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Statistics and set problems

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by BTGmoderatorRO » Sat Sep 30, 2017 7:50 pm
If x > y > z, which cannot be the average (arithmetic mean) of x, y and z?
A. x
B. y
C. x – 1
D. z + 1
E. (z + x)/2
a. Is there a strategic approach to this question?
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by GMATGuruNY » Sun Oct 01, 2017 2:42 am
Roland2rule wrote:If x > y > z, which cannot be the average (arithmetic mean) of x, y and z?
A. x
B. y
C. x – 1
D. z + 1
E. (z + x)/2
Eliminate the four answer choices that CAN be the average of x, y and z.

Let x=3, y=2 and z=1.
Since the values are evenly spaced, the average = the median = 2.
If we plug x=3, y=2 and z=1 into the answer choices, options B, C, D and E are all equal to 2.
Since each of these answer choices CAN be the average of x, y and z, eliminate B, C, D and E.

The correct answer is A.
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by Brent@GMATPrepNow » Sun Oct 01, 2017 6:54 am
If x > y > z, which cannot be the average (arithmetic mean) of x, y and z?

A. x
B. y
C. x – 1
D. z + 1
E. (z + x)/2
Another approach is to apply some number sense.

Since y and z are smaller than x, we can say....
y = a number smaller than x
z = another number smaller than x

So.....
The average of x, y and z = the average of x, a number smaller than x, and another number smaller than x
= (x + a number smaller than x + another number smaller than x)/3
= (a number that's LESS THAN 3x)/3
= a number LESS THAN x

Since the average must be LESS THAN x, the average cannot equal x

Answer: A

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Re: Statistics and set problems

by Scott@TargetTestPrep » Wed Jan 29, 2020 5:13 am
BTGmoderatorRO wrote: ↑
Sat Sep 30, 2017 7:50 pm
 If x > y > z, which cannot be the average (arithmetic mean) of x, y and z?
A. x
B. y
C. x – 1
D. z + 1
E. (z + x)/2
a. Is there a strategic approach to this question?
We are given three different-valued quantities x, y, z, with x having the largest value. Thus, the average of the three quantities can’t be x (the average of any set of distinct numbers is always less than the largest number in the set).

Answer: A

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