Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Not quite understanding this. Am I headed in the right direction at least with this set table?

Hii Anon. the question asks the # of elements in set A. The only fact given is that all elements of set B are in set A. Consider Statement 1) It says that set B has 14 elements. With this 14 elements, how can you deduce the # of elements in set A. Not sufficient at all. consider statement 2) It says 80% elements of set A are not part of set B. What else does this information provides? It tells us that there are 20% elements in set A that are common with set B i.e. the number of elements in set B are 20 % of set A. BUT we don't know the # of elements of set B.

On combining, we know that 20% elements of A=14. Therefore the number of elemements in A can easily be found out. +1C _________________

Re: If set B is a subset of Set A, how many elements are in set [#permalink]

Show Tags

28 Nov 2012, 23:37

1

This post received KUDOS

Marcab wrote:

Hii Anon. the question asks the # of elements in set A. The only fact given is that all elements of set B are in set A. Consider Statement 1) It says that set B has 14 elements. With this 14 elements, how can you deduce the # of elements in set A. Not sufficient at all. consider statement 2) It says 80% elements of set A are not part of set B. What else does this information provides? It tells us that there are 20% elements in set A that are common with set B i.e. the number of elements in set B are 20 % of set A. BUT we don't know the # of elements of set B.

On combining, we know that 20% elements of A=14. Therefore the number of elemements in A can easily be found out. +1C

This need not be true!

I would argue that the OA is incorrect if its C. The answer must be E. There could be a repetition of elements in B (or A).

In this case B is subset of A. Exactly 80% of elements of A are not part of B. (out of 15 unique elements, 12 are not part of B)

But knowing that B has 3 elements would in no way help you to find out number of elements in A.There is no such constraints on A or B that elements could not be repeated.

The answer must be E in my opinion. _________________

Re: If set B is a subset of Set A, how many elements are in set [#permalink]

Show Tags

28 Nov 2012, 23:50

Vips0000 wrote:

Marcab wrote:

Hii Anon. the question asks the # of elements in set A. The only fact given is that all elements of set B are in set A. Consider Statement 1) It says that set B has 14 elements. With this 14 elements, how can you deduce the # of elements in set A. Not sufficient at all. consider statement 2) It says 80% elements of set A are not part of set B. What else does this information provides? It tells us that there are 20% elements in set A that are common with set B i.e. the number of elements in set B are 20 % of set A. BUT we don't know the # of elements of set B.

On combining, we know that 20% elements of A=14. Therefore the number of elemements in A can easily be found out. +1C

This need not be true!

I would argue that the OA is incorrect if its C. The answer must be E. There could be a repetition of elements in B (or A).

In this case B is subset of A. Exactly 80% of elements of A are not part of B. (out of 15 unique elements, 12 are not part of B)

But knowing that B has 3 elements would in no way help you to find out number of elements in A.There is no such constraints on A or B that elements could not be repeated.

The answer must be E in my opinion.

Your opinion would have been true if the question were HOW MANY NUMBERS ARE THERE IN SET A? If we are asked to count # of elements in your set A, then the answer would equal to 17 but if we are asked how many numbers are there in A, then I would answer 15. There is a difference between numbers and elements. _________________

Re: If set B is a subset of Set A, how many elements are in set [#permalink]

Show Tags

29 Nov 2012, 00:45

Marcab wrote:

Vips0000 wrote:

Marcab wrote:

Hii Anon. the question asks the # of elements in set A. The only fact given is that all elements of set B are in set A. Consider Statement 1) It says that set B has 14 elements. With this 14 elements, how can you deduce the # of elements in set A. Not sufficient at all. consider statement 2) It says 80% elements of set A are not part of set B. What else does this information provides? It tells us that there are 20% elements in set A that are common with set B i.e. the number of elements in set B are 20 % of set A. BUT we don't know the # of elements of set B.

On combining, we know that 20% elements of A=14. Therefore the number of elemements in A can easily be found out. +1C

This need not be true!

I would argue that the OA is incorrect if its C. The answer must be E. There could be a repetition of elements in B (or A).

In this case B is subset of A. Exactly 80% of elements of A are not part of B. (out of 15 unique elements, 12 are not part of B)

But knowing that B has 3 elements would in no way help you to find out number of elements in A.There is no such constraints on A or B that elements could not be repeated.

The answer must be E in my opinion.

Your opinion would have been true if the question were HOW MANY NUMBERS ARE THERE IN SET A? If we are asked to count # of elements in your set A, then the answer would equal to 17 but if we are asked how many numbers are there in A, then I would answer 15. There is a difference between numbers and elements.

Probably you'd agree with the definition of subset as: ■Set A is called subset of B if every element of A is also an element of B. We write it as AB (read as "A is a subset of B" or "A is contained in B"). In such a case, we say BA ("B is a superset of A" or "B contains A").

