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very intersting MGMAT set thery question

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very intersting MGMAT set thery question

by ritz » Tue Apr 08, 2008 2:29 pm
I came across this question..
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?
19
13
10
8
7

Let me know how do you solve it.

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by xilef » Tue Apr 08, 2008 3:10 pm
total registrations 25+25+34=84

84 registrations with 68 students = 16 repeats

3 students take 3 classes that means you have 6 duplicate entries

16-6=10 duplicate entries so 10 students registered for 2 classes
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by Stuart@KaplanGMAT » Tue Apr 08, 2008 4:33 pm
xilef wrote:total registrations 25+25+34=84

84 registrations with 68 students = 16 repeats

3 students take 3 classes that means you have 6 duplicate entries

16-6=10 duplicate entries so 10 students registered for 2 classes

Great solution!

We could also apply a general formula for triple overlap questions:

True # of items = #group1 + #group2 + #group3 + #nogroup - 2(#in all 3) - (#in groups 1&2) - (#in groups 2&3) - (#in groups 1&3)

In this question, we know that:

True # = 68
# in all 3 = 3
#group1 = 25
#group2 = 25
#group3 = 34
#nogroup = 0 (everyone is registered in at least one class)
#in two groups = x (in this question we don't care about specific pairs, just pairs in general)

So:

68 = 25 + 25 + 34 + 0 - 2(3) - x
68 = 84 - 6 - x
68 = 78 - x
-10 = -x
x = 10

You can also solve using Venn diagrams, which are a visual representation of the forumla.
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by lunarpower » Thu Apr 10, 2008 2:37 am
Stuart Kovinsky wrote:We could also apply a general formula for triple overlap questions:

True # of items = #group1 + #group2 + #group3 + #nogroup - 2(#in all 3) - (#in groups 1&2) - (#in groups 2&3) - (#in groups 1&3)
dude!
no!

well ok, we could apply this formula ... if we wanted to get wrong answers.

this formula is especially nasty because it's inconsistent with itself (!). namely, it's equivocal:
** '#group1', '#group2', and '#group3', in this formula, are inclusive: they include all members of those groups, regardless of whether they also belong to other groups.
** (#in groups 1&2), (#in groups 2&3), (#in groups 1&3), though, are exclusive: they include only the members who are in JUST the two specified groups, and not in the other group.

you can't use a formula that equivocates like this; it's a recipe for disaster. it happens to work on this particular problem, but that's an artifact of the way the problem is written.

--

here's the actual formula:

if you use INCLUSIVE definitions (e.g., '#group1' includes people in group 1 and other groups), as is the norm:
True # of items = #group1 + #group2 + #group3 + #nogroup + (#in all 3) - (#in groups 1&2) - (#in groups 2&3) - (#in groups 1&3)
note that the plus sign, replacing the minus two, is the only change.
** if you apply this formula to the problem at hand, then you have to write (x + 9) where stuart wrote x, because, under these definitions, you are now counting the 3 people who are in all three groups, three times each.

if you use EXCLUSIVE definitions, you just add them all up, because then your categories correspond exactly to the different regions of the venn diagram.
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by lunarpower » Thu Apr 10, 2008 2:44 am
it's also worth noting that, while it's worth knowing the corresponding formula for two variables

# total = (# in group 1) + (# in group 2) + (# in no group) - (# in both groups),

it's not really worth your time to learn any of the above formulas for three groups. on any official problem featuring three groups, the odds are excellent that you'll be able to solve the problem perfectly well with the combination of wits + venn diagram.

--

i've also noticed that '#nogroup', in official problems, is practically always zero.
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by dextar » Thu Apr 10, 2008 3:42 am
lunarpower wrote:
Stuart Kovinsky wrote:We could also apply a general formula for triple overlap questions:

True # of items = #group1 + #group2 + #group3 + #nogroup - 2(#in all 3) - (#in groups 1&2) - (#in groups 2&3) - (#in groups 1&3)
dude!
no!

well ok, we could apply this formula ... if we wanted to get wrong answers.

this formula is especially nasty because it's inconsistent with itself (!). namely, it's equivocal:
** '#group1', '#group2', and '#group3', in this formula, are inclusive: they include all members of those groups, regardless of whether they also belong to other groups.
** (#in groups 1&2), (#in groups 2&3), (#in groups 1&3), though, are exclusive: they include only the members who are in JUST the two specified groups, and not in the other group.

you can't use a formula that equivocates like this; it's a recipe for disaster. it happens to work on this particular problem, but that's an artifact of the way the problem is written.

