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Re: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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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

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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Re: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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
:-D
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Re: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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.
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k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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ajit257 wrote:
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...

Originally posted by KarishmaB on 06 Dec 2010, 22:46.
Last edited by KarishmaB on 08 Oct 2022, 02:03, edited 1 time in total.
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Re: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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.
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k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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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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Re: k, n, 12, 6, 17 What is the value of n in the list above? [#permalink]
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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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