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Sets A,B and C have some elements in common.If 16 elements

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Sets A,B and C have some elements in common.If 16 elements [#permalink] New post 27 Jun 2006, 02:53
Sets A,B and C have some elements in common.If 16 elements are in both A&B,17 in A&C ,18 elements are in B&C, how many elements do all the three A,B,C have in common?
1) Of the 16 elements in A&B,9 are in C.
2) A has 25 elements,B has 30 and c has 35 elements

Please explain.
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 [#permalink] New post 27 Jun 2006, 03:51
A it is
(1) alone : the number of common elements of all 3 sets is 9
(2) is insufficient
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 [#permalink] New post 27 Jun 2006, 04:47
Edited

Got B with Venn Diagram. Thanks to Yurik79.

Last edited by humans on 27 Jun 2006, 07:52, edited 1 time in total.
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 [#permalink] New post 27 Jun 2006, 05:02
Is it B.

St1 seems incorrect to me. If 16 elements are common how can just 9 be in C? All 16 should be in B also
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 [#permalink] New post 27 Jun 2006, 06:25
B for me
using venn diagram
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 [#permalink] New post 27 Jun 2006, 06:27
It is got to be A.
There are 9 elements common to all sets.

B is insuff for want of n(AUBUC)
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 [#permalink] New post 27 Jun 2006, 06:59
Should be A. 9 common elements in A,B, and C.
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 [#permalink] New post 27 Jun 2006, 07:04
Lets slightly change the statement; what if we were given the total # of elements?
Then the ans would be D right?
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Re: Another Set theory? [#permalink] New post 27 Jun 2006, 08:17
amansingla4 wrote:
Sets A,B and C have some elements in common.If 16 elements are in both A&B,17 in A&C ,18 elements are in B&C, how many elements do all the three A,B,C have in common?
1) Of the 16 elements in A&B,9 are in C.
2) A has 25 elements,B has 30 and c has 35 elements

Please explain.


(A) it is.
I implies 9 elements are common between A, B, C
For II, we still need the total # of elements..
This is what I could deduce from the venn diagram
Total = A + B + C - AB - BC - CA - 2ABC
We have everything but Total and ABC.
Cannot calculate ABC without knowing hte total.
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 [#permalink] New post 27 Jun 2006, 12:05
shampoo wrote:
Lets slightly change the statement; what if we were given the total # of elements?
Then the ans would be D right?


Yes.

There is a formula for this. Let us say Common elements of sets A and B is denoted by Int(A,B) [It is called 'Intersection' in Set Theory], combined elements of sets A and B are denoted by Un(A,B) ['Union' in Set Theory], and number of elements in set A is denoted by n[A], then

n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+n[Int(A,B,C)]

So here you have n[Un(A,B,C)]=25+30+35-16-17-18+n[Int(A,B,C)]

If we have the total number of elements in A, B, C combined then the answer will be D.

In the absence of that vital piece of data, however, we have A. I cannot see how a Venn diagram can help us with that. Can someone explain?
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 [#permalink] New post 27 Jun 2006, 12:50
paddyboy wrote:
shampoo wrote:
Lets slightly change the statement; what if we were given the total # of elements?
Then the ans would be D right?


Yes.

There is a formula for this. Let us say Common elements of sets A and B is denoted by Int(A,B) [It is called 'Intersection' in Set Theory], combined elements of sets A and B are denoted by Un(A,B) ['Union' in Set Theory], and number of elements in set A is denoted by n[A], then

n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+n[Int(A,B,C)]

So here you have n[Un(A,B,C)]=25+30+35-16-17-18+n[Int(A,B,C)]

If we have the total number of elements in A, B, C combined then the answer will be D.

In the absence of that vital piece of data, however, we have A. I cannot see how a Venn diagram can help us with that. Can someone explain?


A venn diagram does help with visualizing what is needed, but does not help solving. Statement A provides this information upfront.
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 [#permalink] New post 27 Jun 2006, 13:34
paddyboy wrote:
shampoo wrote:
Lets slightly change the statement; what if we were given the total # of elements?
Then the ans would be D right?


Yes.

There is a formula for this. Let us say Common elements of sets A and B is denoted by Int(A,B) [It is called 'Intersection' in Set Theory], combined elements of sets A and B are denoted by Un(A,B) ['Union' in Set Theory], and number of elements in set A is denoted by n[A], then

n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+n[Int(A,B,C)]

So here you have n[Un(A,B,C)]=25+30+35-16-17-18+n[Int(A,B,C)]

If we have the total number of elements in A, B, C combined then the answer will be D.

In the absence of that vital piece of data, however, we have A. I cannot see how a Venn diagram can help us with that. Can someone explain?


Shouldn't it be :
n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+2*n[Int(A,B,C)]
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 [#permalink] New post 27 Jun 2006, 17:03
Have to agree with majority -- I think it is (A)

(I) is obvious, but got confused a little bit by (II). But once you write the formula

Total = A + B + C - AB - AC -BC + ABC

everything gets into places. Assuining statement (II), in the above formula we know everything but Total and ABC, so don't have enough info to determine ABC.
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 [#permalink] New post 27 Jun 2006, 18:25
paddyboy wrote:
shampoo wrote:
Lets slightly change the statement; what if we were given the total # of elements?
Then the ans would be D right?


