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10 Dec 2021, 02:30
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E
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Difficulty:
95%
(hard)
Question Stats:
46% (02:34) correct
54%
(02:36)
wrong
based on 192
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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. \(-(\frac{1}{3})^{(\frac{-1}{2})}\)
\(-(\frac{1}{2})^{(\frac{-1}{3})}\); \(-(\frac{1}{4})^{(\frac{-1}{3})}\); \(-(\frac{1}{4})^{(\frac{-1}{2})}\); \(-(\frac{1}{4})^{(\frac{-2}{3})}\); \(-(\frac{1}{3})^{(\frac{-1}{2})}\)
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. \(-(\frac{1}{3})^{(\frac{-1}{2})}\)
15 Jan 2022, 00:53
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Bunuel
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)
General Discussion
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?
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?
