Math questions from any Manhattan GMAT Computer Adaptive Test.
Khalid
 
 

Overlapping Sets :

by Khalid 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. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four 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+b-3)

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
 
 

Re: Overlapping Sets :

by shaji 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. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four 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+b-3)

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
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by RonPurewal Fri Dec 21, 2007 4:44 am

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
 
 

by Guest 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
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by RonPurewal Sun Jun 15, 2008 10:44 pm

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
 
 

two ways to look at it

by Nick 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
 
 

by Nick Sat Dec 20, 2008 11:49 am

I meant second formula will get you there in one step
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by JonathanSchneider Thu Dec 25, 2008 9:08 pm

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
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Re:

by anjali.gmat Thu Nov 17, 2011 4:57 pm

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. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four 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
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Re: Overlapping Sets :

by tim Tue Dec 13, 2011 2:45 am

The formula always works. The problem is you have set up an impossible scenario..
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Re: Overlapping Sets :

by vikramsumer Sun Apr 01, 2012 8:12 am

[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
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Re: Overlapping Sets :

by tim Sun Apr 08, 2012 4:13 pm

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..
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Re: Overlapping Sets :

by vikramsumer Mon Apr 09, 2012 4:56 am

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.
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Re: Overlapping Sets :

by la2ny Mon Apr 09, 2012 3:55 pm

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
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Re: Overlapping Sets :

by tim Tue Apr 17, 2012 7:37 pm

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

Follow this link for some important tips to get the most out of your forum experience:
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