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Is the median of set A bigger than the mean of set B(same

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CEO
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Is the median of set A bigger than the mean of set B(same [#permalink] New post 13 Sep 2003, 19:34
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A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Is the median of set A bigger than the mean of set B(same number
of digits)?

1) set A is consecutive even numbers and Bis consecutive odd numbers
2) the sum of A is bigger than that of B

please explain ?
Intern
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Joined: 22 May 2003
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 [#permalink] New post 14 Sep 2003, 10:38
C ?

Question stem
All that we know is that we have two sets each with equal numbers of terms. To answer the question we need either the details of the terms in each series or some way to determine the median in each.

Statement 1:

let A = 2, 4, 6 Median is 4
let B = 3, 6, 9 Median is 6 B is higher than A and satisfies

let A = 20, 22, 24 Median is 22
let B = 3, 6, 9 Median is 6 A is higher than B and satisfies

Therefore, statement A is insufficient and we should eliminate choices A & D

Statement 2:

let A = 300, 1, 50 Median is 50 Sum is 351
let B = 3, 6, 9 Median is 6, Sum is 18 A is higher than B and satisifies

let A = 300, 1, 5 Median is 5, Sum is 306
let B = 3, 6, 9 Median is 6, Sum is 18 B is higher than A and satisfies

Therefore, statement B is insufficient and we should eliminate choice B>

Together:


Given the number of terms are the same, any set with a larger sum (A) will have a higher median than the set with a lower sum

let A: 4, 6, 8 sum 18 Median 6
let B: 3, 5, 7 sum 15 Median 5 A higher than B

Just to prove that it doesn't matter weather the number of terms is odd or even

let A: 4, 6, 8, 10 sum 28 median 7
let B: 3, 5, 7, 9 sum 24 median 6.5

Therefore together they are sufficient and we should eliminate choice E
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Re: DS : Median [#permalink] New post 14 Sep 2003, 21:02
praetorian123 wrote:
Is the median of set A bigger than the mean of set B(same number
of digits)?

1) set A is consecutive even numbers and Bis consecutive odd numbers
2) the sum of A is bigger than that of B

please explain ?


I think answer is C.

Statement I - Insufficient.

case (1)Imagine A is a set of all negative even consecutive numbers. So, median would be negative. B is a set of all positive consecutive odd numbers. So, mean would be positive. So, median (A) < mean (B)
case (2) Imagine A is a set of all positive even consecutive numbers. And B is a set of all negative odd consecutive numbers. Then, median (A) > mean (B)

Statement II - Insufficient

From this statement, all we can derive is that mean (A) > mean (B) because sum of numbers in set A is greater than that of numbers in B. However, median (A) can be greater than mean (A) i.e. median (A) > mean (B). Or way less than mean (A) so that median (A) < mean (B)

Now consider both the statements together. In a set of even numbers, mean is always equal to median. And since mean (A) > mean (B) from statement II, median (A) > mean (B)
Re: DS : Median   [#permalink] 14 Sep 2003, 21:02
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