What is the median of the list of numbers above ?

Question

-(\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}} Screenshot 2024-01-01 181734.png

chetan2u · Accepted Answer

-(\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}}

KarishmaB · Answer

Here is the video solution to this question: https://youtu.be/PQMxKCld65M 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

kpop1234567890 · Answer

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  
-(4)^1/2=-2
-(16)^1/3=~-2.5  
-(3)^1/2=~-1.8  
-(4)^1/3=~-1.4  
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

sayan640 · Answer

KarishmaB   MartyMurray   gmatophobia  Can you please provide a short and quick solution for this ?

Kinshook · Answer

@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

MartyMurray · Answer

-(\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

sayan640 · Answer

Thank you  Kinshook  ..for providing the most elegant solution...very clear explanation indeed

ManishaPrepMinds · Answer

This question is testing both Statistics & Exponents .

Here we can observe that all the powers are negative .

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

Hence first convert them to positive by taking the reciprocal of the base.
\(-(2)^\frac{1}{3}\), \(-(4)^\frac{1}{2}\), \(-(4)^\frac{2}{3}\), \(-(3)^\frac{1}{2}\), \(-(4)^\frac{1}{3}\)

Simplify by converting the powers to integers, for that lets multiply all the powers by 6
\(-(2)^\frac{1*6}{3}\), \(-(4)^\frac{1*6}{2}\), \(-(4)^\frac{2*6}{3}\), \(-(3)^\frac{1*6}{2}\), \(-(4)^\frac{1*6}{3}\)

=> \(-(2)^2\), \(-(4)^3\), \(-(4)^4\), \(-(3)^3\), \(-(4)^2\)

=> -4, -64, -256, -27, -16

To find median, Arrange in order

=> -256, -64, -27, -16, -4

Median is -27, which corresponds to \(-(\frac{1}{3})^\frac{-1}{2}\)
Hence, Answer D

Watch the related video here:
https://youtu.be/sH6KjULAHlQ

rmahe11 · Answer

GMATNinja  when we are calculating median for this question, I assumed the ones with power -1/2 and there whole number were negative so thought they have to be ignored since they will be complex numbers . Hence we have to consider only 3 numbers . what is wrong in my approach ?

Bunuel · Answer

There won't be complex numbers on the GMAT, nor are they involved in this question. If we had  (-\frac{1}{4})^{-\frac{1}{2}}  and  (-\frac{1}{3})^{-\frac{1}{2}} , you'd be rigt. However, we have  -(\frac{1}{4})^{-\frac{1}{2}}  and  -(\frac{1}{3})^{-\frac{1}{2}}  instead:
 
 -(\frac{1}{4})^{-\frac{1}{2}}=-(4)^{\frac{1}{2}}=-\sqrt{4}=-2

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

lawyeroffduty · Answer

can anyone share some more similar type questions?

Bunuel · Answer

Similar questions to practce: https://gmatclub.com/forum/if-1-x-0-wha ... 29350.html https://gmatclub.com/forum/if-the-media ... 18607.html https://gmatclub.com/forum/the-arithmet ... 68731.html https://gmatclub.com/forum/in-the-list- ... 84777.html https://gmatclub.com/forum/if-the-media ... 44072.html Hope it helps.

What is the median of the list of numbers above ?

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What is the median of the list of numbers above ?

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