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# If 0 < x < 1, what is the median of the values x, x^-1, x^2,

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Senior Manager
Joined: 22 Sep 2005
Posts: 273
If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink]

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14 Nov 2006, 13:41
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If 0 < x < 1, what is the median of the values x, x^-1, x^2, x^1/2 and x^3?

A. x
B. x^-1
C. x^2
D. x^1/2
E. x^3
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Sep 2013, 14:16, edited 1 time in total.
Edited the question and added the OA
Manager
Joined: 29 Aug 2006
Posts: 156

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14 Nov 2006, 13:46
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Rearranging the values for 0<x<1 in ascending order,

x^3,x^2, x,x^1/2,x^-1

Median will be x.
Manager
Joined: 10 Jul 2006
Posts: 71

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14 Nov 2006, 20:28
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A for me too. To make the problem less confusing, I picked a number between 0 and 1 such as x = 1/4 and calculate:
x = 1/4
x^-1 = 4
x^2 = 1/16
x^(1/2) = 1/2
x^3 = 1/8.

From this, rearrange the number and the median is 1/4, which is x. Answer A
Senior Manager
Joined: 24 Oct 2006
Posts: 337

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14 Nov 2006, 20:35
enola wrote:
A for me too. To make the problem less confusing, I picked a number between 0 and 1 such as x = 1/4 and calculate:
x = 1/4
x^-1 = 4
x^2 = 1/16
x^(1/2) = 1/2
x^3 = 1/8.

From this, rearrange the number and the median is 1/4, which is x. Answer A

Yup. I thought picking numbers would be easier and picked 0.5. Though I got the answer (a), I think fractions are easier than decimals.
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Joined: 07 Jul 2004
Posts: 5021
Location: Singapore

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14 Nov 2006, 22:25
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x is a positive fraction. Pick any number, say 1/4

then

x = 1/4
x^-1 = 1/x = 4
x^2 = 1/16
x^1/2 = 1/2
x^3 = 1/64

Rearranging in ascending order:

x^3, x^2, x, x^1/2, x^-1

Median = x

Ans A
Manager
Joined: 13 Dec 2009
Posts: 244
Re: Help required in find the median of a given set [#permalink]

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26 Jun 2010, 04:19
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gautamsubrahmanyam wrote:
If 0<x<1, what is the median of the values x,(x^-1),(x^2),(x^1/2),and (x^3)?

1) x
2) x^-1
3) x^2
4) x^1/2
5) x^3

Easy way - pick a number for x. Let x = $$\frac{1}{4}$$ (Taking $$\frac{1}{4}$$because we need to do a square root here).
1) $$x = \frac{1}{4}$$
2) $$x^-1 = \frac{1}{x} = 4$$
3) $$x^2 = \frac{1}{16}$$
4) $$\sqrt{x} = \frac{1}{2}$$
5) $$x^3 = \frac{1}{64}$$
From the above 5 choices you can see that choices 2 and 4 are greater than x and choices 3 and 5 are less than x. Median must be the middle term when arranged in ascending order. x itself is the median as it is the middle value in ascending order
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Intern
Joined: 11 Aug 2013
Posts: 5
Re: If 0<x<1, what is the median of the values x. x^-1, [#permalink]

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02 Sep 2013, 13:43
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netcaesar wrote:
If 0<x<1, what is the median of the values x. x^-1, x^2, x^1/2 and x^3?

A) x

B) x^-1

C) x^2

D) x^1/2

E) x^3

In info given by stem, basically we have a fraction.
So fraction value is increasing when it is under the root and decreasing when it is in power.
x^3, x^2, x, x^1/2, x^-1.
The mean is the middle number, hence x.
QA is A
Manager
Joined: 24 Oct 2013
Posts: 153
Location: India
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WE: Information Technology (Computer Software)
Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink]

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24 Sep 2016, 21:28
for x between 0 and 1, we alwasy have a standard rule

X^3 < X^2 < X < root(X) < 1/x

following the above rule , when the above given terms can be arranged in the order and the mean is X
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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink]

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02 Nov 2017, 16:57
netcaesar wrote:
If 0 < x < 1, what is the median of the values x, x^-1, x^2, x^1/2 and x^3?

A. x
B. x^-1
C. x^2
D. x^1/2
E. x^3

To find the median, we can let x be any value (fraction or decimal) between 0 and 1. So, let’s let x = ¼. Therefore:

x^-1 = (1/4)^-1 = 1/(1/4) = 4

x^2 = (1/4)^2 = 1/16

√x = √(1/4) = 1/2

x^3 = (1/4)^2 = 1/64

From the smallest to the largest, the order is:

x^3 = 1/64, x^2 = 1/16, x = 1/4, √x = 1/2, x^-1 = 4

Therefore, the median is x.

Note: We used ¼ as the value of x since it’s easy to take the square root of ¼, but any value between 0 and 1 will behave similarly, and x will always be the median.

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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2,   [#permalink] 02 Nov 2017, 16:57
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