Given numbers in ascending order:
4, 5, 6, 7, 9

Statement 1:
Smallest 3 numbers are 4, 5, and 6.
Greatest 3 numbers are 7, 9 and x, where x is any number greater than 7.
Thus:
Median = average of the 2 middle numbers \(= \frac{6+7}{2} = 6.5\)
SUFFICIENT.

Rule:
For any set that is SYMMETRICAL ABOUT THE MEDIAN, average = median.

Statement 2:
Case 1: x=2, so that the resulting set -- 2, 4, 5, 6, 7, 9 -- is symmetrical about the median of 5.5
Case 2: x=8, so that the resulting set -- 4, 5, 6, 7, 8, 9 -- is symmetrical about the median of 6.5
Since the median can be different values, INSUFFICIENT.


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7, 9, 6, 4, 5, x

If x is a number in the list above, what is the median of the list?

so the list is 4,5,6,7,9 and x can be placed anywhere in between..
Three cases..
(I) \(x\leq{5}\), median is \(\frac{5+6}{2}\).....4,x,5,6,7,9
(II) \(5<x<7\), median is \(\frac{x+6}{2}\)......4,5,x,6,7,9
(III) \(x\geq{7}\), median is \(\frac{7+6}{2}\)....4,5,6,7,x,9

(1) x > 7
x>7, so case III, answer is \(\frac{7+6}{2}\)

(2) The median of the list equals the arithmetic mean of the list.
We can have three cases as mentioned above
(I) 4,x,5,6,7,9..... \(x\leq{5}\), median is \(\frac{5+6}{2}=5.5\).....Mean = \(\frac{4+x+5+6+7+9}{6}=5.5....31+x=33...x=2\)
(II) 4,5,x,6,7,9.....\(5<x<7\), median is \(\frac{x+6}{2}\)......Mean = \(\frac{4+x+5+6+7+9}{6}=\frac{x+6}{2}....31+x=3x+18...2x=13...x=6.5\).
(III) 4,5,6,7,x,9.......\(x\geq{7}\), median is \(\frac{7+6}{2}=6.5\)....Mean = \(\frac{4+x+5+6+7+9}{6}=6.5....31+x=39...x=8\)
Thus, two possible values of x, 2, 6.5 and 8
Insuff

A
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The Logical approach to this question is to figure out where x is positioned in the list, since the median of 6 numbers is the average of the two numbers in the middle of the list.
Statement (1) tells us that x and 9 are the two largest numbers, so the median is the average of 6 & 7. That's enough information!
Statement (2) tells us that that the median equals the average. This is a rare case in DS questions in which we must switch to the Precise approach in order to check whether there can be more than one option. If x is larger than or equal to 7, the median, as we saw, is the average between 6 and 7, which is 6.5, so:
(7+9+6+4+5+x)/6=6.5
31+x=39
x=8
Now, if x is smaller than or equal to 5, then the median is the average between 5 and 6, which is 5.5, so:
(7+9+6+4+5+x)/6=5.5
31+x=33
x=2
Since we've already found two different options, it means that statement (2) is not enough on its own, and the correct answer is (A).

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#1
x>7
x=8
so median of set ; 4,5,6,7,8,9; 6+7/2 ; 6.5 for all values of x>7
sufficeint
#2
The median of the list equals the arithmetic mean of the list.

AM = sum of all digits / 6 ; value of x can be <6 , >6 or =6 so AM will vary
insufficient
IMO A

Hi All,

We're told that X is a number in the list (7, 9, 6, 5, 4, X). We're asked for the MEDIAN of the list. This question can be solved with Stats rules and TESTing VALUES. To start though, since there are six values in the list, the MEDIAN will be the AVERAGE of the two 'middle terms' (once we arrange the list from least to greatest). With a variable, we don't yet know which two values will be the 'middle two' values yet.

