Therefore, Either a) d<=d^3<=d^2 or b) d^2<=d^3<=d
a) from d<=d^3 we get d-d^3<=0 or d(1-d^2)<=0. Note d^2 is >=0, therefore either d is (0,-1) (d will be -ve and 1-d^2 will be positive) or d>1 (d will be positive and 1-d^2 will be negative). from d^3<=d^2 we get d^2(d-1)<=0, therefore, d<1. From both parts combined we get d is (0,-1).

b) from d^2<=d^3 we get d^2(1-d)<=0, therefore d>=1 or 0. from d^3<=d we get d(d^2-1)<=0. Either d<=-1 or d (0,1). From combining b all parts we get d=0,1.

From a and b we get d=0,1 or (0,-1). Therefore II is insufficient.

I and II Combined, we can say d is either (-1,0] or 0 or 1. Therefore Insufficient. Answer is E.

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