So. I realized I love Venn Diagrams (Yes, I'm a nerd, but who cares?). I've explained a lot of problems using Venn Diagrams and a forum user PMed me to explain how I drew my Venn Diagrams since he was getting confused with them. I am posting this as an informal and rough guide to how I visualize problems that make use of Venn Diagrams. Here's the question he asked: At a certain school, each of the 150 students takes between 1 and 3 classes. The 3 classes available are Math, Chemistry and English. 53 students study math, 88 study chemistry and 58 study english. If 6 students take all 3 classes, how many take exactly 2 classes? A. 37 B. 43 C. 45 D. 60 Step 1: Deconstruction This is where you extract the given information from a problem. Use a symbol for each section given to you. I usually use the first letter in caps. Please remember at this point that any number specified without explicitly mentioning that it's "exactly" or "only" for a certain subject is not to be taken so. For instance, in this case, it says 53 math students. This doesn't mean students who are taking only 1 subject (Math). This could include students who are taking Math and Chemistry or Math and English or all three too. So now break down the numbers given to us. Total = 150 M (Total) = 53 C (Total) = 88 E (Total) = 58 MCE = 6 We are asked to find MC + CE + EM. Step 2: Drawing the diagram ALWAYS start from the center of the Venn Diagram wherever information is available. This will make life infinitely easier. In this case, the center is the intersection of all three circles, i.e MCE = 6 So fill that in to the diagram you've drawn. VD1.jpg Now that you've gotten that, let's start filling in a variable for each section not known to us. Here, consider each letter to represent only that specific section and not the entire circle or a larger portion. VD1.jpg So now that you have the parts filled in, what you need to do is write down what you have in the diagram in terms of numbers. So we are given the totals for each subject. Look at math first. There are four types of people taking math (each group of these people mutually exclusive, and not in common with any other group) 1. Only math: a 2. Math and Chemistry: y 3. Math and English: x 4. All three: 6 So now represent this as a sum and you get a+x+y+6 = 53 and hence a+x+y = 47 Similarly for the other subjects you get: b+x+z+6 = 58 and hence b+x+z = 52 c+y+z+6 = 88 and hence c+y+z = 82 And then you have the total: x+y+z+a+b+c+6=150 and hence x+y+z+a+b+c = 144 Step 3: Solution This is perhaps the most intuitive part, but in my experience the first part of this step is the same in all overlapping sets problem. It's only what's asked for that's different. Add all the individual equations together to get a combined equation with all the variables and a number. So you get: 2(x+y+z) + a+b+c = 47+52+82 = 181 Rearranging this to get a+b+c we get a+b+c = 181 - 2(x+y+z) Substitute this into the total equation we derived earlier saying x+y+z+a+b+c = 144 so you get: x+y+z+181 - 2(x+y+z)= 144 Upon rearranging this you get: x+y+z = 181-144 = 37 which is option A the right answer. Hope this helps some of you! Please post a reply if there's anything else you want to know about my explanation or anything else.

I really loved the way how you solved this problem and it has helped me immensely. However I spot a mistake at the end of your post. 171 - 144 = 27, but 27 is not one of the answers. Edit: My bad, didn't see that other people have pointed this out! Sorry~

For overlapping sets check the following posts: ADVANCED OVERLAPPING SETS PROBLEMS Overlapping Sets Made Easy! How to draw a Venn Diagram Data Sufficiency Questions on Overlapping Sets Problem Solving Questions on Overlapping Sets

Data Sufficiency Questions on Overlapping Sets Problem Solving Questions on Overlapping Sets

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How to draw a Venn Diagram for problems

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Thedevils88

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30 Jul 2013, 21:53

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I really loved the way how you solved this problem and it has helped me immensely. However I spot a mistake at the end of your post.

whiplash2411

$x+y+z+171 - 2(x+y+z)= 144$

Upon rearranging this you get:

[m]x+y+z = 171-144 = 37[/m]which is option A the right answer.

Hope this helps some of you!

Please post a reply if there's anything else you want to know about my explanation or anything else.

171 - 144 = 27, but 27 is not one of the answers. :shock:

Edit: My bad, didn't see that other people have pointed this out! Sorry~

anujshah

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Thanks for the post whiplash2411. However, there is an easier and faster way to do this.

Math = 53
Chemistry = 88
English = 58
Total = 199

This total counts x, y, z twice (students taking 2 classes) and 6 thrice (students taking 3 classes). We need to count these only once to arrive at the number of students taking exactly 2 classes.

First get rid of the double counting by subtracting the actual number of students from the total above. Deal with the triple counting in the next step.

Total = 199
No. of students =(150)
Remaining = 49

Since 6 has been counted thrice, subtract 6 twice (6*2 = 12) from the remainder to count it exactly once.

Remaining = 49
Subtract 12 =(12)
Final = 37

Thus, the number of students taking exactly 2 classes is 37.

Hope this helps

manhattan187

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Great explanation!

Its amazing how complex word problems can come down to simple solutions (after you learn the step by step process, of course!)

1 - (88+53+58)-3(6) = 181

2 - (150-6) = 144

(1)-(2) =37.

Bunuel

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19 Nov 2015, 23:45

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manhattan187

For overlapping sets check the following posts:

tringuyenminh293

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Love your work. Can you post for examples of venn diagram since I've found it very often in real test

Bunuel

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Love your work. Can you post for examples of venn diagram since I've found it very often in real test

kaleem765

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Loved the explanation. See the attachment: this is how I approached this question.

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