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?
Statistics and set problems
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Eliminate the four answer choices that CAN be the average of x, y and z.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
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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Another approach is to apply some number sense.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
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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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).BTGmoderatorRO wrote: ↑Sat Sep 30, 2017 7:50 pmIf 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?
Answer: A
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