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Updated on: 15 Jul 2020, 08:06
Originally posted by QuantMadeEasy on 15 Jul 2020, 05:12.
Last edited by QuantMadeEasy on 15 Jul 2020, 08:06, edited 1 time in total.
Last edited by QuantMadeEasy on 15 Jul 2020, 08:06, edited 1 time in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (02:16) correct
42%
(02:09)
wrong
based on 215
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1/x^5, 1/x^4, 1/x^3, 0, x^2, x^3
If –1 < x < 0, what is the median of the six numbers in the list above?
A. \(1/x^5\)
B. \(1/x^4\)
C. \(1/2x^3\)
D. \(x^2/2\)
E. \( x^3/2\)
If –1 < x < 0, what is the median of the six numbers in the list above?
A. \(1/x^5\)
B. \(1/x^4\)
C. \(1/2x^3\)
D. \(x^2/2\)
E. \( x^3/2\)
Updated on: 15 Jul 2020, 11:00
Originally posted by CrackverbalGMAT on 15 Jul 2020, 05:51.
Last edited by CrackverbalGMAT on 15 Jul 2020, 11:00, edited 1 time in total.
Last edited by CrackverbalGMAT on 15 Jul 2020, 11:00, edited 1 time in total.
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Since x is a value between -1 and 0, therefore x is a negative value
Let x = -0.5
\(\frac{1 }{ x^5}\) = -32
\(\frac{1 }{ x^4}\) = 16
\(\frac{1 }{ x^3}\) = -8
\(x^2\) = 0.25
\(x^3\) = -0.125
Therefore in ascending order, the terms would be \(\frac{1 }{ x^5}\), \(\frac{1 }{ x^3}\), \(x^3\), 0, \(x^2\), \(\frac{1 }{ x^4}\)
The median would be the average of the middle 2 terms (3rd and 4th terms) which are 0 and \(x^3\) = \(\frac{0 + x^3 }{ 2} \)= \(\frac{x^3}{2}\\
\)
Option E
Arun Kumar
_________________
Key Notes
If the number of observations n is odd, then the median is the \(\frac{n + 1 }{ 2}\) th term
If the number of observations n is even, then the median is the average of the \(\frac{n}{2}\) th and the next terms.
Let x = -0.5
\(\frac{1 }{ x^5}\) = -32
\(\frac{1 }{ x^4}\) = 16
\(\frac{1 }{ x^3}\) = -8
\(x^2\) = 0.25
\(x^3\) = -0.125
Therefore in ascending order, the terms would be \(\frac{1 }{ x^5}\), \(\frac{1 }{ x^3}\), \(x^3\), 0, \(x^2\), \(\frac{1 }{ x^4}\)
The median would be the average of the middle 2 terms (3rd and 4th terms) which are 0 and \(x^3\) = \(\frac{0 + x^3 }{ 2} \)= \(\frac{x^3}{2}\\
\)
Option E
Arun Kumar
_________________
Key Notes
If the number of observations n is odd, then the median is the \(\frac{n + 1 }{ 2}\) th term
If the number of observations n is even, then the median is the average of the \(\frac{n}{2}\) th and the next terms.
15 Jul 2020, 06:02
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QuantMadeEasy
Here, two terms, x^2 and 1/x^4, are positive. One is zero. Three are negative. Of the negative terms, only x^3 is between -1 and 0, so it is the largest of the negative values in the set. So x^3 and 0 are the two "middle terms" in our set of six terms, and their average is the median of the set. So the median is x^3/2. That is not one of the answer choices, so there is something wrong with the question.
CrackVerbalGMAT
I don't follow how you've concluded that x^3 and x^3/2 are equal to each other.
