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

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netcaesar
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If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   Updated on: 24 Apr 2018, 05:19
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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. \(x^3\)
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A

Originally posted by netcaesar on 14 Nov 2006, 13:41.
Last edited by Bunuel on 24 Apr 2018, 05:19, edited 2 times in total.
Edited the question and added the OA
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sidhu4u
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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   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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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   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
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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   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.
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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   14 Nov 2006, 20:35
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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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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   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
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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   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.
Just arrange answers according rules.
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] New post   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] New post   02 Nov 2017, 16:57
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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


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.

Answer: A
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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   06 Jan 2019, 02:42
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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 : \(\frac{-1}{2}\)
\((\frac{-1}{2})^2 = 1/4\)
\((\frac{-1}{2})^3 = -1/8\) -------------- \((\frac{-1}{8}> \frac{-1}{2})\)
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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   17 Oct 2020, 18:59
If 0 < x < 1, what is the median of the values x, x^(-1), x^2, √x and x^3?
If 0 < x < 1, then x^(-1) will be greater than 1, so the largest number among our choices. √x will be greater than x but smaller than 1. x^2 and x^3 will be smaller than x. ----> x^3<x^2<x<√x<x^(-1) So x is the median. A.
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If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   08 Nov 2020, 10:02
Exponent rules are heavily tested on the GMAT... here's quick review of some of them, followed by an explanation of the correct answer on YouTube:

https://www.youtube.com/watch?v=95OHY24 ... uBg2fwokrI
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Re: If 0 < x < 1, what is the median of the values x, x^-1, x^2, [#permalink] New post   03 Feb 2024, 14:51
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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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