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

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Intern
Joined: 03 Aug 2013
Posts: 15
If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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Updated on: 15 Apr 2017, 02:47
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55% (hard)

Question Stats:

70% (02:41) correct 30% (02:41) wrong based on 138 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) $$\frac{x^2}{2x}$$

Originally posted by arakban99 on 04 Sep 2013, 07:09.
Last edited by Bunuel on 15 Apr 2017, 02:47, edited 3 times in total.
RENAMED THE TOPIC.
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Posts: 611
Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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04 Sep 2013, 07:23
1
arakban99 wrote:
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) x^2/2x

Anyone got a GOOD approach

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Comment and KUDOS are appreciated

Hmm. The no of terms is 6. Hence, the median would look like this : $$\frac{a+b}{2}$$. Eliminate all the options except C and D.
Option C = $$\frac{x^3+x^2}{2}$$. Also, we have only 2 terms which are positive, i.e. $$x^2$$ and $$\frac{1}{x^2}$$.
Thus, the median CANNOT contain any positive term.Only option left is D, where both the terms are negative.

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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04 Sep 2013, 07:31
1
My approach: Put any value between -1 and 0. you will get the answer. lets say 0.5. so the list will be as follows:

-8,-2,-0.5,-0.125 and last two numbers will be positive so need to see that.

median will be between -0.5 and -0.125.
So, (x^3+x)/2 (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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04 Sep 2013, 07:56
2
1
arakban99 wrote:
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) x^2/2x

My approach:
Since the absolute value is less than 1, so any inversion of any value will yield much bigger value. Also, square terms are going to be positive so they are going to be the biggest whereas 1/x^3 will be the smallest.
Considering this approach,
the set in ascending order is as follows:
1/x^3, 1/x, x, x3, x^2, 1/x^2

The middle values are x and x^3.

Hence the median will be the mean of these two.
+1 D
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Joined: 03 Aug 2013
Posts: 15
Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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04 Sep 2013, 13:18
mau5 wrote:
arakban99 wrote:
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) x^2/2x

Anyone got a GOOD approach

________________
Comment and KUDOS are appreciated

Hmm. The no of terms is 6. Hence, the median would look like this : $$\frac{a+b}{2}$$. Eliminate all the options except C and D.
Option C = $$\frac{x^3+x^2}{2}$$. Also, we have only 2 terms which are positive, i.e. $$x^2$$ and $$\frac{1}{x^2}$$.
Thus, the median CANNOT contain any positive term.Only option left is D, where both the terms are negative.

D.

How did you deduce that the median cannot be a positive term?
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Joined: 10 Oct 2012
Posts: 611
Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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04 Sep 2013, 19:44
1
arakban99 wrote:
mau5 wrote:
arakban99 wrote:
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) x^2/2x

Hmm. The no of terms is 6. Hence, the median would look like this : $$\frac{a+b}{2}$$. Eliminate all the options except C and D.
Option C = $$\frac{x^3+x^2}{2}$$. Also, we have only 2 terms which are positive, i.e. $$x^2$$ and $$\frac{1}{x^2}$$.
Thus, the median CANNOT contain any positive term.Only option left is D, where both the terms are negative.
D.

How did you deduce that the median cannot be a positive term?

Out of the given 6 terms, we established that only 2 are positive. Thus, in increasing order, the set will look like this :

(-ve term),(-ve term),(-ve term),(-ve term),(+ve term),(+ve term)

Thus, only the middle terms contribute to the median, and both of them are negative.

Hope this helps.
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Joined: 17 Jul 2013
Posts: 76
Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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20 Jun 2014, 02:34
Marcab wrote:
arakban99 wrote:
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) x^2/2x

My approach:
Since the absolute value is less than 1, so any inversion of any value will yield much bigger value. Also, square terms are going to be positive so they are going to be the biggest whereas 1/x^3 will be the smallest.
Considering this approach,
the set in ascending order is as follows:
1/x^3, 1/x, x, x3, x^2, 1/x^2

The middle values are x and x^3.

Hence the median will be the mean of these two.
+1 D

perfect approach ... saved a lots of time for me
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Re: If -1<x<0, what is the median of these six numbers listed be  [#permalink]

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03 Oct 2018, 04:53
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Re: If -1<x<0, what is the median of these six numbers listed be   [#permalink] 03 Oct 2018, 04:53
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