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Twenty-four dogs are in a kennel. Twelve of the dogs are

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Twenty-four dogs are in a kennel. Twelve of the dogs are  [#permalink]

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New post 05 Dec 2009, 02:35
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Twenty-four dogs are in a kennel. Twelve of the dogs are black, six of the dogs have short tails, and fifteen of the dogs have long hair. There is only one dog that is black with a short tail and long hair. Two of the dogs are black with short tails and do not have long hair. Two of the dogs have short tails and long hair but are not black. If all of the dogs in the kennel have at least one of the mentioned characteristics, how many dogs are black with long hair but do not have short tails?

I'll write solution later.
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Re: Overlapping Sets Problems  [#permalink]

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New post 05 Aug 2010, 02:38
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nonameee wrote:
I am completely baffled by this question.

When I solve it using the formula:
# of items total = (a + b + c) - (ab + ac + bc) + (abc)

I get 6

When I solve it the same way as Bullet, I get 3.

Can someone please suggest, where the mistake is?

Thanks.

bibha wrote:
When do we use these formula?>
total = A+B+C - (AnB+BnC+AnC) + Neither + (AnBnC)
total = A+B+C - (sum of exactly two groups) - 2(AnBnC) +Neither



My post from formulae-for-3-overlapping-sets-69014.html?hilit=exactly%20groups

I'd advise to understand the concept rather than to memorize formulas. Look at the diagram below:
Attachment:
Union_3sets.gif
Union_3sets.gif [ 11.63 KiB | Viewed 15279 times ]


FIRST FORMULA
We can write the formula counting the total as: \(Total=A+B+C-(AnB+AnC+BnC)+AnBnC+Neither\).

When we add three groups A, B, and C some sections are counting more than once. For instance: sections 1, 2, and 3 are counted twice and section 4 thrice. Hence we need to subtract sections 1, 2, and 3 ONCE (to get one) and subtract section 4 TWICE (again to count section 4 only once).

Now, in the formula AnB means intersection of A and B (sections 3 and 4 on the diagram). So when we are subtracting \(AnB\) (3 an4), \(AnC\) (1 and 4), and \(BnC\) (2 and 4) from \(A+B+C\), we are subtracting sections 1, 2, and 3 ONCE BUT section 4 THREE TIMES (and we need to subtract section 4 only twice), therefor we should add only section 4, which is intersection of A, B and C (AnBnC) again. That is how thee formula \(Total=A+B+C-(AnB+AnC+BnC)+AnBnC+Neither\) is derived.

SECOND FORMULA
The second formula you are referring to is: \(Total=A+B+C -\){Sum of Exactly 2 groups members} \(- 2*AnBnC + Neither\). This formula is often written incorrectly on forums as Exactly 2 is no the same as intersection of 2 and can not be written as AnB. Members of exactly (only) A and B is section 3 only (without section 4), so members of A and B only means members of A and B and not C. This is the difference of exactly (only) A and B (which can be written, though not needed for GMAT, as \(AnB-C\)) from A and B (which can be written as \(AnB\))

Now how this formula is derived?

Again: when we add three groups A, B, and C some sections are counting more than once. For instance: sections 1, 2, and 3 are counted twice and section 4 thrice. Hence we need to subtract sections 1, 2, and 3 ONCE (to get one) and subtract section 4 TWICE (again to count section 4 only once).

When we subtract {Sum of Exactly 2 groups members} from A+B+C we are subtracting sum of sections 1, 2 and 3 once (so that's good) and next we need to subtracr ONLY section 4 (\(AnBnC\)) twice. That's it.

Now, how this concept can be represented in GMAT problem?

Example #1:
Workers are grouped by their areas of expertise, and are placed on at least one team. 20 are on the marketing team, 30 are on the Sales team, and 40 are on the Vision team. 5 workers are on both the Marketing and Sales teams, 6 workers are on both the Sales and Vision teams, 9 workers are on both the Marketing and Vision teams, and 4 workers are on all three teams. How many workers are there in total?

