T is a set of y integers, where 0 < y < 7. If the average of Set T is

Hi Gmat tiger,
You are right that the median should be either an integer or a multiple of 0.5 ( since the number of integers can be even)
But then, 1/3 is also not an integer and not a multiple of 0.5 !! Couldnt get this

y could be 2, 3, 4, 5, or 6. If y = either 3 or 6, y/3 is an integer.

Can someone break down all the choices as to why they can be median?

Rather than think about whether or not they can be the median, think about whether or not they could be an integer or a multiple of 0.5.

a) 0 integer
b) x could be an integer, all we know is that it's the average
c) –x same as b
d) (1/3)y could be an integer if y=3 or y=6
e) (2/7)y cannot be an integer or multiple of 0.5, as y is 1,2,3,4,5 or 6. Sufficient enough to answer the question

How C) -x could be the median?
Set consists of elements from 1 to 6, the average x is positive, the median is negative which is impossible due to the range from 1 to 6.

It's not always the case that E holds true.
If T contains only one element, let's say 3. The mean is 3. The median OUGHT to be 3, never -3. So C is right.
Can anyone help me out?

The question is "which of the following could NOT be the median of Set T?"
It means we have to find the option for which the given number can NEVER be the median of Set T.

So, \(-3\) cannot be the median in your example, but there are many other cases when it can be.
Set T contains integers, so negative numbers are not excluded.

\(2/7y\) can never be an integer when \(0<y<7\), while the median MUST be an integer, doesn't matter what is \(y\) and what are the numbers in the set.

Dear Bunuel,

Since Set T range from 1 to 6, why the sample of T from part C was T={-1, -1, 5}?

You are asking the same question: The phrase "T is a set of y integers, where 0 < y < 7" does NOT mean that T consist of elements from 1 to 6, it means that the number of elements (the number of terms) in T is from 1 to 6, inclusive.

Hi All,

For this question, you can TEST VALUES to eliminate the 4 answers that COULD be the median of Set T.

We're told that Set T consists of Y integers (where 0 < Y < 7) and that the AVERAGE = X = a POSITIVE INTEGER. We're asked which of the answers could NOT be the median.

Answer A: 0... If Set T is {0,0,3} then the average = 1 and the median = 0. Eliminate A.

Answer B: X... If Set T is {1,1,1} then the average = 1 and the median = 1 = X. Eliminate B.

Answer C: -X... If Set T is {-1,-1,5} then the average = 1 and the median = -1 = -X. Eliminate C.

Answer D: Y/3... If Set T is {1,1,1} then the average = 1 and the median = 1 = Y/3. Eliminate D.

There's only one answer left...

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Average of two integers will always be either an integer (if both are odd or both are even) or Integer/2 in lowest form (if one is odd and the other is even). The average can never be integer/7 in lowest terms.

Since y is between 0 and 7 exclusive, y is not divisible by 7 and hence, (2/7)y will be (integer/7) in lowest terms.

Median of a set of odd number of integers will be one of the integers.
Median of a set of even number of integers will be the average of middle two integers.
Hence, median cannot be of the form (2/7)y.

Answer (E)

Let’s analyze each answer choices using actual examples.

A. 0

Let x = 1 and y = 3, then T = {-1, 0, 4} has a median of 0.

B. x

Let x = 1 and y = 3, then T = {-1, 1, 3} has a median of 1, which is x.

C. -x

Let x = 1 and y = 3, then T = {-2, -1, 6} has a median of -1, which is -x.

D. (1/3)y

Let x = 1 and y = 3, then T = {-1, 1, 3} has a median of 1, which is (1/3)y.

Since we are looking for the one that could NOT be the median of Set T, then the correct answer must be E. Indeed, the median of a set of y integers is either an integer (if y is odd) or an integer divided by 2 (if y is even). For no integer value of y between 0 and 7 will the median be in the form (2/7)y.

Answer: E

Learnings from this question:
[Assumption: all entities of the dataset are integers]
(1) in case of ONE ENTITY or EVERY ENTITY EQUAL data set, Mean=Median.

(2) Median can an Integer or a fraction in the form of multiple of 0.5.

Median is integer if the dataset has
(a) odd nuodd number of entities
(b) even number of entities with the condition that middle twos are either both odd or both even.

Median is integer if the dataset has
even number of entities with the condition that middle twos are odd and even.


(3) Irrespective of the mean (positive, negative or zero), Median can be anything (positive, negative or zero).
Because mean be affected by extreme values, but Median isn’t affected by extremes. Median is based on POSITION middle value in ascending or descending order).

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Number of integers in T could be anything from 1 to 6.
Avg of T is x.

We need the median of T. The median of T will be either the middle integer or the avg of two middle integers. So median will always be either an Integer or Integer/2. When an integer is divided by 2, it gives either an integer or something.5.

Can an Integer/2 give you 3.1428? No.
Even Integer/ 2 = Integer
Odd Integer/2 = a.5

Now look at the given options.

(E) (2/7)y
You know that y will be 1/2/3/4/5/6. It cannot be 7 or a multiple of 7. So you will need to divide by 7.
Upon division by 7, you get a non terminating decimal. This cannot be the median.

Answer (E)

It seems that the question is testing statistics, it is actually testing understanding of decimals.

(A) 0: {-1,0,2}; y=3; median =0; x>0; Possible
(B) x: {-4x,3x,4x}; y=3; median = x; mean = x; Possible for any x>0; Possible
(C) -x: {-x,-x,-x,7x}; y=4; median = -x, mean = x; possible for x>0; Possible

One quicker way to solve between D & E(It took me 2+ minutes to think about this hack - but that is a learning)

Let us think about (D) and (E)
(E) can never be an integer, since y<7. Although that is fine, since median of a set of integers can be a non integer, we need to think about the origin of '7' in the denominator.
Median of a set of integers with even terms can only have 2 in the denominator; So how did 7 end up there?
Let us analyse (D) first:

(D) (1/3)y has the same problem: The origin of 3 in the denominator
A possible case is: y=3 and median = 1;
So, median = 1 = (1/3)*3 = (1/3)y = (D) Therefore, (D) is possible

Thus, the answer is (E)

But why?
In (E), let try to apply the same logic, that we used in (D)
median = 2; and y = some multiple of 7 = 7k
Thus, median = 2 = (2/7)*7k = (2/7)y.
But, according to the question, 0<y<7; Thus, y<7(1)
So, 7 can never be in denominator.

Thus, the answer is (E)

set T consists of y element , 0<y<7 , why you are considering 0 and -ve number.

I think your doubt is already addressed above.

The phrase "T is a set of y integers, where 0 < y < 7" does NOT mean that T consist of elements from 1 to 6, it means that the number of elements (the number of terms) in T is from 1 to 6, inclusive.