S is a set containing a integers, where 0

Question

S is a set containing “a” integers, where 0<a<11. If the arithmetic mean of S is a positive integer “b”, which of the following CANNOT be the median of S?
A. 5a/11
B. -b
C.0
D. b
E. a/5

gmatophobia · Answer

In questions such as this, in which I have no clue on the best possible approach to enter the question, I start by eliminating the options. Using Answer choice elimination We can start with Option B, Option C or Option D. Let's start with Option B. Option B - Can "-b" be the median ? We know the average of the entire set is a positive integer “b”. So let's create a set placing -b at the center. We can add another term (say another value , -2b) to the left of -b. Now, to keep the set balanced at b, we can add a third term 6b to the right of the median. Set A therefore contains the following elements - S = { -2b, \quad -b, \quad +6b } So -b, can be the median of the set A. Eliminate option B. Option D - Can "b" be the median ? We were able to create a set in which the median of the set was -b, hence we can also create a set in which the median is b. We need not necessarily test out this scenario, however it's not a bad idea to do so. Let's start by placing b as the center of the set. We can add another term 2b to the right of b. Now, to keep the set balanced at b, we can add a third term 0 to the left of the median. Set A therefore contains the following elements - S = { 0, \quad +b, \quad +2b } So b, can be the median of the set. Eliminate option D. Option C - Can "0" be the median ? The steps remains the same. We start by placing 0 at the center, and let's add another term say -b to the left of 0. To keep the average at b. we can add a third 4b to the right of the median. Set A therefore contains the following elements - S = { -b, \quad 0, \quad +4b } So 0, can be the median of the set. Eliminate option C. At this stage there is a 50 - 50 chance. If I were running short of time I will select one of the remaining options and move on. 50% is not bad on a question that ends up taking a lot of time. However, if you've come this further, stay along the rest of the explanation

After we have eliminated , B, C and D. Let's pick up E as the option seems less complicated than A. Option E - Can "a/5" be the median ? We are free to choose any value of a as long as the value satisfies the condition 0

Option A

Theprince · Answer

if we consider just option a, we know any integer between 0 and 11 divided by 11 will give recurring numbers and median of any set of integers will have 2 solutions
1. if total no of integers are even then median has to be integer 
2. if total no of integers are odd then it will be a+b/2 and this will never be recurring no.
so option A.

Giallo12 · Answer

Answer A is the right answer, due to the fact that simply the interval of "a" is defined for: 0<a<11, so then 11, being an extreme of such interval is excluded from it and from the possible values of the median too.

GaganB · Answer

- Median of a set is either an integer or of the form (A+B)/2 
- Since in this case the number of terms are less than 11, and the median has to be an integer, therefore 5a/11 can't be the median since it wont be an integer for a < 11.
Hence, ans. A

S is a set containing a integers, where 0<a<11. If the arithmetic

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