Assuming that k and n are integers, st1 tells us nothing about a value. st2 gives us some info, but k can equal 10 and n equal 7 or vice versa.

st1 and 2 combined tells us that n must be the median and 10.

I'll go with C

1) insuffciient (two variables..)
we don't the pattern

2) k, n, 12, 6, 17
re arrange in the ascending order.. following two combinations are possible.
either way n is median.. so n=10
6 ,k, n, 12,17
k,6, n, 12,17


B is the answer

Yeap!! you are right..

for the second statment I combined both and answered B..

obvisiously it is C..
Oh!! God when will I stop the making care less mistakes..

median of x1,x2,x3,x4,x5
iss x3 ( when you arrange them in ascending order)

We have six numbers given.. k, n, 12, 6, 17
(to find the median arrange then in ascending order)
6,12,17 ( we don't know where k and n com in this order)
State 1) says k<n means k alsways comes before n.
series can be..
k,n,6,12,17
k,6,12,17,n
.... (you get more combinations)..

State 2) median is 10 (that means middle numbe must be 10)
__, __, 10,12,17
there are only two posibility with the above combinations
6,_,10,12,17
_,6,10,12,17
here other number_ must be k becaue k<n so 10 must be N.
assume that 10 is k then k<n is contradicting.

Hope this will help you.

How do you know that average(arithmetic mean) must also be 10?

List could be 6,7,10,12,17 (k=7, n=10, k<n) with an average(arithmetic mean) of 10.4.

Mean = median in a series.

We don't know that the numbers listed were a list of series.

Perfect example of how simple looking statistics questions can be a little tricky.

k, n, 12, 6, 17

Stmnt 1: k < n
No idea what n is. Not sufficient.

Stmnt 2: The median of the list is 10.
Since the list has 5 numbers i.e. odd number of numbers, the median must be the middle number i.e. the 3rd number when the numbers in the list are in increasing/decreasing order. Since 10 has to be a number in the list, either k or n has to be 10. But we do not know yet whether k is 10 or n is 10. So we don't know the value of n. Not sufficient.

Using both together, we know one of k and n is 10. We also know that k < n.
If k = 10, n is 10.1/11/12/13/14......etc etc etc
But then the list becomes: 6, k(10), n, 12, 17 (whatever be the arrangement of last 3 elements). k will not be the middle number in this case and hence 10 will not be the median.
If n = 10, k < 10 so the list will look something like:
6, k, n(10), 12, 17 ... Here 10 is the median. n must be 10. Hence, both together are sufficient.
_________________

Practice some questions based on Median. Tag search for some on this forum itself. Median is the most time consuming to figure out in case of missing values or new values. Sometimes range can also be fun. If you get stuck somewhere, let me know...

I intend to write a complete post on statistics concepts soon. Check out http://www.veritasprep.com/blog/ and search for 'Quarter Wit Quarter Wisdom' (That is the tag I use to put my posts.) Look for statistics post in the coming weeks.
_________________

k, n, 12, 6, 17
What is the value of n in the list above?

Given list: {k, n, 6, 12, 17}

(1) k < n. Clearly insufficient.
(2) The median of the numbers in the list is 10 --> now, the list contains odd # of terms, thus its median is the middle term and since no other term is 10 then either n or k must be 10, but we don't know which one. Not sufficient.

(1)+(2) Since from (1) k<n then k=10=median is not possible, because in this case 3 terms will be greater than k (n, 12, and 17) and it won't be the middle term (it'll be the second term), for example {6, k=10, n, 12, 17}. Thus n must be 10. Sufficient.

Answer: C.
_________________