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.
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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

Vote for C

1 insuff
2 insuff

both, let`s start with 2nd statement, it says median is 10, that means 12 and 17 must come after 10, thus three numbers must be 10, 12 and 17. {median by deffinition is the middle number, when the number of numbers is odd, 5 in this case}, so we have: (some number), 6, 10, 12,17
"some number" can be either n or k, "some number" must be less than 10. Since the 1st statement says that k<n, n is 10, and k is "some number"...

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

C for me.

It cant be B since you can have 6,k,n,12,17 or 6,n,k,12,17. Considering both statements together you effectively rule out the second case since k<n. Therefore n=median=10

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..

OA is C.
I still can not understand how did you assess the value of k to find the value of n.
did you use your cognitive abilities to assess values..or do we have a formula to do it.

How did you find that n is 10.

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.

Guys good explanantion. There is one more way to do this problem

S1. As mentioned it is a MAYBE case AD are out

S2. Median is 10. Since there are five numbers in this series the average must also be 10.
{6+12+17+(n+k)}/5 = 10
n+k = 15

Combine: Now we know that wither n or k is 10, but since k<n
n = 10

IMO C

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.

good point..

Its median=10 not the mean=10...

So this approach is not correct.

Mean = median in a series.

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

lets arrange the numbers in increasing order :
6 12 17 with k,n occurring anywhere

hence consider
(1) k < n --------> it does not help in finding n
(2) median of numbers is 10 => k,n can occur anywhere hence either k or n can be 10
(1) & (2) cannot help since k <n does not say whether k=10 or n=10

INSUFFI
IMO E

Karishma,
Thanks a tonne. Awesome. I wasted a quite some time on this question. Could you please give some pointer as to how to deal with type of question in future.

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.
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wow very tricky. I selected the trap answer E by quickly coming to the conclusion that I couldn't tell even with both statements whether k or n was the median so would not N's value.

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.
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Answer: Option C

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