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28 Jul 2006, 23:42
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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.
(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.
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29 Jul 2006, 00:42
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kevincan
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.
S2. At least three elements in S are single-digit negative integers.
Ex1.
set S = {-1,-2,-3} set T = {-2}; -2 = -2 :
but
set S = {-1,-2,-3} set T = {-1}; -2!=-1 :
Thus insufficient.
29 Jul 2006, 03:59
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freetheking
Nice explanation.. Here is another way to solve the problem mentally..
For means of S and T to be same, mean of two integers that are missing from T is equal to mean of set T. Mean of these 2 integers is either integer or has a decimal value of .5.. So A is sufficient.
This means A or D. B itself isn't sufficient.
