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Median

by gmatmachoman » Thu Mar 25, 2010 9:27 pm
Set S consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set S?
A) 1
B) 9
C) 14
D) 17
E) It Cannot Be Determined

This is a 50/51 level problem
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by rockeyb » Thu Mar 25, 2010 10:03 pm
gmatmachoman wrote:Set S consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set S?
A) 1
B) 9
C) 14
D) 17
E) It Cannot Be Determined

This is a 50/51 level problem
n = integers and set S consists of all integers .

Lets arrange the set in ascending order .

S= { -27,-3,-2 , 2, 3, 15, 17 ,n }

Since n is a variable n could be +ve or -ve integer.

lets assume n > 17

So the set contains 8 elements thus median = average of middle 2 in this case 2, 3 .

But (2+3)/2= 2.5

Which is not integer . Hence n can not be greater than 17 in fact n can not be greater than 2.

Now lets assume n = -ve .

Lets say n= -4


S= { -27,-4,-3,-2 , 2, 3, 15, 17 }

Again we see that n could not be less than -2 as (-2+2)/2 = undefined .

Lets say n= -1

Then median =- 1.5

And if n = 1 then also median = 1.5

So only possible values of n are = 0 , 2, -2

If n= 0 median = 1

If n= 2 median = 2

If n= -2 median = -2


So we can not find the value of n

Hence E It Cannot Be Determined.
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by rockeyb » Thu Mar 25, 2010 10:13 pm
An alternate method :

S= { -27,-3,-2 , 2, 3, 15, 17 ,n }

We know the set contains 8 elements .

The median will be always the average of numbers in 4th and 5th place .

Now if we closely look at the numbers the value of n can not be greater than 2 OR less than -2 .

If the number n is greater than 2 OR the number n is less than -2 in either of the cases the median will not be an integer .

Why ? Since integers alternate between odd and even numbers

if n < -2 n will be odd . So the number in number in 4th position will always be odd .

Odd +/- even (-2) = odd .

So average of 2 numbers will not be an integer .

So in order for median to be an integer sum of 4th and 5th place numbers should be even .

This is only possible if both 4th and 5th place numbers are even OR both odd .

But since we have -2 and 2 in 5 th position always the number n has to be even .

But the possible values are 2, -2, and 0.

Hence can not be certain what n is .
Last edited by rockeyb on Thu Mar 25, 2010 10:15 pm, edited 1 time in total.
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by eaakbari » Thu Mar 25, 2010 10:14 pm
When sorted in ascending order without including n for a preliminary view

-27, -3, -2, 2, 3,15, 17

Looking at the choices it evidently cannot be 17, 14 or 9 because the median of this set (excluding n is 2) and it cannot be greater than 3 or less than -2 (neighboring values).
Hence the answer must be 1 and we find that value of n = 0 satisfies this condition.

Hence OA= A

Please give oa
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by pops » Thu Mar 25, 2010 10:22 pm
My approach would be little different:

First sort the 7 given numbers:
S= { -27,-3,-2 , 2, 3, 15, 17 }

now since for median we have to take 2 middle values, one would be 2 for sure.
next, go to answer options:
median 1: for this n has to be 0 so that mean of 0 and 2 is 1. and it rightly fits between -2 and 2.
median 9: not possible. as if n=16 (to get mean of 2 and n=9), we would take 2 and 3, mean of which is 2.5 and not 9
similarly for 14 and 17.
Hence, median 1 is possible. Since, question asks, which 'could' be the median my answer is A.
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by fibbonnaci » Thu Mar 25, 2010 10:24 pm
arranging the integers in ascending order:
-27, -3, -2, 2,3,15,17

now we need to define the value of n. since the statement tells that all the numbers are unique, n cannot take the value -2,-3,2 and 3.
lets say n lies between 3 and 15.
then the median is (2+3)/2 which is not an integer. values above this need not be considered as median will remain 5/2.
checking for values between -27 and -3 show that median is (-2+2)/2 which is zero. zero is an integer and needs to be considered.
when n= 0, 0+2/2 = 1 is the median.
Therefore we have 2 cases. when n < -3 we have median as 0 and when n= 0 we have median as 1.
Thus it is not possible to determine the median. E
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by rockeyb » Thu Mar 25, 2010 10:32 pm
OOPSS!!!!

