Overlapping Sets :

[Khalid]

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]

This question is already dealt with on the forum.

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]

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]

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]

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]

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]

I meant second formula will get you there in one step

[JonathanSchneider]

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]

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]

The formula always works. The problem is you have set up an impossible scenario..

[vikramsumer]

[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]

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..

[vikramsumer]

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]

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]

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..