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Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
27 Jun 2006, 02:53
00:00
A
B
C
D
E
Difficulty:
85% (hard)
Question Stats:
38% (02:10) correct
62% (01:10) wrong based on 65 sessions
Set A, B, C have some elements in common. If 16 elements are in both A and B, 17 elements are in both A and C, and 18 elements are in both B and C, how many elements do all three of the sets A, B, and C have in common?
(1) Of the 16 elements that are in both A and B, 9 elements are also in C (2) A has 25 elements, B has 30 elements, and C has 35 elements.
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#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.
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
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
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
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
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
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
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
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.
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
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 ?
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
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..
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
28 Jun 2006, 16:09
1
This post received KUDOS
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
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
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)
_________________
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.
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
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)
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
30 Jun 2006, 07:56
amansingla4 wrote:
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
Hmmm let me see. Say, total is 100 people, 60 bought Apples, 50 bought Bananas, 35 bought Cranberries. If we know that number of people who bought all three is 10, then we know 25 people bought exactly two of the three, and that 35 people bought more than one fruit (or at least two fruits).
However, in this case we do not know how many people bought A&B, A&C and B&C exactly. It might be the case that 15 people bought A&B, 20 people bought B&C and 20 people bought A&C. The point that needs to be noticed is that when we say 20 people bought B&C they may have or have not bought A as well. In our case 10 of the 20 actually bought all three. I know this sometimes can be very confusing. You just need to make sure if you are talking about "exactly two" or "at least two". Making a Venn gram will help most of the time. _________________
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.
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
10 Dec 2014, 05:49
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