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What is the median of the list of numbers above ?

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gmatophobia
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What is the median of the list of numbers above ? [#permalink] New post   01 Jan 2024, 05:55
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\(-(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{1}{3}}\)

What is the median of the list of numbers above?

A. \(-(\frac{1}{2})^{-\frac{1}{3}}\)

B. \(-(\frac{1}{4})^{-\frac{1}{2}}\)

C. \(-(\frac{1}{4})^{-\frac{2}{3}}\)

D. \(-(\frac{1}{3})^{-\frac{1}{2}}\)

E. \(-(\frac{1}{4})^{-\frac{1}{3}}\)

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D
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chetan2u
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Re: What is the median of the list of numbers above ? [#permalink] New post   01 Jan 2024, 11:11
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gmatophobia wrote:
\(-(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{1}{3}}\)

What is the median of the list of numbers above?

A. \(-(\frac{1}{2})^{-\frac{1}{3}}\)

B. \(-(\frac{1}{4})^{-\frac{1}{2}}\)

C. \(-(\frac{1}{4})^{-\frac{2}{3}}\)

D. \(-(\frac{1}{3})^{-\frac{1}{2}}\)

E. \(-(\frac{1}{4})^{-\frac{1}{3}}\)

Show Spoiler
Attachment:
Screenshot 2024-01-01 181734.png


\(-(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{1}{3}}\)


let us convert the negative power to positive power
\(-(2)^{\frac{1}{3}}, \quad -(4)^{\frac{1}{2}}, \quad -(4)^{\frac{2}{3}},\quad -(3)^{\frac{1}{2}}, \quad -(4)^{\frac{1}{3}}\)

Now let us raise all to same power 1/6
\(-(2^2)^{\frac{1}{6}}, \quad -(4^3)^{\frac{1}{6}}, \quad -(4^4)^{\frac{1}{6}},\quad -(3^3)^{\frac{1}{6}}, \quad -(4^2)^{\frac{1}{6}}\)


\(-(4)^{\frac{1}{6}}, \quad -(64)^{\frac{1}{6}}, \quad -(256)^{\frac{1}{6}},\quad -(27)^{\frac{1}{6}}, \quad -(16)^{\frac{1}{6}}\)

Now all raised to same power. So, if we discard '-' sign for this step, the increasing order will be
\((256)^{\frac{1}{6}}> \quad (64)^{\frac{1}{6}}> \quad (27)^{\frac{1}{6}}>\quad (16)^{\frac{1}{6}}> \quad (4)^{\frac{1}{6}}\)


All though the median will remain the same, let us change the sign by multiplying by - sign
\(-(256)^{\frac{1}{6}}< \quad -(64)^{\frac{1}{6}}< \quad -(27)^{\frac{1}{6}}<\quad -(16)^{\frac{1}{6}}< \quad -(4)^{\frac{1}{6}}\)

Answer: D. \(-(\frac{1}{3})^{-\frac{1}{2}}\)
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What is the median of the list of numbers above ? [#permalink] New post   06 Jan 2024, 04:09
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The key here is to not do messy work. Make a table - just draw column basically.
Step 1: First column - original nos,
Step 2: second column convert nos to positive powers for eg -(1/4)^-1/2 is -(4)^1/2
Step 3: Third column is using estimation to convert the nos with decimals. Dont forget to add negative sign.
-(2)^1/3 = ~-1.1 [ The first perfect cube is 1 and next is 8 so 2 will be a little higher than 1]
-(4)^1/2=-2
-(16)^1/3=~-2.5 [ The second perfect cube is 8 and next is 27 so 16 will be a almost mid way]
-(3)^1/2=~-1.8 [ The first perfect square is 1 and next is 4 so 3 will be a little higher than 1]
-(4)^1/3=~-1.4 [ The first perfect cube is 1 and next is 8 so 4 will be a little higher than 1]
Step 4: Arrange in ascending order - remember we are dealing with negative nos - biggest negative is smallest!
-2.5,-2,-1.8,-1.4,-1.1
Step 5: Identify the corresponding answer i.e -1.8 in in column 1
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Re: What is the median of the list of numbers above ? [#permalink] New post   22 Apr 2024, 09:08
KarishmaB MartyMurray gmatophobia Can you please provide a short and quick solution for this ?
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Re: What is the median of the list of numbers above ? [#permalink] New post   22 Apr 2024, 10:06
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@gmatophobia\(-(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{1}{3}}\)

What is the median of the list of numbers above?

\(-(2^{-1})^{-\frac{1}{3}}, \quad -(4^{-1})^{-\frac{1}{2}}, \quad -(4^{-1})^{-\frac{2}{3}},\quad -(3^{-1})^{-\frac{1}{2}}, \quad -(4^{-1})^{-\frac{1}{3}}\)

\(-2^{\frac{1}{3}}, \quad -4^{\frac{1}{2}}, \quad -4^{\frac{2}{3}},\quad -3^{\frac{1}{2}}, \quad -4^{\frac{1}{3}}\)

\(-2^{\frac{1}{3}}, -2,  -4^{\frac{2}{3}}, -\sqrt{3},  -4^{\frac{1}{3}}\)

Cubing each number

\(-2, -8,  -16, -3\sqrt{3}=-5.19,  -4\)

Arranging them in decreasing order 

-2, -4, -5.19, -8, -16

Median = \( -(\frac{1}{3})^{-\frac{1}{2}}​\)

IMO D
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Re: What is the median of the list of numbers above ? [#permalink] New post   22 Apr 2024, 11:56
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sayan640 wrote:
KarishmaB MartyMurray gmatophobia Can you please provide a short and quick solution for this ?

