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

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Math Expert
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If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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29 Sep 2019, 21:01
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70% (03:08) correct 30% (02:27) wrong based on 37 sessions

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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) $$\frac{x^2}{2x}$$

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Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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29 Sep 2019, 21:31
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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.

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Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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29 Sep 2019, 22:36
1
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
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Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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29 Sep 2019, 23:13
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$$\frac{1}{x^2} >0$$ and $$x^2>0$$
$$\frac{1}{x^3}<-1$$ and $$\frac{1}{x} <-1$$
$$-1<x<0$$ and $$-1<x^3<0$$

Median= $$\frac{x+x^3}{2}$$
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Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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29 Sep 2019, 23:14
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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, $$\frac{x (x^2 + 1)}{2}$$ = $$\frac{-5}{16}$$

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Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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30 Sep 2019, 02:18
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.
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Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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30 Sep 2019, 05:33
Quote:
If $$−1<x<0$$, what is the median of these six numbers listed below

$$1/x^3, 1/x^2, 1/x, x, x^2, x^3$$

(A) $$1/x$$
(B) $$x^2$$
(C) $$x^2(x+1)/2$$
(D) $$x(x^2+1)/2$$
(E) $$x2/2x$$

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

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Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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30 Sep 2019, 07:09
Quote:
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) $$\frac{x^2}{2x}$$

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

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
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Re: If -1<x<0, what is the median of these six numbers listed be   [#permalink] 30 Sep 2019, 07:09
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