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655-705 Level|   Statistics and Sets Problems|                     
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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:

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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) \(5<x<6\), median is \(\frac{x+6}{2}\)......4,5,x,6,7,9
(III) \(x\geq{6}\), median is \(\frac{7+6}{2}\)....4,5,6,7,x,9

(1) x > 7
x>7 means x>6, 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.....\(5<x<6\), median is \(\frac{x+6}{2}\)......Mean = \(\frac{4+x+5+6+7+9}{6}=\frac{x+6}{2}....31+x=3x+18...2x=13...x=6.5\). But out of range from what we had assumed. So, not possible.
(III) 4,5,6,7,x,9.......\(x\geq{6}\), 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 and 8
Insuff

A
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?
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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).

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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.

Bunuel
7, 9, 6, 4, 5, x

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

(1) x > 7
(2) The median of the list equals the arithmetic mean of the list.


DS89950.01
OG2020 NEW QUESTION
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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).

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Can someone explain why D is not the correct answer? Given that the problem indicate Median and Mean are the same which would mean x = 8 is the only option we can have.
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Can someone explain why D is not the correct answer? Given that the problem indicate Median and Mean are the same which would mean x = 8 is the only option we can have.

Hi MrDolo,

When dealing with the information provided in Fact 2, the 'obvious' answer that most people see is X = 8 (and the median would be 6.5) .... but is that really the ONLY possibility?

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 and the MEAN = (33)/6 = 5.5

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Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 1 variable: Let the original condition in a DS question contain 1 variable. Now, 1 variable would generally require 1 equation to give us the value of the variables.

We know that each condition would usually give us an equation, and since we need 1 equation to match the number of variables and equations in the original condition, the equal number of equations and variables should logically lead to answer D.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find the value of 'x' in order to find the median of the list: 7, 9, 6, 4, 5, x


Second and the third step of Variable Approach: From the original condition, we have 1 variable (x). To match the number of variables with the number of equations, we need 1 equation. Since conditions (1) and (2) will provide 1 equation each, D would most likely be the answer.

Let’s take a look at each condition.

Condition(1) tells us that x > 7

=> Arrange is ascending order: 4, 5, 6, 7, x, 9 OR 4, 5, 6, 7, 9, x

=> Median when there are 6 values is calculated by the average of 3rd and 4th value.[an average of 6 and 7 which is 6.5]

Since the answer is unique, the condition is sufficient by CMT 2.


Condition(2) tells us that Median = Average

=> If x > 7, then Median = 6.5 and Average: \(\frac{(4 + 5 + 6 + 7 + 9 + x) }{ 6}\) = 6.5

=> x = 8

=> But if x ≤ 5, then median = 5.5 and Average: \(\frac{(4 + 5 + 6 + 7 + 9 + x) }{ 6}\) = 5.5

=> x = 2

Since the answer is not unique, the condition is not sufficient by CMT 2.


Condition (1) alone is sufficient.

So, A is the correct answer.

Answer: A
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Video solution from Quant Reasoning:
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Bunuel
7, 9, 6, 4, 5, x

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

(1) x > 7
(2) The median of the list equals the arithmetic mean of the list.


DS89950.01
OG2020 NEW QUESTION
Solution:

Question Stem Analysis:

We need to determine the median of a list consisting of 6 numbers, including one unknown value.

Statement One Alone:

Since we know x > 7, arranging the numbers from smallest to largest, we have either:

4, 5, 6, 7, x, 9 or 4, 5, 6, 7, 9, x

We see that either way the median will be (6 + 7)/2 = 6.5. Statement alone is sufficient.

Statement Two Alone:

We can’t determine a unique value for the median, given the information in statement two. For example, if x = 8, then the median and mean are both 6.5. However, if x = 2, then the median and mean are both 5.5. Statement two alone is not sufficient.

Answer: A
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Bunuel
7, 9, 6, 4, 5, x

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

(1) x > 7
(2) The median of the list equals the arithmetic mean of the list.


DS89950.01
OG2020 NEW QUESTION

Answer: Option A


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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?

Since there are 6 terms (even) in the list , the median will be the average of 3 rd and 4th terms when they are arranged in ascending order.

(1) x > 7

When we arrange the terms in ascending order, there are 2 possibilities
4 5 6 7 x 9 or 4 5 6 7 9 x
But in both cases , 3 rd and 4th term will be 6 and 7

The median = (6 + 7 )/2 = 6.5

So , statement 1 is sufficient.

(2) The median of the list equals the arithmetic mean of the list.

mean = ( 4 + 5 + 6+ 7 + 9+ x)/6 = (31 + x)/6


Case 1: x> 7
From statement 1, we know that median = 6.5
if median = mean , 6.5 = (31 + x )/6
39 = 31 + x
x = 8
So when x =8 , median = mean = 6.5

case 2: x <5
median = (5 + 6 )/2 = 5.5

5.5 = (31 + x )/6
33 = 31 + x
x = 2.
If x =2, median = mean = 5.5

With 2 cases itself , we are getting 2 different values for median.
Hence Statement 2 alone is insufficient.

Case 3: 5<x<7

Median = (6+x)/2
Since median = mean, (6+x)/2 = (31 + x )/6

18 + 3x = 31 + x
2x = 13
x = 6.5

median = (6 + 6.5)/2 = 6.25

Option A is the correct answer.

Thanks,
Clifin J Francis
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Bunuel
7, 9, 6, 4, 5, x

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

(1) x > 7
(2) The median of the list equals the arithmetic mean of the list.


DS89950.01
OG2020 NEW QUESTION
Solution:

Question Stem Analysis:


We need to determine the median of a list consisting of 6 numbers, including one unknown value.

Statement One Alone:

Since we know x > 7, arranging the numbers from smallest to largest, we have either:

4, 5, 6, 7, x, 9 or 4, 5, 6, 7, 9, x

We see that either way the median will be (6 + 7)/2 = 6.5. Statement alone is sufficient.

Statement Two Alone:

We can’t determine a unique value for the median, given the information in statement two. For example, if x = 8, then the median and mean are both 6.5. However, if x = 2, then the median and mean are both 5.5. Statement two alone is not sufficient.

Answer: A
Quick Question. Why isn't statement 2 valid if we know that when the mean equals the median, we have an evenly space set. So if it's an evenly space set, wouldn't x=8?­
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jy18

You can say that in a AP, set of evenly spaced elements, the median and mean will be same, but vice versa need not be true.

You can make hundreds of sets given median and mean are equal.

Say both are 10
2,3,10,17,18
2,3,10,15,20
-5,-5,10,10,40
and so on
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Bunuel, can you please share similar questions for practice?
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I received a PM requesting that I explain what it means to be "symmetrical about the median."

Consider the following set:
0, 5, 6, 9, 11, 13, 16, 17, 22
Here, median = 11.

If we calculate the distances between successive terms, we get:
0 <--5--> 5 <--1--> 6 <--3--> 9 <--2--> 11 <--2--> 13 <--3--> 16 <--1--> 17 <--5--> 22

To the left and right of the median, the distances are MIRROR IMAGES:
<--5--><--1--><--3--> <--2--> 11 <--2--><--3--><--1--><--5-->

Any set that satisfies this condition is SYMMETRICAL ABOUT THE MEDIAN.
As noted in my earlier post, for any set that is symmetrical about the median, mean = median.­
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