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Bunuel
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Find the median of the following numbers: 10 Dec 2021, 02:30
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95% (hard)

Question Stats:

46% (02:34) correct 54% (02:36) wrong based on 192 sessions

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Find the median of the following numbers:

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

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

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

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

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

E. \(-(\frac{1}{3})^{(\frac{-1}{2})}\)
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E
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KarishmaB
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Find the median of the following numbers: 15 Jan 2022, 00:53
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Bunuel
Find the median of the following numbers:

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

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

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

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

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

E. \(-(\frac{1}{3})^{(\frac{-1}{2})}\)
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Median of 5 numbers is the third value, whether we arrange them in ascending order or descending.
Since each value is negative, let's ignore negatives for all. Think about it - The middle value for 2, 3, 4 will be the same as middle value for -2, -3, -4.

\((\frac{1}{2})^{(\frac{-1}{3})} = 2^{\frac{1}{3}}\)
\((\frac{1}{4})^{(\frac{-1}{3})} = 4^{\frac{1}{3}}\)
\((\frac{1}{4})^{(\frac{-1}{2})} = 4^{\frac{1}{2}}\)
\((\frac{1}{4})^{(\frac{-2}{3})} = 4^{\frac{2}{3}} = 16^{\frac{1}{3}}\)
\((\frac{1}{3})^{(\frac{-1}{2})} = 3^{\frac{1}{2}}\)

Now raise each to the power 6 to get 4, 16, 64, 256, 27 respectively for the 5 values.

The middle value is 27 i.e. \(-(\frac{1}{3})^{(\frac{-1}{2})}\)

Answer (E)­
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Find the median of the following numbers: 19 Dec 2021, 06:19
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Let us say numbers are a, b, c, d and e.

a^6 = -(1/2)^-2
b^6 = -(1/4)^-2
c^6 = -(1/4)^-3
d^6 = -(1/4)^-4
e^6 = -(1/3)^-3

So,
a^6 = -4
b^6 = -16
c^6 = -64
d^6 = -256
e^6 = -27

So, e^6 is the median.

So, E

Is this the correct approach?
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Find the median of the following numbers: 08 Aug 2024, 09:45
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mpobisetty
Let us say numbers are a, b, c, d and e.

a^6 = -(1/2)^-2
b^6 = -(1/4)^-2
c^6 = -(1/4)^-3
d^6 = -(1/4)^-4
e^6 = -(1/3)^-3

So,
a^6 = -4
b^6 = -16
c^6 = -64
d^6 = -256
e^6 = -27

So, e^6 is the median.

So, E

Is this the correct approach?
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­i used the same thought process to get the solution
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Find the median of the following numbers: 14 Sep 2025, 05:52
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Find the median of the following numbers:

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

To find median, let us raise these numbers to power 6

\(-(\frac{1}{2})^{(\frac{-6}{3})} = -(\frac{1}{2})^{(-2)} = -2^2 = -4 \) ;
\(-(\frac{1}{4})^{(\frac{-6}{3})} = -(\frac{1}{4})^{(-2)} = -4^2 = -16 \);
\(-(\frac{1}{4})^{(\frac{-6}{2})} = -(\frac{1}{4})^{(-3)} = -4^3 = -64\);
\(-(\frac{1}{4})^{(\frac{-12}{3})} = -(\frac{1}{4})^{(-4} = -4^4 = -256\);
\(-(\frac{1}{3})^{(\frac{-6}{2})} = -(\frac{1}{3})^{(-3)} = -3^3 = - 27\)

Arranging the numbers in increasing order
-256, -64, -27, -16, -4

Median = -27; Original median = [m]-(\frac{1}{3})^{(\frac{-1}{2})}

IMO E
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