For integers a, b, c, x, y, and z, x a, b c, z b, y x, and a

Question

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

Luckisnoexcuse · Answer

y>x>a>z>b>c
(a+z)/2
C

varun4s · Answer

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!

Nikkb · Answer

Given that :

x>a
b>c
z>b
y>x
a>z

If we combine all 5 given equations we can write : y>x>a>z>b>c

Now we have 6 terms.. so median will be average of center 2 terms, ie average of 3rd and 4th term
( if no. of terms odd, median is center term, if no. of terms even median is average of center 2 terms)

our series is : y>x>a>z>b>c So center 2 terms are : a and z

Median = (a+z)/2

Answer : C

generis · Answer

x > a b > c z > b y > x a > z To find the greatest and least variables in a sequence, look for the variables that show up only one time on list above, e.g.. Here, those variables are y (the greatest) and c (the smallest). Start on one end, high or low: y > x. Now find x on left side. x > a. Find "a" on left side, etc. Reverse the process on the low end. c is the smallest number. c < b. Look on right side for b. b < z, etc. c < b < z < a < x < y For an even number of terms, the median is the average of the middle two: \frac{a + z}{2} Answer C

JeffTargetTestPrep · Answer

Using the given information, we can determine the following:

c < b < z < a < x < y

Thus, the median is (z + a)/2.

Answer: C

For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >

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