Data Sufficiency

gmatmachoman wrote:
Harsha,
I studied ur reasoning for st1. But mine is also correct right?? If so, I doubt the wordings of the Question

harshavardhanc wrote:

gmatmachoman wrote:
Harsha,
I studied ur reasoning for st1. But mine is also correct right?? If so, I doubt the wordings of the Question

Bhai, that's what I tried to explain in my reasoning that you cannot have odd number of elements in S.
Reason : you first need to find out a number which is 1 less than half of the elements in S.

st 1:Exactly one fewer than half of the integers in S are multiples of 4.

so, any odd number/2 will not give you a perfect integer.

BTW, didn't see you yesterday on BTG.

kevincanspain wrote:
Try this question:

If S is a set of integers, is the median of S an integer?

(1) Exactly one fewer than half of the integers in S are multiples of 4.
(2) Exactly 1/4 of the integers in S are negative.

harshavardhanc wrote:

gmatmachoman wrote:
IMO D
st 1:Exactly one fewer than half of the integers in S are multiples of 4.

example : 4,8,9,10,11( as per the sated condition in st1.)
The series needs to be a Odd numbered series for the condition to be true.Now the median is certainly a integer .
So st 1 alone is sufficient to state median will be a integer for the given Set S.

St2 :Exactly 1/4 of the integers in S are negative

Example set : { -4,1,2,3}(as per the condition in st 2)
Now the median CANNOT be a Integer becox the rule: If there are 2k numbers, the median is the average of the k th and the (k+1)th numbers .
Ok..Let us take one more set for checking consistency:{ -4,-2,1,2,3,4,5,6}
Median is 2.5, certainly NOT a integer. So st2 is sufficient to say a Definite NO.

SO we can choose D.

machoman,

I'm confused .

one fewer than half of the integers in S. Doesn't it mean that number of integers, say n, is divisible by two. So is the set that you have chosen for st1 per the condition?

IMO,

Statement 1 : it just tells us that number in elements in S (n) is even and n/2-1 are multiples of 4.

So median, which is the avg of two mid numbers, may / may not be an innteger.

for e.g { 5, 8, 9, 12, 13, 14} here one fewer than integers in S are multiples of 4. But, the median is not an integer.

if the set is { 4, 7, 6, 10, 12, 13}, here also, one fewer than integers in S are multiples of 4. But the median is an integer.


Statement 2 :

This again tells us that the number of integers in S are even but, doesn't tell anything about the two middle numbers.


so, these two statements on their own are insufficient to answer the question.

Now, let's combine the information.

-Number of elements in S is even.
-one fewer than half of them are multiples of 4.
-1/4 of the total are negative.

again, nothing can be said definitely about the middle two numbers, which decide the median.

hence, IMO E should be the correct choice.

Kevin,

Is the reasoning correct?