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

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

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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) \(\frac{x^2}{2x}\)
[Reveal] Spoiler: OA

Last edited by Bunuel on 15 Apr 2017, 01:47, edited 3 times in total.
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Re: If -1<x<0, what is the median of these six numbers listed be [#permalink]

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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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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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New post 04 Sep 2013, 06:31
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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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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Re: If -1<x<0, what is the median of these six numbers listed be [#permalink]

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New post 04 Sep 2013, 12:18
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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Re: If -1<x<0, what is the median of these six numbers listed be [#permalink]

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New post 04 Sep 2013, 18:44
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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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Re: If -1<x<0, what is the median of these six numbers listed be [#permalink]

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New post 20 Jun 2014, 01:34
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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New post 14 Apr 2017, 13:59
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Re: If -1<x<0, what is the median of these six numbers listed be   [#permalink] 14 Apr 2017, 13:59
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