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Hi everyone, I saw this question in the practice problems of Kaplan GMAT book, but I am not sure if the answer they gave is right....so any help in this regard shall be highly appreciated. The question is as follows:

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

(1) Z = { 18, 28, 11, x, 15, y} (2) The average (arithmetic mean) of the members of set Z is 20.

The book gives the answer as C (BOTH statements TOGETHER are sufficient), but I think the answer is E(BOTH are not sufficient)

The reasoning in the book is as follows: Statements (1) and (2) together: From statement (1), we know the set has 6 elements and that either x or y must be 20. From (2), we know the average is 20. For a set of 6 elements with an average of 20, the sum of the numbers must be 6x20 = 120.

So, it must be the case that 11 + 15 + 18 + 28 + x + y = 120, or x + y = 48. Since we know that either x or y must be equal to 20, let's say that x = 20. So, we have 20 + y = 48 or y=28. Thus the set is {11, 15, 18, 20, 28, 28}. The range must be 28-17 = 11, so the statements together are sufficient. (C) is correct.

Now, my question is.....the answer takes the set to be {11, 15, 18, 20, 28, 28} and comes to the conclusion C. As per the definition of a set, there should not be any repetitive elements. So the set cannot be {11, 15, 18, 20, 28, 28} with the element 28 being repeated. So the set can be {11, 15, 18, 20, 28}, in which case we cannot determine the value of y. So I am not sure if this question is right.

Please let me know if my thought process is wrong....and what should be the right answer?

Thank you for your time. I shall look forward to hearing from all of you.

Re: If set Z has a median of 19 , what is the range of set Z? [#permalink]

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15 Sep 2009, 08:58

good Q..thanks for posting the OE

i have one doubt you saying As per the definition of a set, there should not be any repetitive elements. is it true?

rahularke wrote:

Hi everyone, I saw this question in the practice problems of Kaplan GMAT book, but I am not sure if the answer they gave is right....so any help in this regard shall be highly appreciated. The question is as follows:

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

(1) Z = { 18, 28, 11, x, 15, y} (2) The average (arithmetic mean) of the members of set Z is 20.

The book gives the answer as C (BOTH statements TOGETHER are sufficient), but I think the answer is E(BOTH are not sufficient)

The reasoning in the book is as follows: Statements (1) and (2) together: From statement (1), we know the set has 6 elements and that either x or y must be 20. From (2), we know the average is 20. For a set of 6 elements with an average of 20, the sum of the numbers must be 6x20 = 120.

So, it must be the case that 11 + 15 + 18 + 28 + x + y = 120, or x + y = 48. Since we know that either x or y must be equal to 20, let's say that x = 20. So, we have 20 + y = 48 or y=28. Thus the set is {11, 15, 18, 20, 28, 28}. The range must be 28-17 = 11, so the statements together are sufficient. (C) is correct.

Now, my question is.....the answer takes the set to be {11, 15, 18, 20, 28, 28} and comes to the conclusion C. As per the definition of a set, there should not be any repetitive elements. So the set cannot be {11, 15, 18, 20, 28, 28} with the element 28 being repeated. So the set can be {11, 15, 18, 20, 28}, in which case we cannot determine the value of y. So I am not sure if this question is right.

Please let me know if my thought process is wrong....and what should be the right answer?

Thank you for your time. I shall look forward to hearing from all of you.

Re: If set Z has a median of 19 , what is the range of set Z? [#permalink]

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15 Sep 2009, 09:02

rahularke wrote:

Hi everyone, I saw this question in the practice problems of Kaplan GMAT book, but I am not sure if the answer they gave is right....so any help in this regard shall be highly appreciated. The question is as follows:

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

(1) Z = { 18, 28, 11, x, 15, y} (2) The average (arithmetic mean) of the members of set Z is 20.

The book gives the answer as C (BOTH statements TOGETHER are sufficient), but I think the answer is E(BOTH are not sufficient)

The reasoning in the book is as follows: Statements (1) and (2) together: From statement (1), we know the set has 6 elements and that either x or y must be 20. From (2), we know the average is 20. For a set of 6 elements with an average of 20, the sum of the numbers must be 6x20 = 120.

So, it must be the case that 11 + 15 + 18 + 28 + x + y = 120, or x + y = 48. Since we know that either x or y must be equal to 20, let's say that x = 20. So, we have 20 + y = 48 or y=28. Thus the set is {11, 15, 18, 20, 28, 28}. The range must be 28-17 = 11, so the statements together are sufficient. (C) is correct.

Now, my question is.....the answer takes the set to be {11, 15, 18, 20, 28, 28} and comes to the conclusion C. As per the definition of a set, there should not be any repetitive elements. So the set cannot be {11, 15, 18, 20, 28, 28} with the element 28 being repeated. So the set can be {11, 15, 18, 20, 28}, in which case we cannot determine the value of y. So I am not sure if this question is right.

Please let me know if my thought process is wrong....and what should be the right answer?

Thank you for your time. I shall look forward to hearing from all of you.

regards, Rahul

A set can have repeating values...look up the definition of mode, you will see..

Re: If set Z has a median of 19 , what is the range of set Z? [#permalink]

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15 Sep 2009, 09:44

Good point.

A data set (or dataset) is a collection of data, usually presented in tabular form. In statistics, the mode is the value that occurs the most frequently in a data set or a probability distribution.

Set theory is the branch of mathematics that studies sets, which are collections of objects. There is no repetition of elements in these sets. We have operations like Union, Intersection, Subsets here.

