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Khalid

Post subject: Overlapping Sets : Posted: Wed Dec 19, 2007 3:41 am 


source : MGMAT CAT
In a group of 68 students, each student is registered for at least one of three classes  History, Math and English. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
13
10
9
8
7
I know there is explanation provided for this but I am not getting the steps.
I started from inside out. There are 3 thhat are in all three classes so
a = number of students in both History and Math
b = number of students in both History and English
c = number of students in both Math and English
For brevity I will only show the first step because herein lies my confusion.
Students in Math Only = 25 (a 3+3+b3)
The explantion says Students in Math = 25 (a+b+3) and this is the part that I am missing. Why would we not subtract 3 from the number of students that have both classes?
Can someone please shed some light? Thanks





shaji

Post subject: Re: Overlapping Sets : Posted: Thu Dec 20, 2007 12:29 am 


This question is already dealt with on the forum.
Khalid wrote: source : MGMAT CAT
In a group of 68 students, each student is registered for at least one of three classes  History, Math and English. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes? 13 10 9 8 7
I know there is explanation provided for this but I am not getting the steps.
I started from inside out. There are 3 thhat are in all three classes so
a = number of students in both History and Math b = number of students in both History and English c = number of students in both Math and English
For brevity I will only show the first step because herein lies my confusion.
Students in Math Only = 25 (a 3+3+b3)
The explantion says Students in Math = 25 (a+b+3) and this is the part that I am missing. Why would we not subtract 3 from the number of students that have both classes?
Can someone please shed some light? Thanks





RonPurewal

Post subject: Posted: Fri Dec 21, 2007 4:44 am 


ManhattanGMAT Staff 

Posts: 12809

one extremely efficient way to attack this problem is to use the following formula, which works for all combinations of three sets a, b, c:
# of items total = (a + b + c)  (ab + ac + bc) + (abc)
in the formula, 'a' stands for the # of items in set a (regardless of whether they are in other sets as well), 'ab' for the # of items that are in both sets a and b (again, regardless of whether they're in c), and so forth.
note that you cannot use this formula if you are given information with exclusivity (for instance, the number of items in set a and b but not in set c).
in this problem, you are looking for the value of (ab + ac + bc  3abc). notice that you have to subtract abc 3 times, because those items have been tallied three times (they are in 'ab', they are in 'ac', and they are also in 'bc'). since you already have abc = 3, all you need is the value of (ab + ac + bc).
using the formula,
68 = (25 + 25 + 34)  (ab + ac + bc) + 3
(ab + ac + bc) = 19
answer = 19  3(3) = 10
you can also solve this problem with venn diagrams and associated reasoning. hit up this thread again if you want more details about that.





Guest

Post subject: Posted: Fri Jun 13, 2008 7:29 pm 


RPurewal wrote: one extremely efficient way to attack this problem is to use the following formula, which works for all combinations of three sets a, b, c:
# of items total = (a + b + c)  (ab + ac + bc) + (abc)
in the formula, 'a' stands for the # of items in set a (regardless of whether they are in other sets as well), 'ab' for the # of items that are in both sets a and b (again, regardless of whether they're in c), and so forth.
note that you cannot use this formula if you are given information with exclusivity (for instance, the number of items in set a and b but not in set c).
in this problem, you are looking for the value of (ab + ac + bc  3abc). notice that you have to subtract abc 3 times, because those items have been tallied three times (they are in 'ab', they are in 'ac', and they are also in 'bc'). since you already have abc = 3, all you need is the value of (ab + ac + bc).
using the formula, 68 = (25 + 25 + 34)  (ab + ac + bc) + 3 (ab + ac + bc) = 19 answer = 19  3(3) = 10
you can also solve this problem with venn diagrams and associated reasoning. hit up this thread again if you want more details about that.
How do you get 3(3) in the final step of the problem?
''answer = 19  3(3) = 10''





RonPurewal

Post subject: Posted: Sun Jun 15, 2008 10:44 pm 


ManhattanGMAT Staff 

Posts: 12809

Anonymous wrote: How do you get 3(3) in the final step of the problem? ''answer = 19  3(3) = 10''
quoting from the relevant part of that solution:
notice that you have to subtract abc 3 times, because those items have been tallied three times (they are in 'ab', they are in 'ac', and they are also in 'bc')
i.e., you don't want those items at all, but they've "accidentally" been counted three times in the existing part of the calculation. therefore, if you subtract them out three times, you're good.





