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11 Aug 2006, 05:56
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S is a set of negative integers. T is a subset of S, and there are two fewer elements in T than in S. Is the mean of the elements in S different from that of the elements in T?
(1) The mean of the elements in S is between -4.5 and -5.0.
(2) At least three elements in S are single-digit negative integers.
IMO, A
S1. -5.0<mean<-4.5 tells that the mean can't be the same.
Set S = {...........,X1,X2}
Set T = {...............}
Let mean of set S = s and sum of set S = T1
Let mean of set T = t and sum of set t = T2
If mean of S is equal to mean of T
T1/n = (T1-X1-X2)/n-2
T1 = (X1+X2)n/2 (n is number of elements in set S)
T2 = (X1+X2)n/2-X1-X2 = (X1+x2)(n-2)/2
mean of set S : T1/n = (X1+X2)/2
mean of set T : T2/n-2 = (X1+X2)/2
since X1 and X2 are integers, (X1+X2)/2 is either integer value or .5
but S1. -5.0<mean<-4.5 thus it contradicts assumption. i.e. the mean of set S can't be the same as the mean of set T
Thus, sufficient.
I tried the above reasoning using different sets and it works.
My question is why????
why for the two sets to have the same mean,the two means are equal and are equall to the average of the two missing digits(x1+x2)/2
I need urgent help with this , is this a general rule....then when is it not Valid)
Help please
(1) The mean of the elements in S is between -4.5 and -5.0.
(2) At least three elements in S are single-digit negative integers.
IMO, A
S1. -5.0<mean<-4.5 tells that the mean can't be the same.
Set S = {...........,X1,X2}
Set T = {...............}
Let mean of set S = s and sum of set S = T1
Let mean of set T = t and sum of set t = T2
If mean of S is equal to mean of T
T1/n = (T1-X1-X2)/n-2
T1 = (X1+X2)n/2 (n is number of elements in set S)
T2 = (X1+X2)n/2-X1-X2 = (X1+x2)(n-2)/2
mean of set S : T1/n = (X1+X2)/2
mean of set T : T2/n-2 = (X1+X2)/2
since X1 and X2 are integers, (X1+X2)/2 is either integer value or .5
but S1. -5.0<mean<-4.5 thus it contradicts assumption. i.e. the mean of set S can't be the same as the mean of set T
Thus, sufficient.
I tried the above reasoning using different sets and it works.
My question is why????
why for the two sets to have the same mean,the two means are equal and are equall to the average of the two missing digits(x1+x2)/2
I need urgent help with this , is this a general rule....then when is it not Valid)
Help please
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
11 Aug 2006, 07:08
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If the mean of part of a set is equal to the mean of the whole set, then the mean of the rest of the set must be equal to the mean of the whole set as well.
If: (A+B+C+D+E)/5 = (A+B+C)/3
then, by isloating D+E you get
(D+E)/5 = (A+B+C)/3 - (A+B+C)/5
(D+E)/5 = (2/15) (A+B+C)
Therefore (D+E)/2 = (A+B+C)/3
This is a rule as long as the original if statement is satisfied, could be a subset of any number from the original set.
If: (A+B+C+D+E)/5 = (A+B+C)/3
then, by isloating D+E you get
(D+E)/5 = (A+B+C)/3 - (A+B+C)/5
(D+E)/5 = (2/15) (A+B+C)
Therefore (D+E)/2 = (A+B+C)/3
This is a rule as long as the original if statement is satisfied, could be a subset of any number from the original set.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