Yes.

There is a formula for this. Let us say Common elements of sets A and B is denoted by Int(A,B) [It is called 'Intersection' in Set Theory], combined elements of sets A and B are denoted by Un(A,B) ['Union' in Set Theory], and number of elements in set A is denoted by n[A], then

n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+n[Int(A,B,C)]

So here you have n[Un(A,B,C)]=25+30+35-16-17-18+n[Int(A,B,C)]

If we have the total number of elements in A, B, C combined then the answer will be D.

In the absence of that vital piece of data, however, we have A. I cannot see how a Venn diagram can help us with that. Can someone explain?


Thanks for the explanation.
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 [#permalink] New post 27 Jun 2006, 18:32
Its A. Using a Venn Diagram helps a lot to visualize.
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 [#permalink] New post 28 Jun 2006, 14:36
v1rok wrote:
Have to agree with majority -- I think it is (A)

(I) is obvious, but got confused a little bit by (II). But once you write the formula

Total = A + B + C - AB - AC -BC + ABC

everything gets into places. Assuining statement (II), in the above formula we know everything but Total and ABC, so don't have enough info to determine ABC.


I am still confused by the above eqn.
It appears that,
total shd be = A + B + C - AB - AC -BC -2ABC

When we subtract AB, BC, CA from total area, we remove the overlapping part, but still the area common to all three of them is counted thrice .. We need to subtract it twice to get the real total.

Anyone who can explain which one is correct and why ?
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 [#permalink] New post 28 Jun 2006, 16:07
sgrover wrote:
v1rok wrote:
Have to agree with majority -- I think it is (A)

(I) is obvious, but got confused a little bit by (II). But once you write the formula

Total = A + B + C - AB - AC -BC + ABC

everything gets into places. Assuining statement (II), in the above formula we know everything but Total and ABC, so don't have enough info to determine ABC.


I am still confused by the above eqn.
It appears that,
total shd be = A + B + C - AB - AC -BC -2ABC

When we subtract AB, BC, CA from total area, we remove the overlapping part, but still the area common to all three of them is counted thrice .. We need to subtract it twice to get the real total.

Anyone who can explain which one is correct and why ?



Total = A + B + C - AB - AC -BC + ABC - is correct..
Draw a Venn Diagram, name each part 1-7.. and then verify..
this will give you the above formula..
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 [#permalink] New post 28 Jun 2006, 16:09
sgrover wrote:
v1rok wrote:
Have to agree with majority -- I think it is (A)

(I) is obvious, but got confused a little bit by (II). But once you write the formula

Total = A + B + C - AB - AC -BC + ABC

everything gets into places. Assuining statement (II), in the above formula we know everything but Total and ABC, so don't have enough info to determine ABC.


I am still confused by the above eqn.
It appears that,
total shd be = A + B + C - AB - AC -BC -2ABC

When we subtract AB, BC, CA from total area, we remove the overlapping part, but still the area common to all three of them is counted thrice .. We need to subtract it twice to get the real total.

Anyone who can explain which one is correct and why ?


No this is correct.

In A+B+C we included AB, AC and BC twice so we need to subtract each of these once.
Now we have A+B+C-AB-BC-CA.

But in A+B+C we we also included ABC three times so we need to subtract ABC two times.
Now we have A+B+C-AB-BC-CA-2ABC.

But in subtracting AB, AC and BC we subtracted ABC three times. So need to add 3ABC

Finally we have A+B+C-AB-BC-CA-2ABC+3ABC i.e
A+B+C-AB-BC-CA+ABC

Try it with marking the areas in a Venn Diagram.

Hope this helps.
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 [#permalink] New post 28 Jun 2006, 17:53
See if this helps (from the basic priniciple sticky):

HongHu wrote:
Formula:

Total = N(A) + N(B) + N(C) - N(A n B) - N(A n C) - N(C n B) + N(A n B n C)

If instead of numbers for (A n B) and (A n C) and (C n B), what is given is the total number of people who choose exactly two items, then the formula becomes:

Total = N(A) + N(B) + N(C) - (N(choose exactly two items)) - 2N(choose all three items)

Also,
Total = N(A) + N(B) + N(C) - (N(choose at least two items)) - N(choose all three items)

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 [#permalink] New post 29 Jun 2006, 00:40
HongHu wrote:
See if this helps (from the basic priniciple sticky):

HongHu wrote:
Formula:

Total = N(A) + N(B) + N(C) - N(A n B) - N(A n C) - N(C n B) + N(A n B n C)

If instead of numbers for (A n B) and (A n C) and (C n B), what is given is the total number of people who choose exactly two items, then the formula becomes:

Total = N(A) + N(B) + N(C) - (N(choose exactly two items)) - 2N(choose all three items)

Also,
Total = N(A) + N(B) + N(C) - (N(choose at least two items)) - N(choose all three items)



Can you please explain these two formulas?

Regards,
Aman
  [#permalink] 29 Jun 2006, 00:40
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