median and range

[spoiler]17?[/spoiler]

This is really a good question.

(1) Arranging the sequence gives us: 11 15 18 28 and the unknown x & y.

Knowing that the median is 19 we can tell that we need to have 3 Numbers higher than 19, so x,y are larger than 19. Moreover, the sequence has an even number of objects and so one out of x and y should be 20 (19 is the mean of 18 and 20).
Yet we cannot say anything about the other unknown which can be less than 28 or even higher. Therefore, Insufficient.

(2) Knowing the average by itself does not provide any information to help us to find out the range - Therefore, Insufficient

(1)+(2) now we can calculate the missing known value from the mean equation and know the range. So sufficient.

From here there is no need to calculate and the answer should be C.

But I'll show the calculations: lets assume that x has value of 20 and y is left unknown.

(11+15+18+20+28+y)/6=20
y=28
so, Range = 28-11 = 17

1)Lets arrange these numbers in order.
Median is 19 and we have three numbers less 19 in the set. The fourth known number in the set is 28. If we put 28 in the fourth place then we dont get median 19. so either x or y has to be at the 4th place in the set. Lets assume its x. Value of x has to be 20 in order to get the median 19.
Now we have two possibilities for set Z
z={11,15,18,x=20,28,y} if y>28
z={11,15,18,x=20,y, 28} if y<28

Its not possible to find the range unless we know the last number. So statement 1 is not sufficient.

2)We have no information about how many members are there in the set. All we know is that median is 19 and avg is 20. It doesn't help in finding the range. So insufficient.

Now if we take both 1) and 2) together. We know the value of x and we can find out the value of y using the average.

(11+15+18+20+28+y)/6 = 20
Solve this equation and you get value of y=28

z={11,15,18,x=20,28,y=28}
So range is 28-11=17

Answer is C.