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Re: If set B is a subset of Set A, how many elements are in set [#permalink]
28 Nov 2012, 22:37

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. _________________

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]
28 Nov 2012, 22:50

Expert's post

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]
28 Nov 2012, 23:45

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]
29 Nov 2012, 00:18

Expert's post

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 2723 times ]

Re: If set B is a subset of Set A, how many elements are in set [#permalink]
29 Nov 2012, 00:33

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]
25 Jan 2013, 23:41

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)

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