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

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by EconomistGMATTutor » Tue Nov 14, 2017 7:24 am
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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by Brent@GMATPrepNow » Tue Nov 14, 2017 7:28 am
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

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by Scott@TargetTestPrep » Sun Oct 20, 2019 9:12 am
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

The OA is C.

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

The median of three positive integers must be one of the integers. We see that the quantities in choices A, B, and D can be the median of the integers; since choices A and B are one of the integers, and the quantity in choice D (which is always between x and z) could equal the third integer (which is y). Thus they can't be the correct answer and it narrows down to either choice C or E. Let's analyze them.

If x + z is the median of the three integers, then neither x nor z can be the median since x + z is greater than either of them. This leaves us that the only possible candidate for the median is y. Therefore, y = x + z. However, since y = x + z, y is greater than both of x and z. Thus y is greatest integer and it can't be the median. Therefore, we see that it's impossible for x + z to be the median (and we will leave the readers as an exercise to show that (x + z)/3 can be also the median).

Answer: C

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