If -1

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Competition Mode Question

If  -1<x<0 , what is the median of these six numbers listed below

\frac{1}{x^3} ,  \frac{1}{x^2} ,  \frac{1}{x} ,  x ,  x^2 ,  x^3

(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}{2x}

eakabuah · Answer

We are to find the median of the the following six numbers, 1/x^3 , 1/x^2, 1/x, x, x^2, and x^3, where -1<x<0
Let x=-1/2, Then 1/x^3 = -8
1/x^2 = 4
1/x = -2
x=-1/2
x^2 = 1/4
x^3 = -1/8
Arranging in ascending order yields: 1/x^3, 1/x, x, x^3, x^2, and 1/x^2

The median is therefore average of the third and fourth terms in the list.
=1/2(x+x^3) = x(1+x^2)/2.

The answer is therefore D.

Abhinav kumar · Answer

take x= 0.5 (any number between -1 and 0)
then we will get 1/(x^3) <1/x<x<x^3<x^2<1/x^2
therefore median is (x+x^3)/2
i.e d

nick1816 · Answer

\frac{1}{x^2} >0 and x^2>0 \frac{1}{x^3}<-1 and \frac{1}{x} <-1 -1

unraveled · Answer

If −1 < x < 0, what is the median of these six numbers listed below
 \frac{1}{x^3}  ,  \frac{1}{x^2} ,  \frac{1}{x} ,  x, x^2, x^3 
(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}{2x}

Since there are six numbers then the median would be the average of the middle numbers i.e. 3rd and 4th here. To find the 3rd and 4th numbers first the list has to be arranged in an ascending order.
As x is a fraction, numbers that are fractions would be integers and others would be fractions. Thus leaving  \frac{1}{x^3} ,  \frac{1}{x^2} ,  \frac{1}{x}  aside since they would be at extremes, only  x ,  x^2 ,  x^3  are left. Among the three leaving aside the biggest i.e.  x^2 , the median can be calculated as 
 \frac{(x + x^3)}{2}  =  \frac{x (x^2 + 1)}{2}

It can be verified by taking any value of x. Let  x = \frac{-1}{2}  →  x^3  =  \frac{-1}{8}  then, list becomes -8, -2,  \frac{-1}{2} ,  \frac{-1}{8} ,  \frac{1}{4} , 4 and median = ( \frac{-1}{2}  +  \frac{-1}{8} )/2 =  \frac{-5}{16} 
Also,  \frac{x (x^2 + 1)}{2}  =  \frac{-5}{16}

Answer (D).

Raxit85 · Answer

Let's assume, x = -1/2,
So, the six numbers in ascending sequence of (1/x^3, 1/x^2, 1/x, x, x^2, x^3) will be 1/x^3,1/x,x ,x^3 ,x^2:1/x^2: -8,-2,-0.5,-0.125, 0.25,4.
Median = 3rd + 4th number/2
 = -0.625/2 = -0.3125
A) and B) options are incorrect. Let's check it for other options by placing the value of x as -1/2. 
Option C) gives 1/16 (0.625) which is incorrect.
Option D) matches the above median value (-5/16). So, answer is D.

exc4libur · Answer

x=negative.proper.fraction=-0.5 
 1/x^3, 1/x^2, 1/x, x, x^2, x^3 
 1/(-0.5)^3, 1(-0.5)^2, 1/(-0.5), (-0.5), (-0.5)^2, (-0.5)^3 
 1/(-0.125), 1/(0.25), 1/(-0.5), (-0.5), (0.25), (-0.125) 
 1/(-0.125), 1/(-0.5), -0.5, -0.125, 0.25, 1/(0.25) 
 median=(-0.5+(-0.125))/2=(x+x^3)/2=x(x^2+1)/2

Answer (D)

chetan2u · Answer

Now x is a fraction between 0 and -1..
 First way 
 A quick look at the choice will give you the answer..
Six numbers out of which terms with EVEN power will be positive and rest 4 negative,  so Median will be average of two negative numbers ..
A) Median is average of two different numbers,  so median will NOT be a part of these 7 numbers . Eliminate the choices that contain any of these 7 numbers.. A and B out
B)  Choice will NOT be positive .. x^2(x+1)  will be positive.. Eliminate C
C) E turns out to be  \frac{x}{2}  or  \frac{1}{2}  of x, but  it is supposed to be half of two different numbers of the set ..Eliminate

D Is the answer

Second way 
Properties of negative fractions>-1.... 
 For median, we look for two center values out of 6 numbers 
even power will be the  greatest numbers , so  x^2  and  \frac{1}{x^2}  are out..
Since x is fraction and negative,  \frac{1}{x}  and  \frac{1}{x^3}   will be smallest ..
 Remaining two numbers   x^3  and x..
Average of these = \frac{(x^3+x)}{2}. ...
D

Third way 
Take value as -1/2 and check..  May be better to take fractions than to take decimals

bumpbot · Answer

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If -1<x<0, what is the median of these six numbers listed be

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