--

here's the actual formula:

if you use INCLUSIVE definitions (e.g., '#group1' includes people in group 1 and other groups), as is the norm:
True # of items = #group1 + #group2 + #group3 + #nogroup + (#in all 3) - (#in groups 1&2) - (#in groups 2&3) - (#in groups 1&3)
note that the plus sign, replacing the minus two, is the only change.
** if you apply this formula to the problem at hand, then you have to write (x + 9) where stuart wrote x, because, under these definitions, you are now counting the 3 people who are in all three groups, three times each.

if you use EXCLUSIVE definitions, you just add them all up, because then your categories correspond exactly to the different regions of the venn diagram.
Hi if u apply the formula
True # of items = #group1 + #group2 + #group3 + #nogroup + (#in all 3) - (#in groups 1&2) - (#in groups 2&3) - (#in groups 1&3)
68=25+25+34+3-x
x=19

Where am I wrong?
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by camitava » Thu Apr 10, 2008 4:45 am
In this case, I am agree with Ron. I believe the formula is -
A + B + C - (A n B) - (B n C) - (C n A) + (A n B n C) dextar, too me 19 is coming as the answer to select.
Correct me If I am wrong


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by Stuart@KaplanGMAT » Thu Apr 10, 2008 11:21 am
Dude! No!

I understand taking issue with the specific language that I used, but I don't agree that the equation is wrong.

Why would you add in people who have already been triple counted? We don't want to count them AGAIN, we want to account for duplication (or triplication, as the case may be).

So, for the nitpicky people, let me re-express my equation (which I agree is something you don't really need to know for the GMAT, so feel free to ignore this thread entirely after the first posted solution!):

True # of items = (# of items with quality 1) + (# of items with quality 2) + (# of items with quality 3) + (# of items with none of qualities 1, 2 and 3) - (# of items with only qualities 1&2) - (# of items with only qualities 2&3) - (# of items with only qualities 1&3) - 2(# of items with qualities 1, 2 & 3)

Every item with exactly 2 qualities has been counted twice. Every item with all 3 qualities has been counted 3 times, hence the multiplier of 2 for that term.

The reason that I express the equation in this form is simple: it relates to the kind of information that's usually included in GMAT 3-quality questions!

In the question posted, the question is how many people are in EXACTLY two groups, so my original equation (with languge revisions for those who were looking for an excuse to criticize :D ) applies.
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by lunarpower » Thu Apr 10, 2008 11:41 am
Stuart Kovinsky wrote:Dude! No!

I understand taking issue with the specific language that I used, but I don't agree that the equation is wrong.

Why would you add in people who have already been triple counted? We don't want to count them AGAIN, we want to account for duplication (or triplication, as the case may be).

So, for the nitpicky people, let me re-express my equation (which I agree is something you don't really need to know for the GMAT, so feel free to ignore this thread entirely after the first posted solution!):

True # of items = (# of items with quality 1) + (# of items with quality 2) + (# of items with quality 3) + (# of items with none of qualities 1, 2 and 3) - (# of items with only qualities 1&2) - (# of items with only qualities 2&3) - (# of items with only qualities 1&3) - 2(# of items with qualities 1, 2 & 3)

Every item with exactly 2 qualities has been counted twice. Every item with all 3 qualities has been counted 3 times, hence the multiplier of 2 for that term.

The reason that I express the equation in this form is simple: it relates to the kind of information that's usually included in GMAT 3-quality questions!

In the question posted, the question is how many people are in EXACTLY two groups, so my original equation (with languge revisions for those who were looking for an excuse to criticize :D ) applies.
ok, we're on the same page; it's more a semantics issue than anything else.

camitava & others: note the following text buried in my post above:
** if you apply this formula to the problem at hand, then you have to write (x + 9) where stuart wrote x, because, under these definitions, you are now counting the 3 people who are in all three groups, three times each.
...in other words, the GENERAL formula is much more difficult to apply to this particular problem.

also, let me point out the most important part of what stuart has said:
The reason that I express the equation in this form is simple: it relates to the kind of information that's usually included in GMAT 3-quality questions!

absolutely true.

so, NOTE: stuart's formula works admirably for solving this problem, and indeed many, if not most, 3-set problems on the gmat (whose wordings tend to be restricted in the same bizarre way).

the only danger with stuart's formula is as follows: before using stuart's formula, just verify that 'group1&2', etc. are EXCLUSIVE; in other words, that they refer to the members of EXACTLY two groups. this is something that can't be implied; the evidence must be found explicitly in the problem statement. (in this problem, the evidence is '...exactly two classes')

if you find no such evidence in the problem statement, then apply the general formula found in my post.

sweet
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query

by greatchap » Thu Jul 03, 2008 3:15 am
I have a query guys and will be glad if anyone can correct me.

I am using the following method -

AUBUC = A + B + C -A∩B - B∩C - C∩A + A∩B∩C + none.

ok, here A = history, b=math, c=english
As per question -

68 = 25 + 25 + 34 - a - b - c + 3
68 - 50 - 34 - 3 = -a -b -c
68 - 87 = - a - b - c

thus a + b + c = 19...