Therefore, if A= [1,2,3,4,5] and B= [2,2,2,2,2] will have subset -superset relation. Try to use the calculation done above on this. B is subset. B has 5 elements as per your calculation? right? exactly 80% of elements of A (4 out of 5) are not in B. So How many elements in A? 25?

Anyway, just presented my opinion! _________________

Re: If set B is a subset of Set A, how many elements are in set [#permalink]

Show Tags

29 Nov 2012, 01:18

Thanks for sharing the information. That was helpful. But the answer has to be C.

Vips0000 wrote:

Consider this example:

A: [1,2, 2, 3,3, 4,5,6,7,8,9,10,11,12,13,14,15] The question says that eactly 80% of the elements of set A are not in set B. Also as per the information provided in the previous post, the number of elements in A are 15. So # of elements which are NOt in set B is 12. These 12 numbers comprise the 80 % of entire bigger circle. So the left out 20% is in smaller circle B: [1, 2, 3]

So, on coming to the question, the 14 elements are in set B. These 14 elements are not part of 80% of the elements of set A. So these 14 elements is the 20% of the entire bigger circle. Clear C. Whats the issue?

Attachments

solution anaon.png [ 7.47 KiB | Viewed 4333 times ]

Re: If set B is a subset of Set A, how many elements are in set [#permalink]

Show Tags

29 Nov 2012, 01:33

Marcab wrote:

Thanks for sharing the information. That was helpful. But the answer has to be C.

Vips0000 wrote:

Consider this example:

A: [1,2, 2, 3,3, 4,5,6,7,8,9,10,11,12,13,14,15] The question says that eactly 80% of the elements of set A are not in set B. Also as per the information provided in the previous post, the number of elements in A are 15. So # of elements which are NOt in set B is 12. These 12 numbers comprise the 80 % of entire bigger circle. So the left out 20% is in smaller circle B: [1, 2, 3]

So, on coming to the question, the 14 elements are in set B. These 14 elements are not part of 80% of the elements of set A. So these 14 elements is the 20% of the entire bigger circle. Clear C. Whats the issue?

Yes, figure drawn by you is perfect! But, the conclusion drawn is not. 14 elemenets are not part of 80% doesnt imply that rest 20% =14. It only means that 14 elements of B are made of 20% elements of A. Same as following: A= [1,2,3,4,5] B=[2,2,2,2,2,2,2....,2] (14 times)

Thus it is not sufficient and must be E.

if it is indeed Gmatprep question then also it could be flawed. there are a few incorrect questions in gmatprep. for example check this post by none other than Bunuel. all-the-clients-that-company-x-had-at-the-beginning-of-last-128023.html?hilit=gmatprep#p1049002 In fact I've seen other posts by Bunuel too where he has identified incorrect gmatprep questions and correctly so. problem is finding out that particular post from his 9000+ posts. May be someday I'll stumble upon such a post again _________________

Re: If set B is a subset of Set A, how many elements are in set [#permalink]

Show Tags

26 Jan 2013, 00:41

Vips0000's concern is valid. Repetition of an element in the set can lead to further complexity around this problem. Now, the question whether to count only distinct elements/numbers or count element separately including repetitions remains open and debatable. I see several varying viewpoints around Set problems prepared by Prep companies or even in theories.

Still I believe using the same concept raised by Vips0000, we should arrive at the answer(C).

Lets take a look at below explanation (Courtesy UCSD - Maths dept) -- See attached paper if you want to dig down. Since a set is an un-ordered collection of distinct objects, the following all describe the same 3-element set {a, b, c} = {b, a, c} = {c, b, a} = {a, b, b, c, b}. The first three are simply listing the elements in a different order. The last happens to mention some elements more than once. But, since a set consists of distinct objects, the elements of the set are still just a, b, c. Thus 3 elements.

Another way to think of this is: Two sets A and B are equal if and only if every element of A is an element of B and every element of B is an element of A. Thus, with A = {a, b, c} and B = {a, b, b, c, b}, we can see that everything in A is in B and everything in B is in A. --Read further in the chapter if you wish to.

---------------------------- Back to our problem

Even if we count repeated elements in the set only once (i.e. count distinct) - by applying same concept - we should be able to find out how many elements are set A using statements (1) and (2). A={1, 2, 2, 3, 4, 4, 5, 5, 5} B={2,2,2,2,2}

Here set A has 5 elements (though repeated) and set B contains 1 element (i.e. 2 - though repeated) -> Hence 80% of of elements in set A are not in set B.

Similarly, if (statement 1) "Set B has 14 elements" and (statement 2) "Exactly 80% of the elements in set A are not in Set B" -> A should have 70 (distinct) elements, whether or not they are repeated inside the set. Note that question asks for 'how many elements?' -> 70 (not counting repeats)

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

Sometimes Mom comes into town, you meet her at the airport to surprise her. Shenanigans ensue. You grab dinner and chat. You don’t write a long blog post that...