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Set D is a new set created by combining all the terms of Set [#permalink]

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18 Aug 2013, 13:36

4

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A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

51% (01:07) correct
49% (01:19) wrong based on 132 sessions

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Set D is a new set created by combining all the terms of Sets A, B, and C. No other terms are added to Set D other than those in Sets A, B, and C. What is the median of Set D?

(1) Sets A, B, and C each have a median of 125. (2) Sets A, B, and C each have the same number of terms.

Re: Set D is a new set created by combining all the terms of... [#permalink]

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18 Aug 2013, 14:13

sophietrophy wrote:

Set D is a new set created by combining all the terms of Sets A, B, and C. No other terms are added to Set D other than those in Sets A, B, and C. What is the median of Set D?

(1) Sets A, B, and C each have a median of 125. (2) Sets A, B, and C each have the same number of terms.

I don't know the answer to this, but I think it's E.

Does anyone either know the answer or can explain to me if I am wrong?

Thank you very much.

Note: I haven't a clue as to how to tag this question but it is from the Veritas Prep: X Statistics and Problem Solving book.

I think answer is A. Try with different pairs and you will always end up with 125.
_________________

Re: Set D is a new set created by combining all the terms of... [#permalink]

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18 Aug 2013, 14:47

BangOn wrote:

sophietrophy wrote:

Set D is a new set created by combining all the terms of Sets A, B, and C. No other terms are added to Set D other than those in Sets A, B, and C. What is the median of Set D?

(1) Sets A, B, and C each have a median of 125. (2) Sets A, B, and C each have the same number of terms.

I don't know the answer to this, but I think it's E.

Does anyone either know the answer or can explain to me if I am wrong?

Thank you very much.

Note: I haven't a clue as to how to tag this question but it is from the Veritas Prep: X Statistics and Problem Solving book.

I think answer is A. Try with different pairs and you will always end up with 125.

I don't want to presume that it will hold true in every possible scenario. Is there a rule, by any chance, that will make it true?

Set D is a new set created by combining all the terms of Sets A, B, and C. No other terms are added to Set D other than those in Sets A, B, and C. What is the median of Set D?

(1) Sets A, B, and C each have a median of 125. (2) Sets A, B, and C each have the same number of terms.

i also find A to be inappropriate for instance:

A 10 20 25 28 32 B 5 10 20 (25) 30 40 50 C 10 25 40

D 5 10 20 25 28 30 32 40 50

Set D in this case is {5, 10, 10, 10, 20, 20, 25, 25, 25, 28, 30, 32, 40, 40, 60} --> median=25.
_________________

Re: Set D is a new set created by combining all the terms of... [#permalink]

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03 Mar 2014, 04:54

Bunuel wrote:

darkwraith wrote:

Set D is a new set created by combining all the terms of Sets A, B, and C. No other terms are added to Set D other than those in Sets A, B, and C. What is the median of Set D?

(1) Sets A, B, and C each have a median of 125. (2) Sets A, B, and C each have the same number of terms.

i also find A to be inappropriate for instance:

A 10 20 25 28 32 B 5 10 20 (25) 30 40 50 C 10 25 40

D 5 10 20 25 28 30 32 40 50

Set D in this case is {5, 10, 10, 10, 20, 20, 25, 25, 25, 28, 30, 32, 40, 40, 60} --> median=25.

oh now i get it .... i assumed that the numbers that overlapped were only counted once.... thanks

Here we can see that if we add all of the terms together to form one set, there will still be an equal number of terms on each side of 125, and hence 125 is still the median of the new set.

Re: Set D is a new set created by combining all the terms of Set [#permalink]

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05 Jul 2016, 01:33

Case 1 - Assume all sets have odd number of members Assume Set A - r numbers , 125 , r numbers Assume Set B - k numbers , 125 , k numbers Assume Set C - x numbers , 125 , x numbers

Set D - r + k + x numbers , 125, 125, 125, r + k + x => Median = 125

Case 2 - Even members Assume Set A - r-1 numbers , 125-a,125+a , r-1 numbers Assume Set B - k-1 numbers , 125-b,125+b , k-1 numbers Assume Set C - x-1 numbers , 125-c,125+c , x-1 numbers ( e + f = 125 *2)

Set D - r-1 + k -1 + x-1 numbers , mid numbers , r-1 + k -1 + x-1 numbers now by property that 125-a and 125 +a and others are equidistant from 125 the mid numbers will always arrange such that mid two numbers would be from one set only

median = (125 -a + 125 +a)/2 = 125 ( or same for b , c)

Case 3 - mix of even and odd sets - would play out the same

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