The problem is always better with a Venn Diagram, where there are three circle denoting Action (A), Comedy (C) and Drama (D)
The hunt is to find the potboilers i.e. intersection of A , C & D.
10 action, 20 drama, and 18 comedy.
5 are both action and drama,
3 are both action and comedy,
4 are both drama and comedy
So to find purely Action movies = 10 -(5+3) = 2
purely Drama - 20-(5+4) = 11
purely Comedy = 18 - (3+4) = 11
So total no of movies - movies which are purely of one genre - movies classifiied under 2 categories = Movies falling under all three categories i.e. action-drama-comedies .
So the no of potboilers = 38-5-3-4-2-11-11 = 2

Thank you. I re-read the question again. I thought it was asking for the total and not the intersection.

I still dont get this.... i dont think this can be solved coz if we add 2 in the middle of the venn diagram then the total number of movies which are classified as Action or Drama or Comedy = 32

Only Action = 10-5-3-2 = 0
Action & Drama = 5
Action & Comedy = 3
Only Drama = 20-5-4-2 =9
Drama & Comedy = 4
Only Comedy = 18-4-3-2 = 9
A and D and C = 2
AorDorC = 0+5+3+9+4+9 +2 = 32

Which means 38 - 32 or 6 movies have no classification, however the original problem states that all 38 have been classified into either one of the categories.

dbaner1 wrote:Can someone pls. explain me the solution to the problem. This in an In Action Problem in the Word Translation book in MGMAT. Thanks.

The 38 movies in the video store fall into the following three categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action and drama, 3 are both action and comedy, and 4 are both drama and comedy. How many action-drama-comedies are there?

Tried this problem and I got 24 as the answer

OA 2

Venn diagram happens to be the ultimate resort to such problems. Have three intersecting circles A (= 10), D (= 20), and C (= 18). Start filling it from its interior most part (i.e. take x for A ∩ D ∩ C, which is also the question to be answered here), and have 5 - x for A & D only, 4 - x for D & C only, 3 - x for C & A only, 2 + x for A only, 11 + x for D only, and finally, again have 11 + x for C only, as well. Now, feel free to write

2 + x + 11 + x + 11 + x + 3 - x + 5 - x + 4 - x + x = 38

Or [spoiler]x = 2[/spoiler].