Students choice of subject Problem

wilderness wrote:In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
13
10
9
8
7

I am trying to solve this as part of the set problem using

n(A u B u C) = n(A) + n(B) + n(C) – n(A n B) – n(A n C) – n(B n C) + n(A n B n C)

and assuming that { n(A n B) + n(A n C) + n(B n C) } - n(A n B n C) is my answer.

So 68 = 25 + 25 + 34 - { n(A n B) + n(A n C) + n(B n C) - n(A n B n C) }
Answer : 16.

What is wrong with my approach ? Why can I not do it in the manner I am doing. I have a feeling that I am making a grave mistake because I am not even using the value that 3 people are in all 3 courses. But what is the problem.

Thanks

mbhusal76 wrote:You have used the formula incorrectly:
n(AuBuC)=n(A)+n(B)+n(C)�n(AnB)�n(AnC)�n(BnC)+n(AnBnC)
68 = 25 +25 +34 � (n(AnB)+n(AnC)+n(BnC)) + 3
68 = 84 � (n(AnB)+n(AnC)+n(BnC)) + 3
68 = 87 � (n(AnB)+n(AnC)+n(BnC))

That yields to :
n(AnB)+n(AnC)+n(BnC) = 87 -68 =19

Number of people in exactly two sets is given as = n(AnB)+n(AnC)+n(BnC)-3n(AnBnC)
= 19 � 3*3
= 10

So the correct answer is 10



Better explanation from ManhattanGMAT:
For an overlapping set problem with three subsets, we can use a Venn diagram to solve.

Each circle represents the number of students enrolled in the History, English and Math classes, respectively. Notice that each circle is subdivided into different groups of students. Groups a, e, and f are comprised of students taking only 1 class. Groups b, c, and d are comprised of students taking 2 classes. In addition, the diagram shows us that 3 students are taking all 3 classes. We can use the diagram and the information in the question to write several equations:

History students: a + b + c + 3 = 25
Math students: e + b + d + 3 = 25
English students: f + c + d + 3 = 34
TOTAL students: a + e + f + b + c + d + 3 = 68

The question asks for the total number of students taking exactly 2 classes. This can be represented as b + c + d.

If we sum the first 3 equations (History, Math and English) we get:

a + e + f + 2b +2c +2d + 9 = 84.

Taking this equation and subtracting the 4th equation (Total students) yields the following:

a + e + f + 2b + 2c +2d + 9 = 84
�[a + e + f + b + c + d + 3 = 68]
b + c + d = 10

The correct answer is B.

benjiboo wrote:

mbhusal76 wrote:You have used the formula incorrectly:
n(AuBuC)=n(A)+n(B)+n(C)�n(AnB)�n(AnC)�n(BnC)+n(AnBnC)
68 = 25 +25 +34 � (n(AnB)+n(AnC)+n(BnC)) + 3
68 = 84 � (n(AnB)+n(AnC)+n(BnC)) + 3
68 = 87 � (n(AnB)+n(AnC)+n(BnC))

That yields to :
n(AnB)+n(AnC)+n(BnC) = 87 -68 =19

Number of people in exactly two sets is given as = n(AnB)+n(AnC)+n(BnC)-3n(AnBnC)
= 19 � 3*3
= 10

So the correct answer is 10



Better explanation from ManhattanGMAT:
For an overlapping set problem with three subsets, we can use a Venn diagram to solve.

Each circle represents the number of students enrolled in the History, English and Math classes, respectively. Notice that each circle is subdivided into different groups of students. Groups a, e, and f are comprised of students taking only 1 class. Groups b, c, and d are comprised of students taking 2 classes. In addition, the diagram shows us that 3 students are taking all 3 classes. We can use the diagram and the information in the question to write several equations:

History students: a + b + c + 3 = 25
Math students: e + b + d + 3 = 25
English students: f + c + d + 3 = 34
TOTAL students: a + e + f + b + c + d + 3 = 68

The question asks for the total number of students taking exactly 2 classes. This can be represented as b + c + d.

If we sum the first 3 equations (History, Math and English) we get:

a + e + f + 2b +2c +2d + 9 = 84.

Taking this equation and subtracting the 4th equation (Total students) yields the following:

a + e + f + 2b + 2c +2d + 9 = 84
�[a + e + f + b + c + d + 3 = 68]
b + c + d = 10

The correct answer is B.

This is what I did:

>Total amount of students: 68
>History: 25 students
>Math: 25 students
>English: 34 students
>All three classes: 3 students

That means that there are a total of:
>25-3=22 people taking history
>25-3=22 people taking math
>34-3=31 people taking English
>and 3 people taking all three
>This gives us a total of 78 enrollments (22+22+31+3 = 78)

So we have 78 enrollments, but 68 people. We know that (78-68 = 10) 10 people must be enrolled in at least 2 classes. Therefore, the answer is 10.