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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:

Re: If x > y > z, which cannot be the average (arithmetic mean) of x, y an [#permalink]

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04 May 2017, 09:04

1

This post received KUDOS

Here is what i did on this question -->

Using the definition of arithmetic mean -> Mean is a number that can be used to replace each element of the data set. If mean is X --> Number can be written as X,X,X But the other two numbers are less than X.

Re: If x > y > z, which cannot be the average (arithmetic mean) of x, y an [#permalink]

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04 May 2017, 23:59

GMATPrepNow wrote:

Bunuel 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 + 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:

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
_________________

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

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
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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
_________________

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