If set Z has a median of 19 , what is the range of set Z?

Dear goodyear2013
I'm happy to help with this.
Here's an article about mean & median:
https://magoosh.com/gmat/2012/common-gma ... tatistics/
Here's an article about Data Sufficiency strategies:
https://magoosh.com/gmat/2013/gmat-data- ... ency-tips/

At the outset, we know nothing about Set Z but the median. As usual, we have no way to answer the questions without the statements.

Statement #1: Z = {18, 28, 11, x, 15, y}
We have three numbers less than 19, so they only way we can have a median of 19 is to have one of the variables x = 20, and the other equal something greater than or equal to 20. If x = 20 and y = 3000, then the set would have a median of 19 and a very large range, and because we could pick whatever we want for the value of y, we have no way to determine an exact value of the range. This statement, along and by itself, is not sufficient.

Statement #2: The average (arithmetic mean) of Set Z is 20.
The tricky thing about this statement is that we need to ignore statement #1 completely and just focus on this. Right now, we know median = 19, mean = 20, and absolutely nothing else about the situation. We don't even know how many members are in the set --- 6, or 10, or 500. We know nothing, so we certainly don't know the range. This statement, along and by itself, is not sufficient.

Combined Statements
From the first statement, we know x = 20 and y > 20. If the six numbers have an average of 20, then they have a sum of 120.
18 + 28 + 11 + 20 + 15 + y = 120
We could solve this for y, but we won't be daft enough actually to perform that calculation. This is GMAT Data Sufficiency! We don't actually need to solve for y. All we need to do is determine that y has a unique value, and we are able to create a simple equation for it, then we could solve. At that point, we would know all six numbers in the set, and we would know the range. The combined statements are sufficient.

Answer = (C)

BTW, y = 28, but that's irrelevant for answering the question.

Does all this make sense?
Mike

Hi goodyear.

We are given

{11,15,18,28} and {x,y} and we don't know where x or y is placed in relation to other numbers.

Lets examine the statements

Statement 1

median is 19 and we have 6 numbers in the set, meaning if the numbers are ordered from smallest to greatest, average of 3rd and 4th numbers is 19.
We are already given 3 numbers less than 19, so for S1 to be correct 4th number (say x) must be 20. and y >=20. Any of them less than 20 would make the median smaller than 19. However this does not give us enough information to identify the range of the set. y could be any number greater than 20 making us impossible to determine the range of the set. Not Sufficient.

Statement 2

If mean of the set Z is 20, sum of x+y must be 48. We don't know anything else again there are infinite number of pairs that would satisfy this condition, making us impossible to determine the range of the set. Not sufficient.

Statement 1& 2

We know one of x,y must be 20 and x+y 48, so y must be 28. From this information we can obtain range 28-11 = 17
So Statement 1 and 2 together are sufficient.


Answer C


I hope this helps.

This is how I would approach this question:

Set Z, median 19 - when I think median, I think of placing the numbers in increasing order because the middle number will be median.
Range of Z means I need to know the smallest and largest numbers of set Z.

(1) Z = {18, 28, 11, x, 15, y}
Let's arrange the known numbers in increasing order: 11, 15, 18, 28
There are 6 numbers in the set so the average of 3rd and 4th numbers will be 19. Since the third number 18 is less than 19, the fourth number must be greater than 19 and the average of 3rd and 4th numbers must be 19. The fourth number must be 20 (which will be either x or y, say x). But we still don't know the value of y. If y is 25, the range of Z will be 28 - 11 = 17. If y is 30, the range of Z will be 30 - 11 = 19. If y is 100, the range of Z will be 100 - 11 = 89 and so on...

(2) The average (arithmetic mean) of Set Z is 20.
We only know the median and mean of a set. The extreme values could lie anywhere.

