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

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Joined: 02 Sep 2009
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For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >  [#permalink]

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30 Aug 2017, 23:21
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92% (00:57) correct 8% (01:11) wrong based on 81 sessions

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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

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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >  [#permalink]

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30 Aug 2017, 23:31
Bunuel wrote:
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

y>x>a>z>b>c
(a+z)/2
C
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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >  [#permalink]

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31 Aug 2017, 00:09
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= $$\frac{(3rd item+4th item)}{2}$$= $$\frac{(a+z)}{2}$$

Kudos please if you like my explanation!

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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >  [#permalink]

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31 Aug 2017, 03:58
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

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For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >  [#permalink]

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31 Aug 2017, 16:34
Bunuel wrote:
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

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:

$$\frac{a + z}{2}$$

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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a >  [#permalink]

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02 Sep 2017, 07:20
Bunuel wrote:
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

Using the given information, we can determine the following:

c < b < z < a < x < y

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

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Re: For integers a, b, c, x, y, and z, x > a, b > c, z > b, y > x, and a > &nbs [#permalink] 02 Sep 2017, 07:20
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