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# Of the 40 people who are raise pets, 30 raise dogs, 15 raise

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Of the 40 people who are raise pets, 30 raise dogs, 15 raise [#permalink]

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06 Jan 2005, 14:02
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Of the 40 people who are raise pets, 30 raise dogs, 15 raise cats, and 5 raise birds. Only one people raise all the three kinds of pets. How many people raise two kinds of pets?

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VP
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06 Jan 2005, 14:33
8 !

1. 30-1(raise all the three kinds of pets)=29
2. 15-1(raise all the three kinds of pets)=14
3. 5-1(raise all the three kinds of pets)=4
4. 29+14+4+1=48
5. 48-40=8

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Director
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06 Jan 2005, 15:12
I used the 3 set Venn diagram and got the below Equation

xyz = only 2 types

u v w = only one type

x + y + z + u + v + w = 40 - 1 = 39

x + y + z + 47 -2x-2y-2z = 39

x + y + z = 47 - 39 = 8

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Senior Manager
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06 Jan 2005, 15:18
christoph wrote:
8 !

1. 30-1(raise all the three kinds of pets)=29
2. 15-1(raise all the three kinds of pets)=14
3. 5-1(raise all the three kinds of pets)=4
4. 29+14+4+1=48
5. 48-40=8

from step 4: 29+14+4+1=48
we can tell the 1 person who raise up 3 pets still count as one who raise 2 pets, right?

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Intern
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06 Jan 2005, 20:51
8. It can be easily done with a vein diagram.

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Manager
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08 Jan 2005, 11:21
Use simple formula,

n(AUBUC) equals to

n(A) + n(B) + n(C) - n(A/\B) - n(B/\C) - n(A/\C) + n(A/\B/\C)

where U -- union
/\ -- intersection

therefore

40 equals to 35+15+5 - X + 2

solving for X

I won't recommend drawing venn diagram for these type of questions...
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Bhimsen Joshi

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08 Jan 2005, 12:36
[quote="Bhimsen"]Use simple formula,

n(AUBUC) equals to

n(A) + n(B) + n(C) - n(A/\B) - n(B/\C) - n(A/\C) + n(A/\B/\C)

where U -- union
/\ -- intersection

therefore

40 equals to 35+15+5 - X + 2

solving for X

I won't recommend drawing venn diagram for these type of questions...[/quote]

Not getting ur calculation , how do you get X = 8 from ur eqn ?

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08 Jan 2005, 13:07
I am very sorry...
goofed up totally....

40 eq 30+15+5-X+1

X eq 11 is the ans...
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Bhimsen Joshi

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08 Jan 2005, 15:13
Assume three circles, one for a dog owners, one for cat owners, and one for bird owners.

Then 1 = the place where all three intersect.
Then let b=dog (INTERSECT) cat, a = dog (INTERSECT) bird, and c = CAT (intersect) bird.

Then 30 dog owners = 1 for all + a + b + (exculsively dog owners);

Then.
30 + 15 - (1 + b) + 4 - (1 + a + c ) = 40

(Subtracting out the items we would be double counting, after dogs are done)

Then, a+b+c ( Number of dual owners) = 7

I am getting 7, can someone tell me if this method is valid?

Thanks[/img]

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09 Jan 2005, 10:27
Bhimsen's approach seems to be good. 11 seems right. i wonder whats wrong with 8.

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Manager
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16 Jan 2005, 07:20
Bhimsen,

Correction.The formula should be:
n(AUBUC) equals to

n(A) + n(B) + n(C) - n(A/\B) - n(B/\C) - n(A/\C) - 2* n(A/\B/\C)
+++++
This is so as A/\B/\C happens twice.Thus you need subtract it twice.Per this forumla the answer will be:
40 = 30+15+5 - n(A/\B) - n(B/\C) - n(A/\C) - 2*1
n(A/\B) - n(B/\C) - n(A/\C) = 50-2-40=8
++++

Any thoughts?Rgds,

Anna

Bhimsen wrote:
Use simple formula,

n(AUBUC) equals to

n(A) + n(B) + n(C) - n(A/\B) - n(B/\C) - n(A/\C) + n(A/\B/\C)

where U -- union
/\ -- intersection

therefore

40 equals to 35+15+5 - X + 2

solving for X

I won't recommend drawing venn diagram for these type of questions...

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We can crack the exam together

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16 Jan 2005, 07:20
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