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Bunuel
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If X is the biggest number then X can not be the mean. If you need more evidence, plug in numbers to see.
The answer is A
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answer shud be in between x and z...

only A fits in ,,,

ans A
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Bunuel
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 + y)/2

Nice!

The average (mean) is meant to provide a rough idea of what the numbers look like.
So, since x is the greatest value, it cannot be the average.
Answer:

Dear Brent,

Although it is clear for me the right answer, I wonder about how choice E would be a mean for 3 numbers in the prompt?

Thanks
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Hey chetan2u
There is no doubt in my mind that A is correct option.


But I cannot fathom why E isn't the option too.

Here is what I feel.

For E to be correct =>
x+y+z/3 = y+z/2

Hence y+z=2x

But since y<x and z<x

This can never happen.

Hence E will is correct too.

Hence A and E are both correct.


What am I missing ?


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Stone
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stonecold
Hey chetan2u
There is no doubt in my mind that A is correct option.


But I cannot fathom why E isn't the option too.

Here is what I feel.

For E to be correct =>
x+y+z/3 = y+z/2

Hence y+z=2x

But since y<x and z<x

This can never happen.

Hence E will is correct too.

Hence A and E are both correct.


What am I missing ?


Best
Stone

Hi stone,

I totally agree with you. Your approach is also absolutely correct..
Other wise...
(z+y)/2 is the average of z and y..
So ONLY way that even after adding a new number, the means remain the same is when the new number is SAME as the average...
But in this case x becomes less than y..
Not possible so E is the answer..

So the choice E could be
1) (x+z)/2 or
2) z + y/2

Bunuel pl relook in the Choice E. There is some Typo
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stonecold
Hey chetan2u
There is no doubt in my mind that A is correct option.


But I cannot fathom why E isn't the option too.

Here is what I feel.

For E to be correct =>
x+y+z/3 = y+z/2

Hence y+z=2x

But since y<x and z<x

This can never happen.

Hence E will is correct too.

Hence A and E are both correct.


What am I missing ?


Best
Stone

Hi stone,

I totally agree with you. Your approach is also absolutely correct..
Other wise...
(z+y)/2 is the average of z and y..
So ONLY way that even after adding a new number, the means remain the same is when the new number is SAME as the average...
But in this case x becomes less than y..
Not possible so E is the answer..

So the choice E could be
1) (x+z)/2 or
2) z + y/2

Bunuel pl relook in the Choice E. There is some Typo
______________
Edited. Thank you.
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Bunuel
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

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 numbers is always less than the largest number in the set).

Answer: A
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Mo2men

Dear Brent,

Although it is clear for me the right answer, I wonder about how choice E would be a mean for 3 numbers in the prompt?

Thanks

E) If x = 3, y = 2 and z = 1, then the average of x, y, and z is 2, which is the same as (z+x)/2.
Since the average CAN equal (z+x)/2, we can ELIMINATE E
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Bunuel
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:
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