Now since this 19 has 3 included three times we need to elimate 3 thats included twice. So we do 19 - 3*2 => 19 - 6 or 13
but the answer is 10. Why ??? In the ques (another sets ques) below the above method did work.

[In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?
A) 5
B) 10
C) 15
D) 20
E) 25 ]

[Here 85 = 50 + 30 + 20 -a -b -c + 5 thereby a + b + c = 20 and we did 20 - 5*2 thus 10 answer.] [Here 5 was included 3 times but we subtracted twice only.]
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Re: query

by lunarpower » Thu Jul 03, 2008 3:27 am
one:
greatchap wrote:AUBUC = A + B + C -A∩B - B∩C - C∩A + A∩B∩C + none.
nope.
if you want a formula for A U B U C, then you don't include 'none', because those people/elements/whatever are NOT part of the union of the three sets.

if you want a formula for the total number of people/elements, INCLUDING the ones who aren't part of ANY of the three sets (i.e., the ones who are not part of A U B U C), only then do you add in "none".

two:
greatchap wrote:Now since this 19 has 3 included three times we need to elimate 3 thats included twice. So we do 19 - 3*2 => 19 - 6 or 13
but the answer is 10. Why ???
actually, you want to eliminate all three instances of those 3 people, because you don't want to include those three in your count.
read the problem carefully: it says "EXACTLY two classes", not "at least two classes". this means that the students who are registered for all three classes - i.e., the 3 people you're talking about - shouldn't be included at all.
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Thanks a lot

by greatchap » Thu Jul 03, 2008 6:05 am
Thanks a ton lunarpower, for taking out time to help me out.

From what you mentioned - "if you want a formula for A U B U C, then you don't include 'none', because those people/elements/whatever are NOT part of the union of the three sets."

You mean that till the question does not ask for people/objects who are not in any set, I should not use none even if a figure is given. Thereby if the questions asks how many people are not in any set/whatever then only i should use none.

If that is correct then okay as my answer becomes correct. Without none I can say -

68 - 84 = - a - b - c
so a + b + c = 16 (and not 19 as before)

Thus 16 - 3*2 = 10...correct. But here again I am doing 16 minus 3*2. Means out of 3 people included 3 times, i m removing 6 people and not 9 (all) like you said. Is this okay??
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Re: Thanks a lot

by lunarpower » Thu Jul 03, 2008 10:40 pm
greatchap wrote:If that is correct then okay as my answer becomes correct. Without none I can say -

68 - 84 = - a - b - c
so a + b + c = 16 (and not 19 as before)

Thus 16 - 3*2 = 10...correct. But here again I am doing 16 minus 3*2. Means out of 3 people included 3 times, i m removing 6 people and not 9 (all) like you said. Is this okay??
nope. in fact, ironically, you actually didn't use 'none' in your post above. here's a quote of that post:
greatchap wrote:AUBUC = A + B + C -A∩B - B∩C - C∩A + A∩B∩C + none.

ok, here A = history, b=math, c=english
As per question -

68 = 25 + 25 + 34 - a - b - c + 3
68 - 50 - 34 - 3 = -a -b -c
68 - 87 = - a - b - c

thus a + b + c = 19...
note that you've plugged in for everything except 'none' (the 3 at the end of the formula is A∩B∩C), so this 19 is the formula without 'none'. once you find this quantity, subtract 3x3 to give 10.

in fact, you couldn't include 'none' in this problem if you tried, because that information isn't even given in the problem statement.

hope that clears things up.
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help

by greatchap » Fri Jul 04, 2008 6:08 am
Sorry, I misunderstood A∩B∩C for none.

Infact I am a bit confused now. Can I know the difference between EXACTLY two classes", not "at least two classes or three classes.

Last to last post that you wrote, you mentioned that dont include none. Well I did not include none so what was that for??

I am still unable to figure out the difference between the current ques & the example I gave -

[In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?
A) 5
B) 10
C) 15
D) 20
E) 25 ]

[Here 85 = 50 + 30 + 20 -a -b -c + 5 thereby a + b + c = 20 and we did 20 - 5*2 thus 10 answer.] [Here 5 was included 3 times but we subtracted twice only.]

What I wanna know is how to approach such a ques. Should we use formula AUBUC = A + B + C -A∩B - B∩C - C∩A + A∩B∩C + none.

And why in our current ques we remove 3*3 (9) students and in ques (example one) above we do 20 - 5*2 or remove 10.
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by jsl » Fri Jul 04, 2008 8:16 am
xilef wrote:3 students take 3 classes that means you have 6 duplicate entries
Great solution... however, the above has been driving me crazy for ages... How did you calculate 6? I understand that there were 3 students so in my venn diagram, I have 3 in the centre. However, where does the other 3 come from? I noticed this was one of the trap answers too!

If someone could explain, I'd be really thankful.
Jon
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