What is the median of the nine consecutive even integers in a certain

Question

What is the median of the nine consecutive even integers in a certain list?

(1) The median of the integers in the list is greater than 0.
(2) Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4.

DS11820.02

itspCv2 · Accepted Answer

Making it easy :

9 consecutive even integers

looks nice Median be=X say; 1)X>0 ; clearly insufficient X= 2 or 4 or 6.... 2) Sum of least (-ve & +ve)=-4 Here: they are both even and have signs(+/-) ---> Least +ve even=2(this is a fact) Least -ve even= L (say) L+2=-4(Given)>>>L=-6; Now set becomes set={-6,-4,-2,0,2,4,6,8,10} (since 9 consecutive even integers) And clearly median is =2 Hope this helps

Thanks

chetan2u · Answer

altairahmad , -8, -4, 0... will not fit in as it is an AP but not sequence of consecutive even integers.

There are 9 consecutive integers..

(1) The median of the integers in the list is greater than 0.
 The nine numbers could be anything.. 
2, 4, 6, 8, 10, 12, 14, 16, 18....Median = 10
102, 104, 106, 108, 110, 112, 114, 116, 118....Median = 110
Insufficient

(2) Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4.
Let the least negative integer be  x .
 The least positive integer will be 2, if the set contains positive integers. 
 But let us take there are NO positive integers , then  x+0=-4.....x=-4 
 sequence would be -4, -2, 0, 2, 4, 6, 8, 10, 12  ---  But this contains positive integers too, while we took the least positive integer as none. 
So  x+2=-4...x=-6 , and sequence is -6, -4, -2, 0, 2, 4, 6, 8, 10--- Median = 2. 
Sufficient

B

BrentGMATPrepNow · Answer

Target question :    What is the median of the nine consecutive even integers in a certain list?

Statement 1 : The median of the integers in the list is greater than 0.  
There are several scenarios that satisfy statement 1. Here are two: 
 Case a : The numbers are {0,2,4,6, 8 ,10,12,14,16}. In this case, the answer to the target question is  the median is 8 
 Case b :The numbers are {2,4,6,8, 10 ,12,14,16,18}. In this case, the answer to the target question is  the median is 10 
Since we cannot answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4. 
Since   2   is the  smallest positive even  number, we can write: 
(smallest negative value) +   2   = -4 
This means the smallest negative value = -6
So, the nine consecutive even integers must be {-6, -4, -2, 0, 2, 4, 6, 8, 10}, at which point we can see that   the median is 2  
Since we can answer the  target question  with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

Rebaz · Answer

Hi Brent, donot you think that statement 2 is not worded correctly?

Statement 2 states: Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4.

But you say in your explaination of statement 2:

While statement 2 clearly states that the sum of the LEAST of the negative integers and the LEAST of the positive integers. But you make of the LEAST positive integer biggest value in your explaination. That is what i donot understand, beacuse Least positive is not the same as the biggest value.

This what i think how the list look like. -10 -8 -6 -4 -2 0 2 4 6 and the sum of (-10) + (6) = -4

Acme · Answer

Hi Rebaz,
 You are correct that the above explanation on statement 2 is incorrect.

As per the question statement, the sum of least negative and least positive integer is -4.

So we have only one combination of -6 (least negative) and 2(least positive). Since we know the integers are even . We can identify that 2 is the Median of this set.

Hope this clarifies.

abhish27 · Answer

Your explanation for 2nd statement is incorrect according to me because the 2nd statement clearly states "least of the negative integers and the least of the positive integers". In your explanation of 2nd statement u have smallest+biggestvalue,which according to me is incorrect.Please find my detailed solution below.

abhish27 · Answer

Let the nine consecutive even integers be (2n-8),(2n-6),(2n-4),(2n-2),2n,(2n+2),(2n+4),(2n+6),(2n+8) Statement1: 2n>0 thus, n>0 but can can have any value hence, median cannot be determined. Statement2: Since the statement says"Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4." there must be atleast one negative value and atleast one positive value in the series for the statement to hold.Now calculating the range of n for such condition we get -4

brains · Answer

, .. Not understood the above Solutions . Can you experts please shed some light into this problem

Thanks

brains · Answer

VeritasKarishma  and  Bunuel  .. Can you experts please shed some light into this problem . Could not quite get the solution posted above

Thank you

altairahmad · Answer

Bunuel   chetan2u

How is the OA (B)

For B there are multiple solutions :

Example 1 : -6, -4, -2, 0 ,2 , 4 , 6 , 8 , 10 (-6 +2 = -4  , median = 2)
Example 2 : -8, -4, 0, 4 ,8 , 12, 16, 20, 24 (-8 + 4 = -4  , median = 8)

Please elaborate

robu1 · Answer

Your example 2 is not representative of consecutive even integers. Difference should be two.