\(-(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{1}{3}}\)

What is the median of the list of numbers above?

Since there are five numbers and they are all negative, we can ignore the negative signs because the median will be the third number either way:

\((\frac{1}{2})^{-\frac{1}{3}}, \quad (\frac{1}{4})^{-\frac{1}{2}}, \quad (\frac{1}{4})^{-\frac{2}{3}},\quad (\frac{1}{3})^{-\frac{1}{2}}, \quad (\frac{1}{4})^{-\frac{1}{3}}\)

Then, we can convert all the numbers to positive powers:

\((2)^{\frac{1}{3}}, \quad (4)^{\frac{1}{2}}, \quad (4)^{\frac{2}{3}},\quad (3)^{\frac{1}{2}}, \quad (4)^{\frac{1}{3}}\)

Then, begin ordering them. The following four are easy:

\((2)^{\frac{1}{3}} < (4)^{\frac{1}{3}} < (4)^{\frac{1}{2}} < (4)^{\frac{2}{3}}\)

The following is also clear:

\((2)^{\frac{1}{3}} < (3)^{\frac{1}{2}} < (4)^{\frac{1}{2}}\)

The only question that remains is that of whether \((3)^{\frac{1}{2}}\) is greater than or less than \((4)^{\frac{1}{3}}\).

Since \((3)^{\frac{1}{2}}\) is approximately \(1.7\) and \(1.7^3 = 1.7 × 3 > 4\), we know that \((4)^{\frac{1}{3}} < (3)^{\frac{1}{2}}\).

So, \((2)^{\frac{1}{3}} < (4)^{\frac{1}{3}} < (3)^{\frac{1}{2}} < (4)^{\frac{1}{2}} < (4)^{\frac{2}{3}}\)

A. \(-(\frac{1}{2})^{-\frac{1}{3}}\)

B. \(-(\frac{1}{4})^{-\frac{1}{2}}\)

C. \(-(\frac{1}{4})^{-\frac{2}{3}}\)

D. \(-(\frac{1}{3})^{-\frac{1}{2}}\)

E. \(-(\frac{1}{4})^{-\frac{1}{3}}\)

Correct answer: D­
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Re: What is the median of the list of numbers above ? [#permalink] New post   22 Apr 2024, 14:56
Thank you Kinshook ..for providing the most elegant solution...very clear explanation indeed
Kinshook wrote:
@gmatophobia\(-(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{1}{3}}\)

What is the median of the list of numbers above?

\(-(2^{-1})^{-\frac{1}{3}}, \quad -(4^{-1})^{-\frac{1}{2}}, \quad -(4^{-1})^{-\frac{2}{3}},\quad -(3^{-1})^{-\frac{1}{2}}, \quad -(4^{-1})^{-\frac{1}{3}}\)

\(-2^{\frac{1}{3}}, \quad -4^{\frac{1}{2}}, \quad -4^{\frac{2}{3}},\quad -3^{\frac{1}{2}}, \quad -4^{\frac{1}{3}}\)

\(-2^{\frac{1}{3}}, -2,  -4^{\frac{2}{3}}, -\sqrt{3},  -4^{\frac{1}{3}}\)

Cubing each number

\(-2, -8,  -16, -3\sqrt{3}=-5.19,  -4\)

Arranging them in decreasing order 

-2, -4, -5.19, -8, -16

Median = \( -(\frac{1}{3})^{-\frac{1}{2}}​\)

IMO D

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Re: What is the median of the list of numbers above ? [#permalink] New post   23 Apr 2024, 00:10
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gmatophobia wrote:
\(-(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{1}{3}}\)

What is the median of the list of numbers above?

A. \(-(\frac{1}{2})^{-\frac{1}{3}}\)

B. \(-(\frac{1}{4})^{-\frac{1}{2}}\)

C. \(-(\frac{1}{4})^{-\frac{2}{3}}\)

D. \(-(\frac{1}{3})^{-\frac{1}{2}}\)

E. \(-(\frac{1}{4})^{-\frac{1}{3}}\)

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Screenshot 2024-01-01 181734.png

­I note that each term has a minus in front of it but it is not included in the exponent. So I will ignore all minus signs since we have 5 numbers so median will be the middle number. It will not change whether we put the numbers in increasing or decreasing order. The middle term will be the one whose magnitude is in the middle.
Next I will remove minus signs from the exponents by flipping the bases. I get:

\(2^{\frac{1}{3}} , 4^{\frac{1}{2}} , 4^{\frac{2}{3}} , 3^{\frac{1}{2}} , 4^{\frac{1}{3}} \)

Now, I can simply raise them all to the 6th power. Their relative positioning will stay the same since they all are positive numbers greater than 1

\(2^2, 4^3, 4^4, 3^3, 4^2\)

=\(4, 64, 256, 27, 16\)

Median is 27 which is actually \(-(\frac{1}{3})^{-\frac{1}{2}}\)

 ­Answer (D)

Check discussions on exponents here:
Exponents: https://youtu.be/ibDqnatAMG8
Exponents On number line: https://youtu.be/0rpppnnJNRs
 
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Re: What is the median of the list of numbers above ? [#permalink]
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