(1) X > 7

Fact 1 tells us that X is GREATER than 7, so it will either be the 5th value in the list or the 6th value in the list. For example:
IF....
X = 8, then the list is 4, 5, 6, 7, 8, 9 and the MEDIAN = (6+7)/2 = 13/2 = 6.5
X = 9, then the list is 4, 5, 6, 7, 9, 9 and the MEDIAN = (6+7)/2 = 13/2 = 6.5
X = 10, then the list is 4, 5, 6, 7, 9, 10 and the MEDIAN = (6+7)/2 = 13/2 = 6.5
Etc.
Thus, the answer to the question is ALWAYS 6.5
Fact 1 is SUFFICIENT

(2) The MEDIAN of the list equals the arithmetic MEAN of the list.

The information in Fact 2 requires a bit more work. To start, we can use one of the above TESTs (from Fact 1):
X = 8, then the list is 4, 5, 6, 7, 8, 9 and the MEDIAN = (6+7)/2 = 13/2 = 6.5

However, that's not the only way for the Median to EQUAL the Mean. If the numbers are equally 'spaced out' around the Mean, then the Median will equal the Mean. That can occur with another value of X...
IF... X = 2, then the list is 2, 4, 5, 6, 7, 9 and the MEDIAN = (5+6)/2 = 11/2 = 5.5
Fact 2 is INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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First, rewrite the list in order, remembering we don't know where x falls:
4, 5, 6, 7, 9, x

(1) x > 7 → median = (6+7)/2 sufficient

(2) Mean = Median → x could be 8 since making x=8 creates a set of consecutive integers, and for any evenly spaced set, including any set of consecutive integers, the mean = median. So in that case, the median is 6.5, just as it is in statement (1) since x=8 satisfies statement (1).

Now, look for an easy counterexample. Just as statement (1) made it easy to determine the median because we knew what the two middle numbers were, choosing any x≤5 makes the median = (5+6)/2 = 5.5. To double-check this is possible, the sum = 5.5 x 6 = 33 → x=2. That works, so since two medians are possible, insufficient.

Answer A
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this doesnt take into consideration the case where x=6, and then the median will be 6 as well, though the mean will be still out of range. Am i right?

Hi nakulanand,

In DS questions, you have to pay careful attention to how each of the two Facts 'restricts' the variables involved. With Fact 1, X has to be greater than 7, so X=6 is NOT an option. With Fact 2, you actually have to do a little more work to prove it, but X=6 is NOT an option there either (since Fact 2 tells us that the MEDIAN = MEAN... and X=6 will NOT make that happen).

GMAT assassins aren't born, they're made,
Rich
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Bunuel/chetan2u/veritaskarishma/empowergmatrich

I fall in the trap by choosing option D. Can anyone please verify and correct my logic?

My logic for statement 2, as correct, is that only consecutive numbers have the same mean and median value. So, i saw only digit 8 as x to be consecutive numbers (4,5,6,7,8,9) in the list as other numbers cann't be placed in the list as consecutive numbers. So, i concluded that there will be only median value for the list and i inferred statement 2 to be correct one.

Thanks.

Hi Raxit85,

When dealing with Quant or Verbal concepts, it can often help to think in terms of the simplest examples that you can come up with (since GMAT questions sometimes take concepts that you know and ask you to think about those concepts in ways that you might not have considered).

Here, we're asked to consider how MEAN and MEDIAN might relate to one another. While a group of CONSECUTIVE INTEGERS will have the SAME median and median, that does NOT mean that only consecutive integers fit that pattern.

For example:
1, 2, 3 --> this group of consecutive integers has median = 2 and mean = 2

0, 2, 4 --> this group has the same number of terms as the first group, but they're NOT consecutive; this group ALSO has median = 2 and mean = 2

By thinking about both ideas (all consecutive vs. not all consecutive), you should be able to come up with another group of numbers that fits the information in this prompt. (my explanation - a little higher up in this thread - showcases those ideas).

GMAT assassins aren't born, they're made,
Rich
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Case 3 is wrong. You actually assume that x>7. You may add another two cases, 6<x<7 and x>7 accordingly.

Hi chetan2u,

In case of statement 2, why x = 6.5 is out of scope when we have assume value of x as 5 < x < 7?

Thanks in advance

Hi,

You are correct. That’s a typo as I changed the range from 5<x<6 to 5<x<7, but forgot to change the latter part.
Thanks
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