Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Sets A,B and C have some elements in common.If 16 elements [#permalink]
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

Re: Another Set theory? [#permalink]
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.

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

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

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

(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.

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

(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 ?

(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..

(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

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)

_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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)