In a group of 68 students, each student is registered for at

tonebeeze Bunuel's explanation at the link he provides was very helpful for me. I recommend sticking to the equations as the venn diagrams tend to throw me off.

Can this question be done with the help if matrix?
If yes, then please let me know the procedure.

The matrix method of 2 X 2 works well for two overlapping sets.
For three overlapping sets we would need a 3D matrix of 2 X 2 X 2, which is hard (if not impossible) to visualize on a 2D paper or computer screen.

So, stick with the Venn diagrams or the formulas.

Why can't a 3x3 matrix be used here?

We have 3 types of classes, call them A, B, C.
Each student can be or not in A - 2 possibilities, can be or not in B, also 2 possibilities, can be or not in C, another two possibilities.
So there is a total of 2 x 2 x 2 = 8 different subsets and not 3 x 3 = 9.

If for two overlapping sets A and B you would use a 2 x 2 matrix (A nonA, B nonB), you don't have where to put the information regarding the third set C.
Imagine a cube of 2 X 2 X 2, such that you add the third characteristic related to C on the vertical axis, above the base of 2 x 2 for A and B.
The 3D cube of dimensions 2 X 2 x 2 for three overlapping sets is the parallel of the 2D matrix of dimensions 2 X 2 for two overlapping sets.

Thanks. That has helped.

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Suppose A=all 3
B=History and english
C=history and math
D=math and english
E=english only
F=history only
G=math only

we know that:
A+B+C+D+E+F+G=68
A+B+C+F=25
A+C+D+G=25
A+B+D+E=34
A=3

OK, so we have
B+C+F=22
C+D+G=22
B+D+E=34

add all these 3:
2B+2C+2D+E+F+G=75
we then have:
B+C+D+E+F+G=65

substract from first one the second one:
B+C+D=10
B,C,D - 2 objects only.

The master information collation on all things Overlapping Sets. Thanks to this(and the authors ) for making life simpler
overlapping-sets-made-easy-205636.html

Solution:
Let student who attended exactly
1 course= X, 2 courses =Y, 3 courses = Z (GIVEN Z=3)

Total = X+Y+Z
68= X+Y+3
X+Y=65 ---------(1)

SUM of all the courses :

h + m + e = X +2Y +3Z : Reason : every course considered common between 2 is repeated 2 times, while Z is repeated 3 times : each times whenever we take 2 courses together ( x is in anb , bnc, anc)
25+25+34 = X+2Y +3(3)
84-9 = X+2Y
X+2Y = 75
X+Y=65 FROM EQUATION (1)
solving this we get : Y=10

Hi All,

3-Group Overlapping Sets questions are relatively rare on the Official GMAT (you likely will NOT see this version of Overlapping Sets on Test Day). However, there is a formula that you can use to solve it.

Total = (1st group) + (2nd group) + (3rd group) - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd) - 2(all 3 groups).

In overlapping sets questions, any person who appears in more than one group has been counted more than once. When dealing with groups of people, you're not supposed to count any individual more than once, so the formula 'subtracts' all of the extra times that a person is counted. For example, someone who is in BOTH the 1st group and the 2nd group will be counted twice....that's why we SUBTRACT that person later on [in the (1st and 2nd) group].

In this prompt, we're given the Total, a number for each of the 3 individual groups and the number of people who appear in all 3 groups. The equation would look like this...

68 = 25 + 25 + 34 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)- 2(3)

68 = 84 - 6 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)

68 = 78 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)

(1st and 2nd) + (1st and 3rd) + (2nd and 3rd) = 10

Since the prompt asks for the total number of students that are in exactly 2 classes, we have our answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi guys, I actually got the right answer for this problem, but wanted to confirm if my method of solving it is actually accurate or if I just "got lucky."

History 25
Math 25
English 34

Subtract 3 from each number is 3 people are in all three classes

History 25-3 = 22
Math 25-3 = 22
English 34-3 = 31

Additionally, subtract 3 from the original 68 number since you want to "get rid of them" while you count for the pool.

68-3 = 65

Add up 22+22+31 to get 75

75-65 = 10

Appreciate your help!!

We can use equation approach here.

I(one subject only)+II(2 subjects only)+III(All 3)=68
I(one subject)+2II(two subject)+III(all 3)=25+25+43=84

now put III=3,you will get

I+II=65
I+2II=75

Solve this you will get II=10

Total Students = History + Maths + English - sum of 2 classes - 2 times (all three)

68= 25 + 25 + 34 - x - 2(3)
68 = 84 - x - 6
68 = 78-x
x= 78-68 = 10

Answer B)

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