Translating:
"are placed on at least one team": members of none =0;
"20 are on the marketing team": M=20;
"30 are on the Sales team": S=30;
"40 are on the Vision team": V=40;
"5 workers are on both the Marketing and Sales teams": MnS=5, note here that some from these 5 can be the members of Vision team as well, MnS is sections 3 an 4 on the diagram (assuming Marketing=A, Sales=B and Vision=C);
"6 workers are on both the Sales and Vision teams": SnV=6 (the same as above sections 2 and 4);
"9 workers are on both the Marketing and Vision teams": MnV=9.
"4 workers are on all three teams": MnSnV=4, section 4.

Question: Total=?

Applying first formula as we have intersections of two groups and not the number of only (exactly) 2 group members.

Total=M+S+V-(MnS+SnV+SnV)+MnSnV+Neither=20+30+40-(5+6+9)+4+0=74.

Answer: 74.

Example #2:
Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?

Translating:
"Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs"
Total=59;
Neither=0 (as members are required to sign up for a minimum of one);
"22 students sign up for the poetry club": P=22;
"27 students for the history club": H=27;
"28 students for the writing club": W=28;
"6 students sign up for exactly two clubs": {Exactly 2 groups members}=6, so sum of sections 1, 2, and 3 is given to be 6, (among these 6 students there are no one who is the member of ALL 3 clubs)

"How many students sign up for all three clubs": question is \(PnHnW=x\). Or section 4 =?

Apply second formula: \(Total=P+H+W -\){Sum of Exactly 2 groups members}\(-2*PnHnW + Neither\) --> \(59=22+27+28-6-2*x+0\) --> \(x=6\).

Answer: 6.

Similar problem at: ps-question-94457.html#p728852

BACK TO THE ORIGINAL QUESTION:

Twenty-four dogs are in a kennel. Twelve of the dogs are black, six of the dogs have short tails, and fifteen of the dogs have long hair. There is only one dog that is black with a short tail and long hair. Two of the dogs are black with short tails and do not have long hair. Two of the dogs have short tails and long hair but are not black. If all of the dogs in the kennel have at least one of the mentioned characteristics, how many dogs are black with long hair but do not have short tails?

Total dogs: \(T=24\);
Black dogs: \(B=12\);
With short tails: \(S=6\);
With long hair: \(L=15\);
Only one dog that is black with a short tail and long hair: \(B&S&L=1\);
Only \(B&S=2\);
Only \(S&L=2\);
Only \(B&L=x\).

As we are given members of exactly 2 groups then use second formula:

\(Total=T+B+S -(Sum \ of \ Exactly \ 2 \ groups \ members)- 2*TnBnS + Neither\) --> \(24=12+6+15 -(x+2+2)- 2*1+0\) --> \(x=3\).

Hope it helps.
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Re: Overlapping Sets Problems  [#permalink]

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New post 05 Dec 2009, 03:15
Bullet wrote:
Twenty-four dogs are in a kennel. Twelve of the dogs are black, six of the dogs have short tails, and fifteen of the dogs have long hair. There is only one dog that is black with a short tail and long hair. Two of the dogs are black with short tails and do not have long hair. Two of the dogs have short tails and long hair but are not black. If all of the dogs in the kennel have at least one of the mentioned characteristics, how many dogs are black with long hair but do not have short tails?

I'll write solution later.


OA- 3
given Black (B) dogs =12
Dogs with short tails (ST) = 6
Dogs with long hair (LH) = 15
n(B ∩ ST ∩ LH) = 1
n(B ∩ ST) = 2
n(ST ∩ LH) = 2

let x be n(B ∩ LH) which we need to find out
n(only ST) = 1
n(only B) = 9-x [ n(B) - n(B ∩ ST ∩ LH) - n(B ∩ ST) - n(B ∩ LH)]
n(only ST) = 1 [n(ST) - n(B ∩ ST ∩ LH) - n(B ∩ ST) - n(ST ∩ LH)]
n(only LH) = 12-x [n(LH) - n(B ∩ ST ∩ LH) - n(B ∩ ST) - n(ST ∩ LH)]

now we know all of the dogs in the kennel have at least one of the mentioned characteristics so

n(only B) + n(only ST) + n(only LH) + n(B ∩ ST ∩ LH) + n(B ∩ ST) + n(B ∩ LH) + n(ST ∩ LH) = 24
9 -x + 1 + 12 -x + 1 + 2 + x + 2 = 24
27-x = 24
x=3
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Re: Overlapping Sets Problems  [#permalink]

