GMAT CLUB OLYMPICS: If x 0, is x

Question

If x ≠ 0, is x < 1 ?

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

(2)  -\frac{1}{x^2}  is the median of  -\frac{1}{x^2} ,  -x  and  x^5 .

M36-16

Bunuel · Accepted Answer

Official Solution:

If  x ≠ 0 , is  x < 1  ? 
 
(1)  \frac{1}{x}  is the median of  \frac{1}{x} ,  -x^2  and  -x^3  
 
If  x  were a positive number, then  \frac{1}{x}  would be positive, but both  -x^2  and  -x^3  would be negative. This case is not possible because  \frac{1}{x}  would be the largest number in the list and thus cannot be the median. Therefore,  x  must be negative. Sufficient.
 
(2)  -\frac{1}{x^2}  is the median of  -\frac{1}{x^2} ,  -x  and  x^5  
 
We can have the following cases:
 
A.  -x \leq -\frac{1}{x^2} \leq x^5 . Multiply all three parts by  x^2  (we can safely do that since  x^2  will be positive):  -x^3 \leq -1 \leq x^7 . From  -x^3 \leq -1  we get  1 \leq x^3  and finally  1 \leq x .
 
B.  x^5 \leq -\frac{1}{x^2} \leq -x . Multiply all three parts by  x^2  (we can safely do that since  x^2  will be positive):  x^7 \leq -1 \leq -x^3 . From  -1 \leq -x^3  we get  x^3 \leq 1  and finally  x \leq 1 .
 
Two different answers. Not sufficient.

Answer: A

SChatrg · Answer

S1: assuming x to be +ve, no way a +ve number can be the median along with 2 other -ve numbers. Thus, x has to be < 0 . Suff.

S2: 
take x as 2, we have -1/4 is the median of -1/4, -2, 32 - true.
take x as -2, we have -1/4 is the median of -1/4, 2, -32 - also true.
hence S2 is insuff.

Option A

JonShukhrat · Answer

I tried to explain everything in detail in the attached file. 
My answe is A

chetan2u · Answer

x ≠ 0 (1) \frac{1}{x} is the median of \frac{1}{x} , -x^2 and -x^3 . -x^2 is always negative, while \frac{1}{x} and -x^3 will have opposite sign. When x>0, \frac{1}{x}>0 , while -x^2 and -x^3 will be negative. This case is not possible. Hence x<0. Sufficient (2) -\frac{1}{x^2} is the median of -\frac{1}{x^2} , -x and x^5 . When x>0, -\frac{1}{x^2}<0 , while x^5>0 and -x<0 . Always true for x>1. When x<0, -\frac{1}{x^2}<0 , while x^5<0 and -x>0 . Always true for x<-1. Hence, Insufficient A

chetan2u · Answer

x ≠ 0 (1) \frac{1}{x} is the median of \frac{1}{x} , -x^2 and -x^3 . When x>0, \frac{1}{x}>0 , while -x^2 and -x^3 will be negative. This case is not possible. Hence x<0. Sufficient (2) -\frac{1}{x^2} is the median of -\frac{1}{x^2} , -x and x^5 . -\frac{1}{x^2} is always negative, while -x and -x^5 will have opposite sign. Thus, the signs of three in increasing order would be (-, -, +), which is possible both when x>0 and when x<0. Insufficient A

GMAT CLUB OLYMPICS: If x 0, is x < 1 ?

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