Set D is a new set created by combining all the terms of Set

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

i also find A to be inappropriate
for instance:

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

D 5 10 20 25 28 30 32 40 50 (odd)

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

I just kind of imagined this in my head and it worked out:



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.

Therefore option 1) is sufficient.

Answer: A

1. median is the same for every set. when arranged, the median of all 3 will be 125. A sufficient.
2. alone not sufficient.

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

Other cases -

­Answer cant be A.
For example
A=[124 126]
B=[1 125 3]
C=[3 125 200]
Then D=[1 3 3 124 125 125 126 200] and the median=124.5
 

­you have not included the repetitions of 10,20 and 40. Add those and you will get 25 as median. No instructions were given not to use repetitive numbers in the set

I hate to spoil the party 10 years later, but darkwraith is right and (1) is insufficient. By definition, a set is composed of unique elements, and thus cannot contain repeats. A *list* can have as many repeats as it wants. For instance, if we talk about the set of all primes, we can't say it contains "2,2,2, . . . 3,3,3, etc." Each unique prime is in the set; it wouldn't mean anything to have a repeated element.

From here, we can determine the answer to be E. Once we realize that Set D will not have repeats of any elements that recur across sets, we can manipulate this to weaken one side and move the median. For instance, let's say we're using both statements and testing 3 sets with 4 elements each, with a median of 125 each time. To get 125 out of the median, we can use a recurring element less than 125. I've chosen 120:

Set A: 110, 120, 130, 140 (Median = 125)
Set B: 120, 124, 126, 133 (Median = 125)
Set C: 120, 123, 127, 190 (Median = 125)

As a check on accuracy, notice that 120 occurs 3 times, but will still only show up once in Set D. That means that Set D will have only 10 elements, rather than 12. Its median will be the average of the middle two terms. Now let's build it:
Set D: 110, 120, 123, 124, 126, 127, 130, 133, 140, 190 (Median = 126.5)

The set got weighted toward the high end, since the repeated use of 120 left us with fewer elements below 125 than above it. You can construct many other versions of Set D using the same method for A/B/C. You can also push the median below 125 by going the other way (for instance, by making the last term of each set 140). And of course, we could make each set odd with an actual 125 in the middle, and then the median of D would definitely be 125. So 1&2 together are still Insufficient.