A certain list consists of 3 different numbers. Does the median of the

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Hi Bunuel !

Thanks a lot for the explanation, it's concise and to the point - and these help a lot!

Could I ask one thing, though? In the second statement, you mention that the sum cannot be three times the smallest or largest number.
Why not? Is there a rule for series/sets that forbids this, or is this some logical deduction?

All the best,

Gil

We have that x < y < z.

The sum = x + y + z.

Obviously 3x < x + y + z < 3z because x < y < z (x + x + < x + y + z < z + z + z).

Does this make sense?
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Oh wow when you put it like that it does. Thanks a lot!

Hi BrentGMATPrepNow, question asked Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers? So it means are these numbers evenly spaced set?
Does this mean we should only assing even numbers here e.g 8 ,10 ,12 if number testing? Thought assign 1,2,3 work too but is this still correct though? Thanks Brent

Since we know a list consists of three different numbers, then we can reward the target question as "Are the three numbers equally spaced?"
Don't forget that, when testing numbers for each statement, the values must satisfy the information in each statement.

Noted thanks BrentGMATPrepNow

Could you please go through my reasoning for the 2nd statement and let me know if it is correct?
Let the 3 nos be a,b,c where a<b<c
Say 3 times the mean = 3 times smallest no. Thus,
a+b+c=3a
Thus b+c=2a
This cannot be possible until b and c are less than or equal to a. But the qs states that these 3 numbers are distinct. Hence not possible. Same would be the case if the sum is 3 times the greatest number. a+b=2c... not possible becasue of the above mentioned reason

We have that x < y < z.

The sum = x + y + z.

Obviously 3x < x + y + z < 3z because x < y < z (x + x + < x + y + z < z + z + z).

Does this make sense?[/quote]

Oh wow when you put it like that it does. Thanks a lot![/quote]

Hi, I tried some derivation- may be helpful thus posting it here:
For the 2nd statement, I tried solving and deriving- 3nos: a,b,c where a<b<c
Say 3 times the mean = 3 times smallest no. Thus
a+b+c=3a
Thus b+c=2a
This cannot be possible until b and c are less than or equal to a. But the qs states that these 3 numbers are distinct. Hence not possible. Same would be the case if the sum is 3 times the greatest number. a+b=2c... not possible because of the above mentioned reason

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