Nick

Post subject: two ways to look at it Posted: Sat Dec 20, 2008 11:47 am 


You can look at this in two ways
Total = H + M + E  (H intersect M)  (H intersect E)  (M intersect E) + (H intersect M intersect E)
Total = H + M + E  only (H intersect M)  only (H intersect E)  only (M intersect E)  2 * (H intersect M intersect E)
only (H intersect M) implies
(H intersect M)  (H intersect M intersect E)
The first one will get you there in one step.
In this question "only" two is asked
Often on GMAT exam, "only" two is given, so keep formula two in mind





Nick

Post subject: Posted: Sat Dec 20, 2008 11:49 am 


I meant second formula will get you there in one step





JonathanSchneider

Post subject: Posted: Thu Dec 25, 2008 9:08 pm 


ManhattanGMAT Staff 

Posts: 373

Yes, Nick, if you can memorize that second formula, go for it. It includes the same info, just in a different way. For those of you who don't want to memorize it, no worries; memorizing the formula Ron layed out will get you there almost as quickly and in more cases.





anjali.gmat

Post subject: Re: Posted: Thu Nov 17, 2011 4:57 pm 


Course Students 

Posts: 2

RonPurewal wrote: one extremely efficient way to attack this problem is to use the following formula, which works for all combinations of three sets a, b, c:
# of items total = (a + b + c)  (ab + ac + bc) + (abc)
in the formula, 'a' stands for the # of items in set a (regardless of whether they are in other sets as well), 'ab' for the # of items that are in both sets a and b (again, regardless of whether they're in c), and so forth.
note that you cannot use this formula if you are given information with exclusivity (for instance, the number of items in set a and b but not in set c).
in this problem, you are looking for the value of (ab + ac + bc  3abc). notice that you have to subtract abc 3 times, because those items have been tallied three times (they are in 'ab', they are in 'ac', and they are also in 'bc'). since you already have abc = 3, all you need is the value of (ab + ac + bc).
using the formula, 68 = (25 + 25 + 34)  (ab + ac + bc) + 3 (ab + ac + bc) = 19 answer = 19  3(3) = 10
you can also solve this problem with venn diagrams and associated reasoning. hit up this thread again if you want more details about that. Hi Ron, let's say that the qs is: In a group of 68 students, each student is registered for at least one of three classes  History, Math and English. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour students are registered for English. If 10 students are registered for all ' two classes', how many students are registered for exactly three classes? Why doesn't the formula work here? I get the following! 68 = 84 10 + abc abc = 6





tim

Post subject: Re: Overlapping Sets : Posted: Tue Dec 13, 2011 2:45 am 


ManhattanGMAT Staff 

Posts: 4923 Location: Southwest Airlines, seat 21C

The formula always works. The problem is you have set up an impossible scenario..
_________________ Tim Sanders Manhattan GMAT Instructor





vikramsumer

Post subject: Re: Overlapping Sets : Posted: Sun Apr 01, 2012 8:12 am 


Forum Guests 

Posts: 9

[color=#FF0080] I am facing a challenge with the same question. The equation I made was as follows.
25  (a+b+3) + 25  (a+c+3) + 34  (b+c+3) = 68
This equation seems incorrect, but I do not understand why. Can anyone please help !
a = intersection between H and M b = intersection between H and E c= intersection between E and M





tim

Post subject: Re: Overlapping Sets : Posted: Sun Apr 08, 2012 4:13 pm 


ManhattanGMAT Staff 

Posts: 4923 Location: Southwest Airlines, seat 21C

if by "the same question" you mean the question i already said doesn't work, then the reason you're having trouble with it is the same reason the last person had trouble  the question doesn't work. if you're referring to a different question, please identify it and explain your steps as well so we can help you find the error..
_________________ Tim Sanders Manhattan GMAT Instructor





vikramsumer

Post subject: Re: Overlapping Sets : Posted: Mon Apr 09, 2012 4:56 am 


Forum Guests 

Posts: 9

tim wrote: if by "the same question" you mean the question i already said doesn't work, then the reason you're having trouble with it is the same reason the last person had trouble  the question doesn't work. if you're referring to a different question, please identify it and explain your steps as well so we can help you find the error.. The question is the same, but can you elaborate on what you mean by the question doesn't work. Do you mean to say that I am using the incorrect formula ?.. (As mentioned, the question is form CAT 1) My steps are as follows Total no of Students =68 History = 25 Math = 25 English = 34 History and Math only = a History and English only = b English and maths only = c (a,b,c does not contain the 3 common for all) The last step,the equation 68 = 25  (a+b+3) + 25  (a+c+3) + 34  (b+c+3) This is what I did, and I need your help to find the flaw in this approach.





la2ny

Post subject: Re: Overlapping Sets : Posted: Mon Apr 09, 2012 3:55 pm 


Course Students 

Posts: 14

Ron,
Can you please show the method using Venn diagrams, as that is the way I normally attack these problems dealing with 3 sets.





tim

Post subject: Re: Overlapping Sets : Posted: Tue Apr 17, 2012 7:37 pm 


ManhattanGMAT Staff 

Posts: 4923 Location: Southwest Airlines, seat 21C

Hi Vikram, i see that you have reverted to the original question; that's what i was asking about, since when you said the "same" question i initially thought you referred to the impossible question someone else had posed. The problem here is you're just using the wrong formula. Reread the thread and use the right formula, and you should be fine..
la2ny, the formula discussed in this thread is exactly the Venn diagram approach. Draw the Venn diagram and you'll see..
_________________ Tim Sanders Manhattan GMAT Instructor





