I know this one has been posted before by Kevincan but i read the xplanation over and over and still cant get the sense of it.

I will highly appreciate if someone can explain it to me with Just logic and not rocket science math.

Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.

Thanks

OA is B foe some weired reason.

Here you are another one that i didn't get the explanation to it too.

Please if you understands the explanation give me a ring

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.


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.

Here is the explanation that was posted ( i see it rocket science)
Those folks know things i ve never heared about in my entire life


Proof of stmt B.

difference between Y 9th term and X 9th term

1. Y9-X9 > 2

let
F(x) = x - x^2
F(x) = -(x-1/2)^2 + 1/4 (parabola, concave down)
Thus, when x = 1/2 we got maximun value of F(x) = 1/4

Assuming that sum of differences between each term of Set X and each term of Set Y, 1st through 8th term, has the maximun value
2. (1/4)*8 = 2

Y9-X9 is always greater than 2. (Y9-X9 >2 and (X1-Y1)+(X2-Y2)..+(X8-Y8) <= 2)

Sufficient.

Hey yezz,

Thanks alot. I understand your explanation. I just told myself that the increase in the whole integer would offset that average totally; however, my mistake nonetheless.

In regards to B - I guess once you know the range of the highest and lowest numbers, you can figure whether or not if it is greater.

Dahiya, I perceive you are a lady.

Thanks a million if i were to choose an instructor to guide me with GMAT , IT WOULD BE YOU .

i ve been going through many of the problems you solved.

and believe me when i tell you , you are the simplest logical analysis i have seen attacking most of the problems and if you d allow me .... we think on the same wave lenght.

Thanks a million once again.

What if 8 numbers are less than 1?

hmm..

st1 says median of X is greater than median of Y..

ur example...X median is b, which is 1.01 while the median of Y is 1.02...kind of defies wht st1 says...

Sorry if i offended you (though being a lady is a wonderfull thing to be )......We dont want ladies to make a strick or something.


Still i stick to the same openion about your logical reasoning to maths.

By the way ( great adventurous photos)