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).

Consider this example:

A: [1,2, 2, 3,3, 4,5,6,7,8,9,10,11,12,13,14,15]
B: [1, 2, 3]

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.
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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
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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!
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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
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Sometimes the student writes the question wrong, then it is unuseful in some how.

But for me as the question is posed per se, the answer is C

Basically it says: 14 = 0.2 x........I do not see a particular assumption here or inference
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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)

Hence choice(C) should be the answer.

Feel free to correct if anything wrong!


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