Since the question is related to mean and median, do we assume that the set mentioned here is a dataset? Is this fine?

The GMAT will never try to trick you about technicalities of the definition of a set. It is exceptionally rare for a GMAT question to even mention a 'set' at all; most stats questions are about 'lists', or provide real world data, so that it's clear that repetition of elements is permitted. See Questions 133, 134 and 136 from the DS section of OG12 for typical examples of GMAT-style stats questions. Among all of the 452 questions in the Official Guide, only one (Q67 in the PS section) simply mentions a 'set', and in that question, the sets are given to you, so it doesn't matter if you know whether repetition is permitted in a set. Note that Q199 mentions 'data sets' (though again, the sets are provided, and clearly repetition is allowed), and Q9 and Q31 in the Diagnostic Test at the beginning of the book mention a 'set of performance scores' and a 'set of measurements', respectively, each phrase just another way of saying 'data set'. Again, repetition of elements is clearly allowed in Q31 (it's the only way Statement 2 can be true), and isn't relevant to consider in Q9.

In general, on the GMAT, you're extremely unlikely to see a question even mention a 'set' without some sort of qualification, and in general you should assume that repetition of elements *is* allowed in sets on the GMAT, since in pretty much every GMAT stats question, the question will describe a list or real world situation in which repetition of elements is permitted. The question quoted above from the Kaplan book is not worded in the way a real GMAT question would be; a real GMAT question would call the set 'a set of test scores' or something similar, to make clear that they don't intend the definition of 'set' used in mathematical set theory.
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(1) Average of 3rd and 4th members must be 19. First 3 members of arranged set must be {11, 15, 18…} and either x or y must equal 20, 4th member. Remaining members include 28 and the other variable. Other variable can be between 20 and 28, or it can be greater — nothing in the statement or stem places a restriction on this value. Since we do not know the value of the largest member, Insufficient. (2) Knowing that the median is 19 and that the average is 20 tells us nothing about the value of the highest and lowest members. They could be anything at all. Insufficient (1) & (2) Remaining variable from (1) must make the sum of all the members of Z equal 20 when divided by 6 (by using the average formula). We have one remaining variable for which we can solve. Given this information, we can determine the smallest and largest members, and from there find the range. Sufficient.

Hi, I want to request the clarification of the solution as OE is not very clear to me..

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.

Re: If Set Z has a median of 19, what is the range of Set Z? [#permalink]

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05 Feb 2014, 15:54

1

This post received KUDOS

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.

(1) Average of 3rd and 4th members must be 19. First 3 members of arranged set must be {11, 15, 18…} and either x or y must equal 20, 4th member. Remaining members include 28 and the other variable. Other variable can be between 20 and 28, or it can be greater — nothing in the statement or stem places a restriction on this value. Since we do not know the value of the largest member, Insufficient. (2) Knowing that the median is 19 and that the average is 20 tells us nothing about the value of the highest and lowest members. They could be anything at all. Insufficient (1) & (2) Remaining variable from (1) must make the sum of all the members of Z equal 20 when divided by 6 (by using the average formula). We have one remaining variable for which we can solve. Given this information, we can determine the smallest and largest members, and from there find the range. Sufficient.

Hi, I want to request the clarification of the solution as OE is not very clear to me..

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

(1) Average of 3rd and 4th members must be 19. First 3 members of arranged set must be {11, 15, 18…} and either x or y must equal 20, 4th member. Remaining members include 28 and the other variable. Other variable can be between 20 and 28, or it can be greater — nothing in the statement or stem places a restriction on this value. Since we do not know the value of the largest member, Insufficient. (2) Knowing that the median is 19 and that the average is 20 tells us nothing about the value of the highest and lowest members. They could be anything at all. Insufficient (1) & (2) Remaining variable from (1) must make the sum of all the members of Z equal 20 when divided by 6 (by using the average formula). We have one remaining variable for which we can solve. Given this information, we can determine the smallest and largest members, and from there find the range. Sufficient.

Hi, I want to request the clarification of the solution as OE is not very clear to me..

Merging similar topics. Please search before posting. Thank you.
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(1) Average of 3rd and 4th members must be 19. First 3 members of arranged set must be {11, 15, 18…} and either x or y must equal 20, 4th member. Remaining members include 28 and the other variable. Other variable can be between 20 and 28, or it can be greater — nothing in the statement or stem places a restriction on this value. Since we do not know the value of the largest member, Insufficient. (2) Knowing that the median is 19 and that the average is 20 tells us nothing about the value of the highest and lowest members. They could be anything at all. Insufficient (1) & (2) Remaining variable from (1) must make the sum of all the members of Z equal 20 when divided by 6 (by using the average formula). We have one remaining variable for which we can solve. Given this information, we can determine the smallest and largest members, and from there find the range. Sufficient.

Hi, I want to request the clarification of the solution as OE is not very clear to me..

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...

...

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) ?

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.

If set Z has a median of 19 , what is the range of set Z? [#permalink]

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25 Feb 2015, 00:24

EMPOWERgmatRichC wrote:

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.

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).

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.

Re: If set Z has a median of 19 , what is the range of set Z? [#permalink]

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26 Feb 2015, 07:13

EMPOWERgmatRichC wrote:

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

I never do that. Great advice! will start working on it from now on.

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If set Z has a median of 19 , what is the range of set Z? [#permalink]

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15 Jul 2016, 00:47

rahularke wrote:

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

(1) Z = {18, 28, 11, x, 15, y} (2) The average (arithmetic mean) of Set Z is 20.

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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