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Statistics- can't be median question

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by vishalwin » Sat Nov 28, 2015 11:43 am
q1) Which of the following cannot be the median of the three positive integers x,y and z?

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

Q2) Which of the following cannot be the median of the three ordered positive integers x,y and z?

A) x
B) z
C) x+z
D) x+z/2
E) x+z/3
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vishalwin
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Source: — Problem Solving |

by DavidG@VeritasPrep » Sat Nov 28, 2015 12:36 pm
vishalwin wrote:q1) Which of the following cannot be the median of the three positive integers x,y and z?

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

Q2) Which of the following cannot be the median of the three ordered positive integers x,y and z?

A) x
B) z
C) x+z
D) x+z/2
E) x+z/3
Use a bit of logic. Clearly, any of the three variables could be the median, so A and B are out. Now consider C.

If z were the median then, x + z = z. Solving, we get x = 0. But the variables must all be positive, so this is out.
If x were the median, then, x + z = x. Solving, we get z = 0. Same problem
If y were the median, then x + z = y. But this makes no sense. If y is equal to the sum of the other two numbers, and those numbers are positive, then y must be the largest of the set, and therefore, not the median.

Because no scenario can satisfy the constraints of this problem, the answer is C.
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