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For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a > z. Which of the following is the median of the six integers?
A. z−a
B. (x+y)/2
C. (a+z)/2
D. z
E. (c−y)/2
A. z−a
B. (x+y)/2
C. (a+z)/2
D. z
E. (c−y)/2
Luckisnoexcuse
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30 Aug 2017, 23:31
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Bunuel
For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a > z. Which of the following is the median of the six integers?
A. z−a
B. (x+y)/2
C. (a+z)/2
D. z
E. (c−y)/2
A. z−a
B. (x+y)/2
C. (a+z)/2
D. z
E. (c−y)/2
y>x>a>z>b>c
(a+z)/2
C
31 Aug 2017, 00:09
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x>a and a>z
Thus, x>a>z
x>a>z and y>x
Thus, y>x>a>z
y>x>a>z and z>b
Thus, y>x>a>z>b
y>x>a>z>b and b>c
Thus, y>x>a>z>b>c
Since there are 6 items in the series
Median= \(\frac{(3rd item+4th item)}{2}\)= \(\frac{(a+z)}{2}\)
Answer: C.
Kudos please if you like my explanation!
Thus, x>a>z
x>a>z and y>x
Thus, y>x>a>z
y>x>a>z and z>b
Thus, y>x>a>z>b
y>x>a>z>b and b>c
Thus, y>x>a>z>b>c
Since there are 6 items in the series
Median= \(\frac{(3rd item+4th item)}{2}\)= \(\frac{(a+z)}{2}\)
Answer: C.
Kudos please if you like my explanation!
