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Which of the following CANNOT be the median of the 3 . . .

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Which of the following CANNOT be the median of the 3 . . .

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Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. (x+z)/2
E. (x+z)/3

The OA is C.

Experts how can I show that C can not be the median? Is there a formula?

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GMAT/MBA Expert

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Hello M7MBA.

Let's try with some examples.

The median of y, x, z and z, x, y is x. So, option A is possible.

The median of x, z, y and y, z, x is z. So, option B is possible.

If x=1, y=2 and z=3, then the median is 2. By the other hand, $$median\ 2=\frac{1+3}{2}=\frac{x+z}{2}.$$
So, option D is possible.

Finally, if we set x=3, y=4 and z=9, then the median is 4, and $$median\ 4=\frac{3+9}{3}=\frac{x+z}{3}.$$
So, option E is possible.

In conclusion, option B is NOT possible.

The correct answer is B.

I hope this may help you.

Feel free to ask me if you have a doubt.

Cheers.

_________________
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GMAT/MBA Expert

Post
M7MBA wrote:
Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. (x+z)/2
E. (x+z)/3
When the 3 positive integers are arranged in ascending order, the median will be the middle value.
Answer choice C suggests that the median will be greater than x AND greater than z
Since the median of 3 numbers cannot be greater than 2 of the numbers, the correct answer is C

Cheers,
Brent

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