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-comedy are there?
I set it up correctly with a Venn Diagram. I am definitely missing something here. My understanding was 5 ,4, and 3 are representing the actual quanities of those sections. However, in the MGMAT book it shows it listed as " 5-x" , 3-x, and 4-x. can someone explain to me why it is an expression and not the actual values provided in the problem?
Thank you!
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MGMAT Chapter 7 word translations # 7 ( video category)
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Anurag@Gurome
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Let us assume that x = no. of movies classified under action, drama and comedy.ashah627 wrote: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-comedy are there?
I set it up correctly with a Venn Diagram. I am definitely missing something here. My understanding was 5 ,4, and 3 are representing the actual quanities of those sections. However, in the MGMAT book it shows it listed as " 5-x" , 3-x, and 4-x. can someone explain to me why it is an expression and not the actual values provided in the problem?
Thank you!
Action = 10 -(3 - x + 5 - x + x) = 2 + x
Drama = 20 -(5 - x + x + 4 - x) = 11 + x
Comedy = 18 -(4 - x + x + 3 - x) = 11 + x
Therefore, 2 + x + 11 + x + 11 + x + 5 - x + 3 - x + 4 - x + x = 38
36 + x = 38
x = 2
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everything's eventual
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Hey Bro...when dealing with these kind of problems you should always start with number of movies classified as all three (i.e. Action, Drama and Comedy).
Let's assume this as x
Now 5 are both in action and drama. Out of these 5 , x are in all three. So number of movies classified as Action and drama but not as comedy = 5-x
Similarly, movies classified as action and comedy but not as drama = 3-x.
Movies classified as drama and comedy but not as action = 4-x
Now assume the following :
Number of movies classified as only action = a
Number of movies classified as only drama = d
Number of movies classified as only comedy = c
So now you have following :
a + c + d + (5 - x) + (4 -x) + (3 - x) +x = 38
a + c + d - 2x = 26 - (i)
You also have following :
a + 8 - x = 10
c + 7 - x = 18
d + 9 - x = 20
Add all these equations and you get,
a +c +d + 24 - 3x = 48
a + c + d - 3x = 24 - (ii)
Solve for equations (i) and (ii) and you get x = 2
Therefore answer is 2.
Let's assume this as x
Now 5 are both in action and drama. Out of these 5 , x are in all three. So number of movies classified as Action and drama but not as comedy = 5-x
Similarly, movies classified as action and comedy but not as drama = 3-x.
Movies classified as drama and comedy but not as action = 4-x
Now assume the following :
Number of movies classified as only action = a
Number of movies classified as only drama = d
Number of movies classified as only comedy = c
So now you have following :
a + c + d + (5 - x) + (4 -x) + (3 - x) +x = 38
a + c + d - 2x = 26 - (i)
You also have following :
a + 8 - x = 10
c + 7 - x = 18
d + 9 - x = 20
Add all these equations and you get,
a +c +d + 24 - 3x = 48
a + c + d - 3x = 24 - (ii)
Solve for equations (i) and (ii) and you get x = 2
Therefore answer is 2.
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alexander.vien
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Much easier way to do this - use the 3 Venn Diagram method that MGMAT teaches. Draw 3 circles each with an overlap for both and with a area for "all 3" in the middle.
Then, simply work from inside/out. Fill in 5, 3, and 4 for all the "both" sections. Then, for each individual circle, just subtract both of the "both" fields in each circle.
So, Drama = 20 - 9 = 11
Comedy = 18 - 7 = 11
Action = 10 - 8 = 2
Now, simply add everything up.
11 + 11 + 2 + 5 + 3 + 4 = 36
We know that there are 38 movies, so "all 3" must be 2.
Then, simply work from inside/out. Fill in 5, 3, and 4 for all the "both" sections. Then, for each individual circle, just subtract both of the "both" fields in each circle.
So, Drama = 20 - 9 = 11
Comedy = 18 - 7 = 11
Action = 10 - 8 = 2
Now, simply add everything up.
11 + 11 + 2 + 5 + 3 + 4 = 36
We know that there are 38 movies, so "all 3" must be 2.
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Solution:ashah627 wrote: ↑Sun Sep 09, 2012 5:30 pmThe 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-comedy are there?
I set it up correctly with a Venn Diagram. I am definitely missing something here. My understanding was 5 ,4, and 3 are representing the actual quanities of those sections. However, in the MGMAT book it shows it listed as " 5-x" , 3-x, and 4-x. can someone explain to me why it is an expression and not the actual values provided in the problem?
Thank you!
We can use the following formula:
Total = Group A + Group B + Group C - Both A and B - Both A and C - Both B and C + All three + None
In this problem, Total is 38, Group A (action) is 10, Group B (drama) is 20 and Group C (comedy) is 18. There are 5, 3 and 4 movies in the double overlaps, and None = 0. Let’s denote “All three” by n and fill in this information in the formula:
38 = 10 + 20 + 18 - 5 - 3 - 4 + n + 0
38 = 48 - 12 + n
n = 38 - 36 = 2
Answer: B
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