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New post 05 Dec 2009, 04:25
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Bullet wrote:
Twenty-four dogs are in a kennel. Twelve of the dogs are black, six of the dogs have short tails, and fifteen of the dogs have long hair. There is only one dog that is black with a short tail and long hair. Two of the dogs are black with short tails and do not have long hair. Two of the dogs have short tails and long hair but are not black. If all of the dogs in the kennel have at least one of the mentioned characteristics, how many dogs are black with long hair but do not have short tails?

I'll write solution later.


The answer is indeed 3.

12+6+15-4-x-2(1) = 24
which implies x = 3.
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Re: Overlapping Sets Problems  [#permalink]

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New post 05 Dec 2009, 05:04
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Attachment:
24-Dogs in Kannel.GIF
24-Dogs in Kannel.GIF [ 10.16 KiB | Viewed 16216 times ]


9 - x + 2 + 1 + 1 + 2 + x + 12 - x = 24

27 - x = 24

x = 3
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Re: Overlapping Sets Problems  [#permalink]

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New post 08 Dec 2009, 09:22
I am curious to know why the answer is not 7.

Total = Group1 + Group2 + Group3 - Group1/2 - Group1/3 - Group2/3 + 2(Group1/Group2/Group3)

24 = 12+6+15-2-2-x+2
24 = 33-4-x+2
24 = 29-x+2
22 = 29-x

x=7

Can someone point out my mistake?
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Re: Overlapping Sets Problems  [#permalink]

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New post 08 Dec 2009, 21:44
ManUnited27 wrote:
I am curious to know why the answer is not 7.

Total = Group1 + Group2 + Group3 - Group1/2 - Group1/3 - Group2/3 + 2(Group1/Group2/Group3)

24 = 12+6+15-2-2-x+2
24 = 33-4-x+2
24 = 29-x+2
22 = 29-x

x=7

Can someone point out my mistake?


ManU,

I reckon the formula that you are using is not correct -

The formula to use is -
Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither

If you put the values in the above, the answer comes out to be 3.
See my earlier post above.
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Re: Overlapping Sets Problems  [#permalink]

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New post 10 Dec 2009, 14:16
This wasnt complex but very time consuming. Good problem, though.
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Re: Overlapping Sets Problems  [#permalink]

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New post 05 Aug 2010, 01:18
I am completely baffled by this question.

When I solve it using the formula:
# of items total = (a + b + c) - (ab + ac + bc) + (abc)

I get 6

When I solve it the same way as Bullet, I get 3.

Can someone please suggest, where the mistake is?

Thanks.
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Re: Overlapping Sets Problems  [#permalink]

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New post 05 Aug 2010, 01:28
When do we use these formula?>
total = A+B+C - (AnB+BnC+AnC) + Neither + (AnBnC)
total = A+B+C - (sum of exactly two groups) - 2(AnBnC) +Neither
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Re: Overlapping Sets Problems  [#permalink]

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New post 07 Aug 2010, 07:07
Bunuel,
This is great. Thank you soooo much :-)
I hope not to make mistakes again with overlapping sets problem.
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Re: Overlapping Sets Problems  [#permalink]

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New post 08 Aug 2010, 23:15
Quote:
As we are given members of exactly 2 groups then use second formula:


Thank you, Bunuel. I knew how to derive the formula. But I've only now realized that we have the number of exactly two groups and not two groups including the intersection of three sets. Very careless of me.

Thank you again.
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Re: Twenty-four dogs are in a kennel. Twelve of the dogs are  [#permalink]

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