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

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k, n, 12, 6, 17 What is the value of n in the list above? [#permalink] New post 05 Sep 2007, 07:41
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A
B
C
D
E

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

(1) k < n
(2) The median of the numbers in the list is 10.
[Reveal] Spoiler: OA
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Re: DS: Value of n [#permalink] New post 05 Sep 2007, 07:52
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

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

(1) k < n
(2) The median of the numbers in the list is 10.
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 [#permalink] New post 05 Sep 2007, 08:17
yup I get C too...

in a question like this..always write out the numbers in a list of increasing or decreasing order

6, K, N, 12, 17

we know K<N then N is the only option that can be 10...
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 [#permalink] New post 05 Sep 2007, 12:06
C for sure,

if median is 10 then n cannot be higher then 10 (for example 11 is not possible), k could be any smaller number 7, 8 ...

Ans: C
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Re: DS: Value of n [#permalink] New post 05 Sep 2007, 13:48
sidbidus wrote:
k, n, 12, 6, 17
What is the value of n in the list above?

(1) k < n
(2) The median of the numbers in the list is 10.


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"...
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 07:34
chan4312 wrote:
k, n, 12, 6, 17
What is the value of n in the list above?

(1) k < n
(2) The median of the numbers in the list is 10.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


IMO C)

Combining both statements you get

k or 6 k or 6 n=10 12 17
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 07:36
chan4312 wrote:
k, n, 12, 6, 17
What is the value of n in the list above?

(1) k < n
(2) The median of the numbers in the list is 10.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


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
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 07:41
nmohindru wrote:
chan4312 wrote:
k, n, 12, 6, 17
What is the value of n in the list above?

(1) k < n
(2) The median of the numbers in the list is 10.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


IMO C)

Combining both statements you get

k or 6 k or 6 n=10 12 17


How did you solve can u explain it.
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 07:51
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
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 08:11
pmenon wrote:
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.. :twisted: :twisted:
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 08:30
x2suresh wrote:
pmenon wrote:
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.. :twisted: :twisted:


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.
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 09:02
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chan4312 wrote:
x2suresh wrote:
pmenon wrote:
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.. :twisted: :twisted:


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.
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 09:14
x2suresh wrote:
chan4312 wrote:
x2suresh wrote:
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.. :twisted: :twisted:


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.


Now..I got it.

Thanks for the explanation Suresh.!

Appreciated..
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 09:31
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
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 11:04
x97agarwal wrote:
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.
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 11:08
new2gmat wrote:
x97agarwal wrote:
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.
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 11:13
Thanks Suresh, I am kind of new on GC but this place sure seems interesting :)
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 14:09
Mean = median in a series.

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

x97agarwal wrote:
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
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Re: List : DS Question [#permalink] New post 06 Aug 2008, 16:26
chan4312 wrote:
k, n, 12, 6, 17
What is the value of n in the list above?

(1) k < n
(2) The median of the numbers in the list is 10.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


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
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Re: NEED SOME Help on this DS question [#permalink] New post 06 Dec 2010, 17:41
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ajit257 wrote:
k, n, 12, 6, 17
What is the value of n in the list above?

(1) k < n
(2) The median of the numbers in the list is 10.


I thought n have numerous value here ? Can someone explain where i am going wrong


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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Re: NEED SOME Help on this DS question   [#permalink] 06 Dec 2010, 17:41
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