Using both together, Z = {11, 15, 18, 20, 28, x}
Mean of set Z is 20 = (11 + 15 + 18 + 20 + 28+x)/6
x = 28

Range of Z = 28 - 11 = 17
Sufficient

Answer (C)

In this question, is it alright to assume that 18 will be the third number without knowing x and y ?
Also, the assumption will hold true for similar questions (questions with sets containing unknown numbers) ?

Hi b2bt,

I assume that you're looking at Fact 1.

In this Fact, we know that there are 6 numbers in Set Z: {11, 15, 18, 28 and......X and Y.....}. We just don't know what X and Y are YET....Once we include the information from the prompt (that the Median of the Set = 19), then we KNOW that 18 has to be the "third" value in the group because the Median = 19....

Since the set has 6 values, the Median will be the AVERAGE of the "third" and "fourth" numbers (once they're all put in order from least to greatest). For that Median to hold true, both X and Y MUST be greater than 18 (specifically, one of them is 20 and the other is 20 or greater), so by extension, 18 MUST be the "third" number.

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

Hi Rich,
This is how I approached option (A)
The numbers are 11,15,18,28,x,y
1.) So I first took 15 and tried to check what number could be paired up with it to make the median 19, assuming that the arrangement could be x,11,15,y,18,28. I soon realized that would not be possible as that number has to be 23.
2.) Then I realized that the I'll have to pair up 18 with another number. The only two arrangement possible would be 11,15,18,y,x,28 or 11,15,18,y,28,x
The second arrangement could change the range so option (A) is not sufficient.

I want to know whether step 1 was required or there is something glaringly obvious to avoid that. I waste a lot of time on such questions (question with set containing unknown number and about median/mean/mode) as I'm extra cautious while solving them. The reason is that I fail to understand how the values of median/mode/mean change as the numbers change.

Hi b2bt,

Since we're dealing with 6 numbers, and the Median = 19, we have to remember the "rules" of statistics. The Median will be the Average of the "third" and "fourth" numbers (once you put them in order from least to greatest). There are only two possibilities that we have to consider...

1) The "third" and "fourth" numbers are BOTH 19
2) The "third" and the "fourth" numbers AVERAGE to 19

So we could have....
_ _ 19 19 _ _

OR

Some variation on the Average = 19....

_ _ 18 20 _ _
_ _ 17 21 _ _
Etc.

Once you recognize the that you have an 11, a 15 and an 18 in the Set, then the first option isn't possible (the "third" term CAN'T be 19, since there are 3 other terms than are smaller than that).

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

Hmm... And for the same reason
_ _ 15 23 _ _
Won't be possible. Liked your approach!

Sent from my Nexus 4 using Tapatalk

Hi b2bt,

That small example that you just typed represents the EXACT type of work and thinking that you have to do to maximize your score on Test Day. The simple act of writing down an example and noting that it's NOT possible for that to be the answer is part of the process of "zero-ing in" on the correct answer. That little bit of 'extra work' (on the pad, NOT in your head) could easily lead to a score increase on the Official GMAT.

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

Range = {biggest element - smallest element}
(1) Z = {18, 28, 11, x, 15, y}
Median is given to us = 19
Median is the middle value
For set with even number of element median =sum of two middle element /2
In this case Z = {11,15,18,28, X, Y}
median is given as 19 but\(\frac{18+28}{2}\) is not 19, therefore X or Y should come after 18,
so either X or Y must be 20, but the remaining unknown can have any value from greater than 20 and thus range cannot be determined
{11,15,18,20,28, 111] range = 111-11 =100
or
{11,15,18,20,28,31] range = 31-11 =20
or
{11,15,18,20,22,28] range = 28-11 =17

INSUFFICIENT

(2) The average (arithmetic mean) of Set Z is 20.
Meaning \(\frac{11+15+18+28+X+Y}{6}=20\)
72+X+Y=120
X+Y=48
We cannot find what is X and Y and thus cannot find the range
INSUFFICIENT

Merging both we know, that either X or Y is 20 therefore the remaining will be 48-20=28 and thus range will be 28-11=17
SUFFICIENT

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