Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the

Bunuel.... is it possible if you could point me to more questions like these on the forum.

DS: search.php?search_id=tag&tag_id=34
PS: search.php?search_id=tag&tag_id=55

thanks a ton.... Bunuel

Arranging the numbers, except x, the series is 1,2,3,9,11

Its a 6 term series. The median would be calculated by taking the mean of the middle two terms i.e. the 3rd and the 4th term.

so (3 + x)/2 is the median. Also, at the same time the number should be less than 9. The numbers could be 4,5,6,7,8. (distinct integers) Had 6 been an option, then x would be any other term but in the first 4.

Only one of the options suffice the conditions given.

Answer: B

This is one great Question on the median.
Here is my solution

Number of terms = 6
Hence median = 3rd+4th/2

Lets use Brute Force here.

Option 1 =>
Median =3
Hence 3rd term + 4th term = 6

This can only occur when x=3
But if x= 3 => This will violate the original condition that all elements are distinct.
Hence rejected.

Option 2-->

Median = 5
Hence 3rd term +4th term = 10

This can happen when x=7

Hence Acceptable case

Hence B

Since the integers are positive and distinct, x must be greater than 3. However, since we don’t know the exact value of x, we can start by ordering the given integers from least to greatest in the following possible scenarios:

Scenario 1: If the ordering is 1, 2, 3, x, 9, 11, then x is between 3 and 9

Scenario 2: If the ordering is 1, 2, 3, 9, x, 11, then x must be 10.

Scenario 3: If the ordering is 1, 2, 3, 9, 11, x, then x is greater than 11.

In the first scenario, the median is (3 + x)/2. In the latter two scenarios, the median is (3 + 9)/2 = 6. Since 6 is not an answer choice, the median must be (x + 3)/2 in which x is an integer between 3 and 9. Because x is an integer between 3 and 9, the median, (x + 3)/2, must be some value between 3 and 6. Thus:

3 < x < 9

6 < x + 3 < 12

3 < (x + 3)/2 < 6

The only number in the choices that satisfies this condition is 5, which is answer choice B.

Answer: B

Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?

A. 3
B. 5
C. 7
D. 8
E. 9

\(Solution: \\
\)
Possible combinations:

\(1 | 2 | 3 | 9 | x | 11 \)-> Median \(\frac{(3+9)}{2} = 6\) Not in your answer choices

\(1 | 2 | 3 | x | 9 | 11\) Note that x must be \(3 <= x <= 9\)

X = 8-> \(\frac{(3+8)}{2}= \frac{11}{2} = 5.5 \)

X = 7->\( \frac{(3+7)}{2} = 5\) \(POSSIBLE\)

Answer B.

In the question stated what if the value of the x is lying between 9 and 11?

I think this is addressed in my post here: https://gmatclub.com/forum/given-distin ... ml#p876372

P.S. If x is more than 9 so 10 or more then the median will be the average of 3 and 9 so (3+9)/2=6 (the maximum median possible).

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