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

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Joined: 22 Sep 2005
Posts: 262
If 0 < x < 1, what is the median of the values x, x^-1, x^2,  [#permalink]

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Updated on: 24 Apr 2018, 04:19
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35% (medium)

Question Stats:

66% (00:55) correct 34% (00:51) wrong based on 788 sessions

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If 0 < x < 1, what is the median of the values $$x$$, $$x^{-1}$$, $$x^2$$, $$\sqrt{x}$$ and $$x^3$$?

A. $$x$$

B. $$x^{-1}$$

C. $$x^2$$

D. $$\sqrt{x}$$

E. $$x^3$$

Originally posted by netcaesar on 14 Nov 2006, 12:41.
Last edited by Bunuel on 24 Apr 2018, 04:19, edited 2 times in total.
Edited the question and added the OA
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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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26 Jun 2010, 03:19
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1
4
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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Joined: 29 Aug 2006
Posts: 155
Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2,  [#permalink]

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14 Nov 2006, 12:46
1
Rearranging the values for 0<x<1 in ascending order,

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

Median will be 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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14 Nov 2006, 19:28
4
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
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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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14 Nov 2006, 19:35
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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Posts: 4804
Location: Singapore
Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2,  [#permalink]

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14 Nov 2006, 21:25
2
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
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Joined: 11 Aug 2013
Posts: 5
Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2,  [#permalink]

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02 Sep 2013, 12:43
1
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
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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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24 Sep 2016, 20:28
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, 15:57
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]

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06 Jan 2019, 01:42
Important takeaway - (especially important for DS questions)

<--------I---------I---------I---------->
..........-1..........0..........1

Between 0 to 1 -
As the exponent increases, the value decreases.

Example : $$\frac{1}{2}$$
$$(\frac{1}{2})^2 = 1/4$$
$$(\frac{1}{2})^3 = 1/8$$

Between -1 to 0 -
As the exponent increases, the value increases

Example : $$\frac{-1}{2}$$
$$(\frac{-1}{2})^2 = 1/4$$
$$(\frac{-1}{2})^3 = -1/8$$ -------------- $$(\frac{-1}{8}> \frac{-1}{2})$$
Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, &nbs [#permalink] 06 Jan 2019, 01:42
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