Sorry guys I missed the UNIQUE part of the question .


fibbonnaci is correct as always .

Thanks fibo .
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by fibbonnaci » Thu Mar 25, 2010 10:42 pm
hey rockey but ur answer matches mine though :)
and one more thing in ur post: 0/2 is not undefined. 0/2 will lead to 0 as the answer. any number when divided by zero leads to undefined.

lets see what gmatmachoman has as OA for us.
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by rockeyb » Thu Mar 25, 2010 11:01 pm
fibbonnaci wrote:hey rockey but ur answer matches mine though :)
and one more thing in ur post: 0/2 is not undefined. 0/2 will lead to 0 as the answer. any number when divided by zero leads to undefined.

lets see what gmatmachoman has as OA for us.
Yup I guess so but the reasoning is not correct .

ye ok 0/2 = 0 . Point noted.

Whats the OA machoman ?
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by Phirozz » Fri Mar 26, 2010 9:10 pm
fibbonnaci wrote:arranging the integers in ascending order:
-27, -3, -2, 2,3,15,17

now we need to define the value of n. since the statement tells that all the numbers are unique, n cannot take the value -2,-3,2 and 3.
lets say n lies between 3 and 15.
then the median is (2+3)/2 which is not an integer. values above this need not be considered as median will remain 5/2.
checking for values between -27 and -3 show that median is (-2+2)/2 which is zero. zero is an integer and needs to be considered.
when n= 0, 0+2/2 = 1 is the median.
Therefore we have 2 cases. when n < -3 we have median as 0 and when n= 0 we have median as 1.
Thus it is not possible to determine the median. E
@fibo

u solved it correctly but answer is A and not E.
Question is which of the following could be. So 1 can be the answer which is option A.
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by sanju09 » Sat Mar 27, 2010 6:43 am
gmatmachoman wrote:Set S consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set S?
A) 1
B) 9
C) 14
D) 17
E) It Cannot Be Determined

This is a 50/51 level problem
It's a bit ambiguous to read choice E as It Cannot Be Determined. I bet it is a carelessly translated DS. But again it's a good question, anyone can fall into the could be/must be trap. If you find a could be sort of problems, always give the eyecatching options a chance to prove its fitness. [spoiler]1[/spoiler] fits and so is hit!
The mind is everything. What you think you become. -Lord Buddha



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by Fiver » Sat Mar 27, 2010 9:27 am
Agree with [spoiler]A
If n=0 then median = 1[/spoiler]
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by indiantiger » Sat Mar 27, 2010 12:03 pm
Given to us :
S={-27,-3,-2,2,3,15,17,n} (in sorted order and lets for a second assume n is the largest)
S= has all unique integers
total elements in the set = 8
median for this set = ( V5+V4 )/2
V4 is the fourth value and V5 is the fifth.

My approach,
I picked the every answer choice from a-d

Case of a(1), says median can be 1
then
v4+v5 = 2
the smallest middle values from the given set could be 2,3,n
so
2+n =2 , possible for n =0

lets put in the set and the set S is {-27,-3,-2,0,2,3,15,17}

you can consider the case of n+3 = 2, n = -1

then set becomes s ={-27,-3,-2,-1,2,3,15,17}
does not work

Case b (9)
S={-27,-3,-2,2,3,15,17,n}
v4+v5 = 18
smallest possible middle values from the set 2,3 (there can be other cases , I am considering these for simplicity) from the given set as it is

in any case of the smallest value the values of n (16 or 15) dont give the median = 9
{-27,-3,-2,2,3,15,15,17}
{-27,-3,-2,2,3,15,16,17}

The same can be done for other choices c, d.

Please do feel free to correct me if I am doing anything wrong.
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by newton9 » Sat Mar 27, 2010 6:50 pm
I think the answer is A.

- 27, -3, -2, 2 , 3 , 15 , 17

If n = 0 the Median is 1 So 1 can be Median.
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by sandy_online » Sat Mar 27, 2010 9:12 pm
i think here better to work with answer choices ( as language is could be.. and could be type questions contain a lot variation in answers, so better to work with answer choices )

arrange set numbers in increasing order.
keeping in mind n to be unique integer, we just need to see the combination of n and set numbers which gives any of the answer choices.

if we get more than one answer choice then answer will be E, otherwise which ever choice we get..

so doing little bit of mind joggling choice A comes out.
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