T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?
A 0
B x
C -x
D (1/3)y
E (2/7)y
Thanks
Median of set T
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Median of the set has to be either an integer or an
integer when multiplied by 2. Since 0 < y < 7, E doesn't satisfy
either condition. Hence, E has to be the answer.
Just to quote examples
For option A
-1, 0, 4 satisfies
For option B
1, 2, 3 satisfies
For option C
-2, -1, 6 satisfies
For option D
1, 2, 3 satisfies
HTH
integer when multiplied by 2. Since 0 < y < 7, E doesn't satisfy
either condition. Hence, E has to be the answer.
Just to quote examples
For option A
-1, 0, 4 satisfies
For option B
1, 2, 3 satisfies
For option C
-2, -1, 6 satisfies
For option D
1, 2, 3 satisfies
HTH
alex.gellatly wrote:T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?
A 0
B x
C -x
D (1/3)y
E (2/7)y
Thanks
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I understand by trying out different values to make it satisfy. However, that approach is a little tedious and dangerous. Is there an algebraic approach or a rule that one should be aware of?gmat_and_me wrote:Median of the set has to be either an integer or an
integer when multiplied by 2. Since 0 < y < 7, E doesn't satisfy
either condition. Hence, E has to be the answer.
Just to quote examples
For option A
-1, 0, 4 satisfies
For option B
1, 2, 3 satisfies
For option C
-2, -1, 6 satisfies
For option D
1, 2, 3 satisfies
HTH
alex.gellatly wrote:T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?
A 0
B x
C -x
D (1/3)y
E (2/7)y
Thanks
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Alex,
The only thing that you need to know is in the first two lines of my reply i.e median has to
be an integer or an integer when multiplied by 2. You do not have to try different values to
solve the problem.
HTH
The only thing that you need to know is in the first two lines of my reply i.e median has to
be an integer or an integer when multiplied by 2. You do not have to try different values to
solve the problem.
HTH
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That's a good rule!gmat_and_me wrote:Alex,
The only thing that you need to know is in the first two lines of my reply i.e median has to
be an integer or an integer when multiplied by 2. You do not have to try different values to
solve the problem.
HTH
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The Answer is [E]
T --> Set of Y Integers, meaning T can contain 1 or 2 or 3 or 4 or 5 or 6 Integers.
[A] The Set (-1,0,4) --> SATISFIED
(B) The Set (2,3,4) --> SATISFIED
[C] The Set (-2,-3,-4) --> SATISFIED
[D] Take values of y which are divisible by "3" so 3,6
y=3 --> (-1,1,3) --> SATISFIED
[D] The only possible value to have Integer value for (2y/7) is y=7
BUT, y <7 NOT SATISFIED
T --> Set of Y Integers, meaning T can contain 1 or 2 or 3 or 4 or 5 or 6 Integers.
[A] The Set (-1,0,4) --> SATISFIED
(B) The Set (2,3,4) --> SATISFIED
[C] The Set (-2,-3,-4) --> SATISFIED
[D] Take values of y which are divisible by "3" so 3,6
y=3 --> (-1,1,3) --> SATISFIED
[D] The only possible value to have Integer value for (2y/7) is y=7
BUT, y <7 NOT SATISFIED
R A H U L
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The set contains 1-6 elements.
A. Consider a set {0, 0 ,6}. Here, mean = 2 and median = 0.
B. Consider a set { x }. Here, mean = x and median = x
C. Consider a set {-1, -1, 5}. Here mean(x) = 1 and median = -1= -x
D.Consider a set {1, 1, 1}. Here mean = 1, y = 3 and median = 1 = y/3.
E. T is a set of y integers where 0 < y < 7. All the elements in the set should be integers. So, the median also needs to be an integer.
Now, if 2y/7 become the median it also needs to be an integer which is not possible for any possible value of y i.e. 1,2,3,4,5 or 6. Hence, 2y/7 can't be the median.
A. Consider a set {0, 0 ,6}. Here, mean = 2 and median = 0.
B. Consider a set { x }. Here, mean = x and median = x
C. Consider a set {-1, -1, 5}. Here mean(x) = 1 and median = -1= -x
D.Consider a set {1, 1, 1}. Here mean = 1, y = 3 and median = 1 = y/3.
E. T is a set of y integers where 0 < y < 7. All the elements in the set should be integers. So, the median also needs to be an integer.
Now, if 2y/7 become the median it also needs to be an integer which is not possible for any possible value of y i.e. 1,2,3,4,5 or 6. Hence, 2y/7 can't be the median.
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Let's analyze each answer choices using actual examples.alex.gellatly wrote:T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?
A 0
B x
C -x
D (1/3)y
E (2/7)y
Thanks
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
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If we have a set of y INTEGERS, there are two possible cases when it comes to the MEDIAN.alex.gellatly wrote:T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?
A 0
B x
C -x
D (1/3)y
E (2/7)y
Thanks
Case a: y is an ODD number, in which case the MEDIAN equals the one middle integer (when all of the integers are arranged in ascending order). In this case, the median must be an integer.
Case b: y is an EVEN number, in which case the MEDIAN equals the average of the two middle-most integers (when all of the integers are arranged in ascending order). To find the average of the two middle-most integers, we'll add them together and divide the SUM by 2. So, if their SUM is even, then we'll get an integer value for the median. If their SUM is odd, then the median will be something.5. For example, if the two middle-most integers are 3 and 6, the median will be 4.5.
CONCLUSION: In this scenario, the median of Set T will be EITHER an integer OR something.5.
When we check the answer choices, answer choice 2y/7 can equal NEITHER an integer NOR something.5. We know this because y must be an integer AND 0 < y < 7. So, there's no way for 2y/7 to ever evaluate to be an integer or something.5.
Answer: E
Cheers,
Brent