KarishmaB · Answer

If answer is to be (B), the question needed to specify that these 9 consecutive even integers are the only integers in the list, that there are no odd integers.

The list has 9 consecutive even integers.
So the list would be something like:
-8, -6, ... 0, 2, 4, 6 ...
or
-6, -4, -2, 0, 2, 4, ....
or
-2, 0, 2, 4, 6, ...
or
-40, -38, -36 ...
etc

(1) The median of the integers in the list is greater than 0.

The median of all integers is greater than 0. Just tells us that median is positive. Not sufficient.

(2) Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4.

If the list has both negative and positive even numbers that are consecutive, it must have 0 too. So the least positive even integer in the list must be 2.
Then the least negative integer in the list must be -6 (because if L is the least negative number, L + 2 = -4 then L = -6)
Then the list must be 
-6, -4, -2, 0, 2, 4, 6, 8, 10
Median = 2

Answer (B)

matteotchodi · Answer

What is the median of the nine consecutive even integers in a certain list?

(1) The median of the integers in the list is greater than 0.
(2) Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4.

stated question: what is the median of the set

1) clearly insufficient.... we can make an infinite number of sets with this statement in which the median may differ

2) the wording is confusing:

Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4.

note*: the LEAST NEGATIVE INTEGER is the integer that is negative but has the highest absolute value:

just for the sake for clarification: compare -100000 and -1...… -100000 is smaller (often our mind just thinks in absolute terms and tricks us to erroneous thinking)

Least negative integer + Least positive integer = - 4

least positive even integer per definition is 2

n + 2 = - 4

n = - 6

so the first element in ascending order will be negative six

so we can find the median

B

ueh55406 · Answer

Hi  Bunuel   chetan2u

can you share if this is the correct approach for second statement?

n+8 is median (middle term of the 9 terms, hence, our median)

Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4.

n+(n+2)+(n+4)+(n+6) + (n+10)+(1+12)+(n+14)+(n+16) = -4

when we solve this, we do get a median, but it's -1/2. , so inadvertently I did get the right answer. but I am not sure if it's the right approach.

CEdward · Answer

What is the median of the nine consecutive even integers in a certain list?

9 consecutive integers: 
n, n + 2, n + 4, n + 6, n + 8, n + 10, n + 12, n + 14, n + 16

(1) The median of the integers in the list is greater than 0.
This tells us that n + 8 > 0 ---> n > -8 
Insufficient

(2) Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is –4.
We can get the median from this because there will only be two integers that will get us -4 in a list of 9 consecutive.

Sufficient.

B.

flapjack29 · Answer

We are looking for the median

Statement 1:

This is clearly not Suff. This basically means our median could be ANY  Positive Even  integer to infinity.

Statement 2:

Were dealing with consecutive even integers here, so remember, the least positive integer HAS to be 2 and which leaves us with only one NEGATIVE integer that could be added to 2 allowing us to achieve a sum of of -4. Therefore, our first number in the sequence would be -6. Now that we have our first number, we can list the 9 consecutive even integers from there and find our Median. Sufficient.

Great Problem!

Doryhihii · Answer

Hi  BrentGMATPrepNow , about (2) sum of the least of negative integers and least of positive integers are not the biggest and smallest number of the list, I don't understand why we can have median? Kindly help to explain. Thank you.

BrentGMATPrepNow · Answer

My bad. I read statement 2 as " Of the integers in the list, the sum of the least of the negative integers and the GREATEST of the positive integers is –4" I've edited my response accordingly at https://gmatclub.com/forum/what-is-the- ... l#p2507096 . Kudos for you!!

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What is the median of the nine consecutive even integers in a certain

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