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ds - mean of 2 sets

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ds - mean of 2 sets [#permalink] New post 25 Oct 2005, 10:45
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A
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C
D
E

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If the two sets have an equal number of
numbers, is the mean of set Q lower than the
mean of set P?

(1) Set Q consists of consecutive even integers
and set P of consecutive odd integers.
(2) The median of Q is higher than the mean of P.
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 [#permalink] New post 25 Oct 2005, 11:06
Premise, sets have equal number of terms.

(1) example consecutives
Q 2, 4, 6 and P 3, 5, 7 mean Q<P
Q 4, 6, 8 and P 3, 5, 7 mean Q>P
insuff.

(2) med Q > mean P
Q 1, 9, 10 and P 2, 4, 6 mean Q>P
Q -5, 5, 6 and P 3, 4, 5 mean Q<p
insuff.

(1)&(2)
from (1) we know that the sets are consecutive and therefore the mean = median!
from (2) we know that the median of Q is larger than mean of P and therefore the mean of Q is larger then the mean of P. suff.
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 [#permalink] New post 25 Oct 2005, 11:33
I am getting C.

Given both sets have equals #items.
Q: Qmean (Qm) < Pmean (Pm) ?

1) Take sets P=consec even int's, Q= consec odd int's
P{-4,-2,0,2} and Q{1,3,5,7} result in Qm>Pm
However, P{2,4,6} and Q{-5,-3,-1} result in Qm<Pm
NOT SUFF => BCE

2) says Qmed>Pm
P{1,2,3} and Q{1,3,5} satisfy the condition Qmed>Pm (3>2) and result in Qm>Pm (3>2)
P{-1,0,1} and Q{-3,1,1} satisfy the condition Qmed>Pm (1>0) and result in Qm<Pm (-1/3<0)
NOT SUFF => CE

1+2)
Median and Mean of consecutive integers (odd or even) is the same
E.g. {1,3,5,7} med=4, mean=4
So if Median of a set Q of consec int's is > than mean of another set P(doesn't matter what it's content) then it follows that Mean of such set Q must be > than mean of P.
Effectively saying Qm>Pm when Qmed>Pm given Q is set of consec int's
Some examples: P{-4,-2,0}&Q{1,3,5}; P{-4,-2,0,2}&Q{-3,-1,1,3}
SUFF => C

I realized this Q was a time killer!
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Re: ds - mean of 2 sets [#permalink] New post 26 Oct 2005, 00:57
christoph wrote:
If the two sets have an equal number of
numbers, is the mean of set Q lower than the
mean of set P?

(1) Set Q consists of consecutive even integers
and set P of consecutive odd integers.
(2) The median of Q is higher than the mean of P.


Q = {2,4,6} mean = 4, and P = {301,303,305} mean = 303
=> Mean of Q < Mean of p
But if q={302,304,306} mean = 304, and p = {1,3,5} mean = 3, then mean of Q > Mean of p.

So statement 1 is not suff.

Statement 2. Let Q = {1,2,3} median = 2 and mean = 2
Let P = {3,1,2} medain =1 and mean = 2 , statement 2 is not suff.

Combining both. P = Set of consecutive odd int.
Q = Set of consecutive even integers.
No of terms in p and q are same.
Median of p is greater than median of Q
If the no of terms are same, and the middle value of p is greater than q and p being consecutive odd and q being consecutive even, then each corresponding elements of p is greater than its corresponding element in Q. So the mean of p is also greater than mean of Q.Hence C.
Re: ds - mean of 2 sets   [#permalink] 26 Oct 2005, 00:57
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