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A) Insuff. It can be (-12, -12, 0 , 12, 12) or (-4, -6, -14, 1,23) ...
B) Insuff. Similar to above...

If there are equal positive and negative values, |mean| of each would be same. From A and B, it can be inferred it is not the case.
if |mean of positive| > |mean of negative|
The |value per positive| is more. Which means fewer positive numbers than negative.
So from Above, the answer to the question is 'No'.

Re: Above Average Averages Question [#permalink]
16 Oct 2005, 09:40

gsr wrote:

HIMALAYA wrote:

suppose if negative numbers = 3, positive numbers = 2 suppose if negative numbers = 4, positive numbers = 1 in both of the conditions, we cannot have a mean 0.

1. (-6,-7,-8,10,11) -> mean is 0 2. (-6,-7,-8,-10,31) -> mean is 0

that is not the solution. donot forget the constraints given in statement i and ii.

If the constraint also said "all numbers in the sequence are integers", then what you say is true (N=4,P=1 or N=3,P=2) and maybe, we can say question/answer is impossible, forget A or B or....

Any one possibility is enough. Like this one -> (-6, -6, 0, 0, 12)

If the constraint also said "all numbers in the sequence are integers", then what you say is true (N=4,P=1 or N=3,P=2) and maybe, we can say question/answer is impossible, forget A or B or....

Any one possibility is enough. Like this one -> (-6, -6, 0, 0, 12)

no, its not possible and your example is flawed because 0 is not positive and the mean of 0, 0 and 12 is also not 12..

still do not forget the constraints gievn in statements 1 and 2.

Quote:

The mean value of 5 numbers is 0. Are there more positive than negative values in this group of numbers?

1) The mean of positive numbers in the group is 12 2) The mean of negative numbers in the group is -6